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ELECTROSTATICS & CAPACITORS

&

CURRENT ELECTRICITY

CONTENTS

I ELECTROSTATICS & CAPACITORS

1 Theory 3

2 IIT-JEE Objectives 11

3 IIT-JEE Subjectives 22

4 Board Exam Questions 42

II CURRENT ELECTRICITY

1 Theory 49

2 IIT-JEE Objectives 52

3 IIT-JEE Subjectives 64

4 Board Exam Questions 86

III Model Papers

1 Board Exam Model Paper 92

2 JEE – Main Model Paper 94

3 NEET Model Paper 97

4 CET Model Paper 100

FORUM FOR EXCELLENCE IN EDUCATION

ELECTROSTATICS & CAPACITORS 3


Coulomb’s law; Electric field and potential; Electrical Potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field, Electric field lines; Flux of electric field; Gauss’s law and its application in

simple cases, such as, to find filed due to infinitely long straight wire, uniformly charge infinite plane sheet and

uniformly charged thin spherical shell. Capacitance; Parallel plate capacitor with and without dielectric; Capacitors

in series and parallel; Energy stored in a capacitor.

ELECTROSTATICS

Electrostatics is that branch of physics which deals

with electric charges at rest.

Material Particles like Electrons and Protons

possess a characteristic property (like mass) called

electric charge, which is responsible for the electric force

between them. There are two types of charges – positive

charge (ex: protons) and negative charge (ex: electrons )

Properties of Charges:

1. Electric charge: Charges with the same electric sign

repel each other and charges with opposite electrical sign

attract each other.

In the figures (a) Two charged rods of the same signs

repel each other. & in (b) Two charged rods of opposite

signs attract each other.

2. Charge is quantized: The charge on a body is

always an integral multiple of the charge (denoted

as ‘e’) of an electron. q = ne, n = 1, 2, 3, ……. e = 1.60 x 10

–19 C

3. Charge is conserved: For an insolated system, the

net charge always remains constant. Charge can

neither be created nor be destroyed.

4. Charge is additive in nature

Coulomb’s Law: The electrostatic force of attraction or

repulsion between two stationary point charges is directly

proportional to the product of the magnitude of the

charges and inversely proportional to the square of the

distance between them.

2

21 ||||

r

qqkF (Coulomb’s law),

where, unit of charge is coulomb. k = 9 x 109 Nm

2/C

0 = Permittivity = 8.85 10 –12

C2/N.m

2.

Two charged particles, separated by distance r, repel each

other if their charges are (a) both positive and (b) both

negative. (c) They attract each other if their charges are

of opposite signs. In each of the three situations, the force

acting on one particle is equal in magnitude to the force

acting on the other particle but points in the opposite

direction.

Force due to Multiple charges (principle of

superposition): It states that when a number of charges

are interacting the total force on a given charge is the

vector sum of the forces exerted on it by all other

charges.

If we have n charged particles, they interact

independently in pairs and the force on any one of them,

let us say particle 1, is given by the vector sum.

ln15141312net.1 F......FFFFF

Shell Theorems:

1. A shell of uniform charge attracts or repels a charged

particle that is outside the shell as if all the shell’s charge were concentrated at its centre.

2. If a charged particle is located inside a shell of uniform

charge , there is no net electrostatic force on the

particle from the shell.

Electric Field due to a point charge:

The region of space surrounding a charge or system of

charges in which other charged particles experience

electrical forces is called the electric field due to the

charge. It is a vector field.

Below Fig (a) A positive test charge q0 placed at point P

near a charged object. An electrostatic force F acts on the

test charge. (b) The electric field E at point P produced

by the charged object.



To find the electric

field due to a point

charge q at any

point P at a

distance ‘r’ from point charge, put a

positive test charge

q0 at that point.

From Coulomb’s law, the magnitude

of the electrostatic

force acting on ‘q0’ is

q qF

r

The magnitude of

the electric field

vector is

qFE

q r

The principle of superposition applies to electric fields

also, just like for electrostatic forces.

Electric Field lines:

Michel Faraday thought of the space

around a charged body as filled with

lines of force. At any point, the

direction of straight field line, or the

direction of the tangent to a curved

field line gives the direction of E at

that point. The field lines are drawn

so that the number of lines per unit

area,

measured in a

plane that is

perpendicular

to the lines, is

proportional

to the

magnitude of E .

Fig (a) shows

the

electrostatic

force F acting

on a positive test charge near a sphere of uniform

negative charge. Fig a shows a sphere of Uniform

negative charge. If a positive charge is placed anywhere

near the sphere, an electrostatic force pointing towards

the centre of the sphere will act on the test charge as

shown.

In above fig (b) The electric field vector E at the location

of the test charge, and the electric field lines in the space

near the sphere. The field lines extend toward the

negatively charged sphere.

Electric field lines extend away from positive charge

(where they originate) toward negative charge (where

they terminate)

In the adjacent fig (a)

The electrostatic

force F on a Positive

test charge near a

very large, non-

conducting sheet

with uniformly

distributed positive

charge on one side,

(b) The electric field

vector E at the

location of the test

charge, and the

electric field lines in the

space near the sheet.

The field lines extend

away from the positively charged sheet. (c) Side view of

(b).

Field lines for two equal positive point charges. The

charges repel each

other. (The lines

terminate on distant

negative charges.) To

“see” the actual three-

dimensional pattern of

field lines, mentally

rotate the pattern

shown here about an

axis passing through

both charges in the

plane of the page. The

three-dimensional

pattern and the electric field it represents are said to have

rotational symmetry about that axis. The electric field

vector at one point is shown; note that it is tangent to the

field line through that point.

The Electric Field

due to an Electric

Dipole:

Field lines for a

positive and a nearby

negative point charge

that are equal in

magnitude. The

charges attract each



other. The pattern of field lines and the electric field it

represents have rotational symmetry about an axis

passing through both charges. The electric field vector at

one point is shown; the vector is tangent to the field line

through the point.

In fig (a) An electric dipole. The

electric field vectors E(+) and E(-)

at point P on the dipole axis

resulting from the two charges

are shown. P is at distances r(+)

and r(-) from the individual

charges that make up the, dipole

& (b) shows the dipole moment p

of the dipole points from the

negative charge to the positive

charge.

A system of two equal and

opposite charges separated by a

certain distance ‘r’ is called an

electric dipole.

Let us find the electric

field due to the dipole at

a point ‘P’ on the axis of the dipole. Applying the

superposition principle for

electric fields, the magnitude “E’ of the electric field at ‘P’ is E = E(+) – E(-)

= 2

)(02

)(0 r

q

4

1

r

q

4

1

= 2

0

2

0 d2

1z4

q

d2

1z4

q

So

q d dE

z zz

If z >> d , using binomial theorem

q d qdE [ P qd]

zz z

P = qd is known as electric dipole moment of the dipole

and its direction is from the negative to the positive end

of the dipole.

A point charge in an electric field: When a charged

particle is in an electric field E ( produced by other

stationary or slowly moving charges ), an electrostatic

force F

act on the particles as given by

EqF

It has the direction of E , if the charge q of the particle

is positive and has the opposite direction if q is negative.

A dipole in an electric field :Fig (a) An electric dipole

in a uniform electric field E. Two centers of equal but

opposite charge

are separated by

distance d. Their

center of mass

cm is assumed

to be midway

between them.

The bar between

them represents

their rigid

connection. (b)

Field E causes a

torque on

dipole moment

p.

As shown in figure the electrostatic force in the opposite

direction acts on the charged ends of the dipole and with

same magnitude F = q E. The forces on the charged ends

do produce a net torque on the dipole about its

centre of mass.

The centre of mass lies on the line connecting

the charge ends, at some distance x from end and thus a

distance (d – x) from the other end.

τ = Fx sin + F ( d – x) ( = r x F i.e = rF sin )

= Fd sin

substituting F = q E and d = P/q = PE sin or

Ep

The torque acting on a dipole tends to rotate p (hence

the dipole) into the direction of field E i.e in clockwise

direction. So by including a minus sign the torque is

= - pE sin

Potential Energy of an Electric dipole: Potential

energy can be associated with the orientation of an

electric dipole in an electric field.

Let the potential energy to be zero, when = 900 , with

the aid of equation , dW , potential energy U at

any angle is is

U W d pEsin

Evaluating the integral

leads to U = - PE cos

In Vector form, U P . E

When a dipole rotates from initial orientation L to

another Orientation F , the total work W done on the

dipole by the electric field is W = - U = - (Uf – Ui).

The Electric Flux: The electric Flux through a surface

placed inside electric field represents the total number of

electric lines of force crossing the surface in a direction

normal to the surface.



Fig: (a) A

Gaussian

surface of

arbitrary

shape

immersed in

an electric

field. The

surface is

divided into

small

squares of

area A.

The electric

field vectors

E and the

area vectors

A for three

representativ

e squares,

marked 1, 2,

and 3.

The flux of the electric field for the Gaussian surface is

A.E

When the sum of above equation becomes an integral, we

have for the definition of electric flux

Ad.E

The flux of the electric field is a scalar with SI unit

Nm2/C.

Gauss’ Law: It states that total normal electric flux E

over a closed surface is ( 1/ ) times the total charge

enclosed within the surface. It tells us that

0 = qenc

By substituting value of

dA.E , the definition of

flux, we can also write Gauss’ law as

enc0 qAd.E

(Gauss’ law). If qenc is positive, the net flux is outward. If qenc is

negative, the net flux is inward. Guass law and

Coulomb;s law

are equivalent

and each can

be derived

from the other.

Fig:A spherical

Gaussian

surface

centered on a

point charge q.

Electric Potential Energy : It is a physical quantity

which determines the amount of work done when the

configuration of a given charge distribution is changed.

If the system changes its configuration from an initial

state ‘L’ to a different final state ‘f’ the electrostatic forces work W on the particles. Then the resulting

change U in the potential energy of the system.

U = Uf – Ui = W

Electric Potential : The potential energy per unit

charge at a point in an electric field is called the electric

potential ‘V’ at that point. Thus, q

UV

Electric Potential is a scalar. We can define the potential

difference between points ‘L’ and ‘f’ as

f iW

V V Vq

i.e Negative of the work done by the electrostatic

force to move a unit charge from the initial point to the

final point.

If we set UL = 0 at infinity as our reference

potential energy , then the electric potential must also be

zero . We can define the electric potential ‘V’ at any

point in an electric field to be q

WV

Equipotential surfaces:

Adjacent points that have the same electric potential form

an equipotential surface, which can be either an

imaginary surface or a real physical surface.

No net work is done on a charged particle by an electric

field when the particle is moved between two points on

the equipotential surface.

Electric field

lines and

cross sections

of equi-

potential

surfaces for

(a) a uniform

field, (b) the

field of a

point charge,

and (c) the

field of an

electric

dipole.

Potential due to a Point Charge:

The positive point charge q produces an electric field E

and an electric potential V at point P. We find the

potential by moving a test charge q0 to P from infinity.



The test charge is shown at distance r' from the point

charge, undergoing differential displacement ds.

Consider a point ‘P at a distance ‘R’ from a fixed particle of positive charge ‘q’

The dot product dscosEsd.E

represents the

work done during different displacement ds.

The above equation can be written as

f iR

V V Edr [Work done – in moving the particle

from R to ]

set Vf = 0 (at ) and Vi = V (at R) & substitute

2r

q

4

1E

With these changes in earlier equations, we get

R R

q qV dr

rr

= R

q

4

1

0

.

Solving for V and switching R to r, we then have

r

q

4

1V

0

A positively charged particle produces a positive electric

potential. A negatively particle produces a negative

electric potential.

Potential due to an Electric dipole:

Fig: (a) Point P is a distance r from the midpoint O of a

dipole. The line OP makes an angle with the dipole

axis-

(b) If P is far from the dipole, the lines of lengths r(+)

and r(-) are approximately parallel to the line of length r,

and the dashed black line is approximately perpendicular

to the line of

length r(-).

The net Potential

at ‘P’ is given by

i ( ) ( )i

V V V V

( )

q q

r r

=

)()(

)()(

0 rr

rr

4

q

When r > > d

r(-) – r(+) d

cos and r(-)

r(+)_ r2

20 r

cosd

4

qV

pcos. [ P qd]

r

Potential due to system of

charges

Using super position

principle, the net potential at

a point due to multiple

charges is equal to sum of

potential due to individual

charges is equal to sum of

potentials due to individual

charges at that particular point.

Vnet=∑ Vini=1 =14π∈ ∑ iipni=1 (for n point charge)

Potential energy of system of charges

The electric potential energy of a system of fixed charges is

equal to the work that must be done by an external agent to

assemble the system, bringing each charge from an infinite

distance.

The potential energy for the pair of charges is

U = W = - q2V= - 14πϵ (for similar charges)

(-ve sign for opposite charges since negative work against

mutual attraction)

Initially W1=0 since no potential exists work done in

bringing q2 from infinity to r2 is,

W2=q2 V(r2)=14πϵ

Potential at any point ‘P’ due to charges q1 & q2 is,

V12 = 14π∈ ( p + p)

Only work done in bringing q3 from r3



W3 = q3V(r3) = π∈ (q qr + q qr )

Hence total work done is

U=W1+W2+W3

= 14π∈ + +

This can be extended to any number of charges

Calculating the field from the potential :

A test charge q0 moves

distance ds from one

equipotential surface to

another. (The

separation between the

surfaces has been

exaggerated for clarity.)

The displacement A

makes an angle with

the direction of the

electric field E.

From fig the work, the electric field does on the test

charge during the move is - q0 dV. It can be written as

0q E .d s

, or q0E (cos) ds. Equating these two

expressions for the work yields

-q0 dV = q0 E(cos ) ds,

or ds

dVcosE

Since E cos is the component E in the direction of d

s , above equation becomes

s

VEs

.

For simple situation ( E is uniform ), the equation

becomes s

VE

,

The component of E in any direction is the negative of

the rate of charge of the electric potential with distance in

that direction.

CAPACITORS

Capacitance: Two conductors, isolated electrically

from each other and from their surroundings, form a

capacitor. When the capacitor is charged, the conductors,

or plates as they called, carry equal but opposite charges

of magnitude q.

The

charge q and the

potential

difference V for

a capacitor are

proportional to

each other, that is

q = CV

The proportion -

ality Constant C is called the capacitance of the capacitor.

Its value depends only on the geometry of the plates and

not on their charge or potential difference.

The SI unit of Capacitance

1 Farad = 1Coulomb per volt

Fig shows two conductors as a Capacitor

(a) A parallel-

plate capacitor,

made up of two

plates of area A

separated by a

distance d. The

plates have

equal and

opposite charges

of magnitude q

on their facing

surfaces. (b) As

the field lines

show, the

electric field is

uniform in the

central region

between the

plates. The

field is not

uniform at the

edges of the plates, as indicated by the “fringing” of the field lines there.

Calculating the Capacitance

1.Assume a charge q on the new plates :

2.Calculate the electric Field E

between the plates in

terms of this charge, using Gauss Law

3.If you Know E

, calculate potential difference ‘V’ between the plates.

4.Calculate ‘C’ from equation, q = CV

Calculating the Electric Field: We shall use Gauss

law, qAd.E0

In all cases that we shall consider, the Guassian surface

will be such that whenever electric flux passes through it

E

and Ad will be parallel and the above equation ,

then will reduce to q = 0 EA

Calculating the potential difference: The potential

difference between the plates of a capacitor is related

to the field, E by

f

f ii

V V E.d s

The E and d s

will have opposite directions, so that

sd.E = – Eds.

Then by above equation

EdsV

in which - and + remind us that our path of integration

starts on the negative plate and ends on the positive plate.

Parallel – Plate Capacitor: We draw a Gaussian

surface that enclose just the charge q on the positive

plate,



Fig shows a charged parallel-plate capacitor. A Gaussian

surface encloses the charge on the positive plate. The

integration is taken along a path extending directly from

the positive plate to the negative plate. So

q = 0EA , ……. (1) where, A is the area of the plate.

Then

d

0

EddsEEdsV ……. (2)

substituting eq(1) and eq.(2) in q = CV , we get

d

AC 0

Cylindrical Capacitor:

Above figure shows a cross section of a long cylindrical

capacitor showing a cylindrical Gaussian surface of

radius r and the radial path of integration along which,

Eq. of V is to be applied. This figure also serves to

illustrate a spherical capacitor in a cross section through

its center.

Let a cylinder of length L and radius r , closed by end

caps and placed as is shown in figure, then

rL2EEAq 00

Solving for E yields Lr2

qE

0

Substitution of this result in to Eq. of V. yields

a

bln

L2

q

r

dr

L2

qEdsV

0

a

b0

From the relation, C = q/V, we then have ds = - dr )

)a/bln(

L2C 0

Spherical Capacitor: From Fig above as a Gaussian

Surface, draw a sphere of radius r concentric with the two

shells, then

q = 0EA = 0E (4r2),

q

Er

a

b

q qV Eds

b ar

= q b a

ab

From the relation, C = q/ v we have

ab

ab4C 0

An Isolated Sphere : from above eqn,

C = 40 b/a1

a

.

If we then let b ( i.e assuming the second spherical

shall to be of infinite radius ) and substitute R for a, we

find C = 40R

Capacitors in Parallel : Capacitors connected in

parallel can be replaced with an equivalent capacitor that

has the same total charge ’q’ and the same potential difference V as the actual capacitors.

Fig (a) Three

capacitors connected in

parallel to battery B.

The battery maintains

potential difference V

across its terminals and thus across each capacitor. (b)

The equivalent capacitance Ceq replaces the parallel

combination. The charge q on Ceq is equal to the sum of

the charges q1, q2, and q3, on the capacitors of (a).

From the fig, q1 = C1V, q2 = C2V, and q3 = C3V

The total charge on the parallel Combination

q =q1 + q2 + q3

= (C1 + C2 + C3)V

Then 321eq CCCv

qC

Capacitors in Series:

In figures three capacitors

connected in series to B. The

battery maintains potential

difference V. The equivalent

capacitance Ceq replaces the

series combination. The

potential difference V across Ceq

is equal to the sum of the

potential differences V1, V2, and

V3 across the capacitors.

When a potential difference V is

applied across several



10

capacitors connected in series , the capacitors have

identical charge q. The

sum of potential

differences across all the

capacitors is equal to the

applied potential difference

V.

11

C

qV ,

qV

C

, and

33

C

qV .

321321

C

1

C

1

C

1qVVVV .

The equivalent capacitance is then

321eq

C/1C/1C/1

1

V

qC

321eq C

1

C

1

C

1

C

1

Energy stored in an Electric Field: At a given instant,

the potential difference V| between the plates is q

|/ C. If

an extra increment of charge dq’ is then transferred, then

'dqC

'q'dq'VdW

The work required to bring the total capacitor up to a

final value of q is

q

0

2

C2

q'dq'q

C

1dWW

This work is stored as potential energy U in the

capacitors.

C2

qU

2

CV

q = CV ]

We can also write this as

DIELECTRIC: AN ATOMIC VIEW

1. Polar Dielectrics: The molecules of some dielectrics,

like water, have permanent electric dipole moments. The

alignment of the electric dipole as shown in fig. produces

an electric field that is directed opposite to the applied

field and smaller in magnitude.

In the left fig molecules with a permanent electric dipole

moment showing their

random orientation in the

absence of an electric

field. In right fig an

electric field is applied, producing partial alignment of

the dipoles. Thermal agitation prevents complete

alignment

Non Polar Dielectrics: Regardless of whether they

have permanent electric dipole moments, molecules

acquire dipole moments by induction, when placed in an

external dielectric field (induced dipole moments)

Capacitor with a Dielectric:

Fig: (a) If the potential difference between the plates of a

capacitor is maintained, as by battery B, the effect of a

dielectric is to increase the charge on the plates. (b) If the

charge on the capacitor plates is maintained, as in this

case, the effect of a dielectric is to reduce the potential

difference between the plates. The scale shown is that of

a potentiometer, a device used to measure potential

difference (here, between the plates). A capacitor cannot

discharge through a potentiometer.

The capacitance of any capacitor can be written in the

form C = 0 …….. (1) in which has the

dimensions of length. With a dielectric completely filling

the space between , above equation becomes

air0 kCkC , …….. (2) where, Cair is the value of the capacitance with only air

between the plates.

Michael Faraday is largely responsible for developing the

concept of capacitance and the SI unit of capacitance is

named after him.



JEE Corner

OBJECTIVES

1. A charged oil drop of mass 9.9 x 10-15

kg is suspended in

a uniform electric field of 3 X 104 Vm

-1. If g= 10 ms

-2, the

charge on the oil drop will be

(a) 3.3 x 10-18

C (b) 3.2 x 10-18

C

(c) 1.6 x 10-18

C (d) 4.8 x 10-18

C

Ans: (b)

Solution: =

⇒ = = . × − × × = . × − , which

is not possible as it not an integral multiple of 1.6 × − .

Hence the correct choice is (b).

2. Let there be a spherically symmetric charge distribution

with charge density varying as = − up to

r=R, and P(r)=0 for r > R, where r is the distance from the

origin. The electric field at a distance r (r < R) from the

origin is given by

(a) − (b) −

(c) − ( d)

−

Ans: (c)

Solution: For r < R = ∫ = ∫ ( − ) ×

= ∫ −

= −

For Gauss’s law

∫ . =

⇒ × = −

⇒ = ( – )

3. A thin semi circular ring of radius r has a positive charge

q distributed uniformly over it as in the Fig. The net field at the centre is

(a) (b)

(c) (d)

Ans: ( c )

Solution: Linear charge density is =

To find the net electric field at , we divide the ring into

a large number of very small elements each of length .

The charge of an element is = = , where

is the angle subtended by the element at centre . The electric field at due to the element at is. ∆ = directed radially outwards

Similarly, the electric field at due to an element

symmetrically opposite to is also ∆E. It is clear that the horizontal components = = ∆ cs cancel

while the vertical components = ∆ add up and

are directed along the negative −direction. Hence the

net electric field at is = ∫∆ = − ∫ sin

= |− cs | = − cs = = = π

4. Let = be the charge density distribution for a

solid sphere of radius R and total charge Q. For a point P



inside the sphere at distance r1 from the centre of the

sphere, the magnitude of electric field is

(a) (b)

(c) Zero (d)

Ans: (a)

Solution: Using gauss’s theorem ∫ . = , where

q is the charge enclosed inside the Gaussian spherical

surface of radius r1, × = (1)

= ∫ = ∫ × =

= ∫ = (2) 5. A tiny spherical oil drop carrying a net charge q is

balanced in still air with a vertical uniform electric field

of strength

× − . When the field is switched off, the

drop is observed to fall with terminal velocity ×− − . Given g = 9.8 − , viscosity of the air . × − − and the density of oil =900kg − ,

the magnitude of q is [IIT JEE] . × − C (b) . × − C c . × − C (d) . × − C

Ans: (d)

Solution: = = g = g

= ⁄

Substituting the value of r we get. ( ) ⁄ =

or =

Again substituting = we get, =

or ⁄ ⁄ = ( ) ⁄ ⁄

Substituting the values we get, = × √ × × . × × √ . × − × × − = . × − C

Correct option is (d).

6. Three infinitely long charge sheets are placed as shown

in figure. The electric field at point P is [IITJEE]

(a) (b) − (c)

(d) −

Ans: (b)

Solution: All the three plates will produce electric field

at P along negative z-axis. Hence, p = [ + + ] (− )

= −

Correct answer is (b).

7. A non-conducting ring of radius 0.5m carries a total

charge of 1.11 x 10-10

C distributed non-uniformly on its

circumference producing an electric field E everywhere

in space. The value of the integral ∫ − ℓ=ℓ=∞ ℓ ℓ = ℎ

in volt is [IIT JEE]

(a) +2 (b) – 1 (c) – 2 (d) zero

Ans: (a)

Solution: −∫ . ℓ = ∫ ℓ=ℓ=∞

ℓ=ℓ=∞ = −

but V(infinity) = 0 − ∫ . ℓ ℓ=ℓ=∞ Corresponds to potential at centre of

ring.

And (centre) = = × . × −. ≈

8. A thin Spherical conducting shell of radius R has a

charge q. Another charge Q is placed at the centre of the

shell. The electrostatic potential at a point P at a

distance R/2 from the centre of the shell is [AIEEE]

(a) πε (b) πε − πε

(c) πε + πε (d) +πε

Solution: Potential at P due to charge q on the shell is

(see Fig).



V1= πε

Potential at P due to charge Q at the centre of the shell is

= ⁄ =

Therefore, total potential at P is

V = V1 + V2 = πε + πε

9. A Parallel plate capacitor with air between the plates

has a capacitance of 9pF. The separation between its

plates is ‘d’. The space between the plates is now filled with two dielectrics. One dielectric has

dielectric constant K1 = 3 and thickness while the

other one has dielectric constant K2=6 and thickness

. Capacitance of the capacitor is now.

[AIEEE]

(a) 40.5 pF (b) 20.25 pF (c) 1.8 pF (d) 45 Pf

Ans: (a)

Solution: Refer to Fig. Given = = .

= ⁄ = ⁄ = =

= ⁄ = ⁄ = = Capacitors

C1 and C2 are in series. The equivalent capacitance is

= + = × = = × = .

10. A parallel plate air filled capacitor shown in figure (a)

has a capacitance of 2 F. When it is half filled with a

dielectric of dielectric constant k = 3 as shown in figure

(b), its capacitance becomes

(a) 4 F (b) 3 F (c) 1.5 F (d) 0.5 F

Ans. (a)

Solution :

If A is the area of each plate, the capacitance of the air –

filled capacitor shown in figure. (a) is

d

AC 0

0

where C0 = 2 F (given)

The capacitance of air capacitor in figure (b) is

2

C

d2

A

d

2/AC 000

1

The capacitance of dielectric field capacitor

in figure (b) is

2

kC

d2

Ak

d

2/AkC 000

2

Since C1 and C2 are in parallel,

)k1(2

C

2

kC

2

CCCC 000

21

= F4)31(2

F2

11. Three parallel conducting plates of equal area A are

placed in such a way that the outer plates are kept

distances d1 and d2 away from the middle plate as

shown in figure. The outer plates are grounded and

the middle plate carries a charge Q. Then the

electric field E1 and E2 between the two pairs of

plates is related as

(a) E1 = E2 (b) E1d2 = E2d1

(d) E1d1 = E2d2 (d) E1d12 = E2 d2

2

Ans. (c)

Solution :

This system is equivalent to two capacitors connected in

parallel.

Q1 + Q2 = Q V1 = V2

2

2

1

1

C

Q

C

Q

2

0

1

0

11 dA

)QQ(

A

dQ

)dd(A

Qd

A

QQd)dd(Q

21

2112211



Electric field,

2

10

0

11

d

d1

A

Q

E

1

20

0

22

d

d1

A

Q

E so E1d1 = E2d2

12. A parallel plate capacitor of area A, plate separation d and

capacitance C is filled with three different dielectric

materials having dielectric constants K1, K2 and K3 as

shown. If a single dielectric material is to be used to have

the same capacitance C in this capacitor then its dielectric

constant K is given by

(a) = + + (b) = + +

(c) = + + (d) K= + + +

Ans: (d)

Solution:

Applying C= − − + + ,

/− ⁄ – ⁄ + / + / =

Solving this equation, we get

K= + + +

13. Let C be the capacitance of a capacitor discharging

through a resistor R. Suppose t1 is the time taken for

the energy stored in the capacitor to reduce to half its

initial value and t2 is the taken for the charge to

reduce to one-fourth its initial value. Then the ratio

t1/t2 will be

(a) 1 (b) (c ) (d ) 2

Ans: (c)

Solutions:

The charge decays according to the equation

Q = − /

Where = = charge at t = 0.

Energy stored is U = = − / = − /

where = Intial nergy at t = . Given U = = .. i.e. = − / ⇒ / =

or = ⇒ = ln

Given Q = at t = . Hence

= /

Which gives = = = , which is choice (c)

14. Seven capacitors each of capacitance 2µF are

connected in a configuration to obtain an effective

capacitance . Which of the following

combination will achieve the desired result?



Ans: (a)

Solution: In series, C = +

= + =

15. The electric field E1, at one face of a parallelepiped is

uniform over the entire face and is directed out of the

face. At the opposite face, the electric field E2 is also

uniform over the entire face and is directed into that face.

The two faces described above are inclined at 30° from

the horizontal, while and are both horizontal.

Given that = 9 × 104 N/C and = 11 × 10

4 N/C.

Given that no other electric field lines cross the surfaces

of the parallelepiped. The net charge contained within is

(in coulomb) (given ε0 = 8.854× 10-12

C2

N-1

m-2, ℓ = 0.06

m, b = 0.05)

(a) +1.82 × 10-5

b) – 2.66 x 10-10

(c) – 0.4 x 10-6

d) zero

Ans: (b)

Solution:

We find the flux through the parallelepiped and then

apply Gauss law to find the net charge within = cs °

Flux through this face is:

∅ = ∫ . = ∫ ° = cs ° ∫ . = cs °

Where = . × . = . m

Similarly, we have: = cs °

Flux through this face is ∅ = ∫ . = ∫ ° = cs ° ∫ = cs °

where A = 0.003 m2 (as earlier)

Net flux = ϕ = ϕ + ϕ = − cs ⟹ ϕ = × − × cs . = − − ′ ⇒ ∮ ∙ =

⇒ ∮ ∙ = ϕ = − . − ⇒ = . × − − = − . × −

The enclosed charge is negative meaning that the ‘flux into’ the parallelepiped is larger than the ‘flux-out’.

16. Two thin wire rings, each of radius R. are placed at a

distance d apart with their axes coinciding. The charges

on the rings are +q and –q. The potential difference

between the centers of the two rings is

[AIEEE]

(a) (b) [ − √ + ]

(c)Zero (d) [ − √ + ]

Solution: Let AC=BD =r = √ + (see Fig.)

Potential at A is VA = potential at A due to charge

+q on ring 1 + potential at A due to charge –q on

ring 2 = − = − Potential at B is

VB = potential at B due to charge –q on ring 2+

potential at B due to charge +q on ring 1 = − + = − − ( − ) − ( − )



= ( − ) = [ − √ + ]

17. Two concentric spheres of radii R and r have similar

charges with equal surface densities (σ). What is the

electric potential at their common centre?

(a) σ/ε0 (b) σε (R – r)

(c) σε (R + r) (d) none of the above

Ans: (c)

Solution: Let Q and q be the charges on the spheres.

The potential at the common centre will be = × + ×

= [ × + × ]

But = =

= [ + ] = + K.E.= − =

18. Two equal point charges are fixed at = − = + on the X-axis. Another point charge Q

is placed at the origin. The change in the electrical

potential energy of Q, when it is displaced by a small

distance along the X-axis, is approximately proportional

[IITJEE]

(a) (b) (c) (d) ⁄

Ans: (b)

Solution: When Q s at position , then

= [ + + – ] = –

= [ − ] ≈ × − ≈ ℎ

∝ ∝ .

19. A charged ball B hangs from a silk thread S, which makes

an angle θ with a large conducting sheet P as shown in

Fig.

The surface charge density of the sheet is proportional

to

(a) cs (b) ct (c) sin (d) tan

Ans: (d)

Solution: Electric field due to sheet is =

T sin θ ℎ = ; = area of sheet. It follows from Fig.

that

T cs = ………. (1) T sin = = ………. (2) Dividing (2) by (1) we get tan =

Since E is uniform and A, m and g are fixed, ∝ tan

20. Two point charge +q and –q are held fixed at (−d,0) and

(d,0) respectively of a − co-ordinate system. Then

[IIT JEE]

a) The electric field E at all points on th -axis has the

same direction.

b) Work has to be done in bringing a test charge from ∞ to the origin.

c) Electric field at all points on -axis is along -axis.

d) The dipole moment is 2qd along the -axis.

Ans: (c)

Solution: The diagrammatic representation of the given

question is shown in figure.



The electrical field at all points on the -axis will not

have the same direction.

For – ≤ ≤ , electric field is along positive -axis

while for all other points it is along negative -axis.The

electric field at all points on the -axis will be parallel

to the -axis (.e., ) [option(c)]. The electrical potential at

the origin due to both the charges is zero, hence, no work

is done in bringing a test charge from infinity to the

origin.Dipole moment is directed from the –q charge to

the +q charge (.e.= ). 21. A parallel plate capacitor of plate area A and plate

separation d is charged to potential difference V and then

the battery disconnected. A slab of dielectric constant K

is then inserted between the plates of the capacitor so as

to fill the space between the plates. If Q,E and W denote

respectively, the magnitude of charge on each plate, the

electric field between the plate (after the slab is inserted),

and work done on the system, then

(a) = (b) =

(c) = (d) = −

Ans: (d)

Solution: When the battery is disconnected, the charge

remains unchanged. Hence

= = ..… (1) Electric field in the dielectric between plates of capacitor = = ….. (2) After the insertion of dielectric, the new capacity C′ is

given by C′ =

Now V′ = Cı = KC = VK work done on the system = − ′ ′ = − = ( – ) = ( ) ( − )

22. A parallel plate capacitor is maintained at a certain

potential difference. When a dielectric slab of thickness 3

mm is introduced between the plates, the plate separation

had to be increased by 2 mm in order to maintain the

same potential difference between the plates. The

dielectric constant of the slab is

[AIEEE]

(a) 2 (b) 3 (c) 4 (d) 5

Ans: (b)

Solution: The capacitance before the introduction of the

slab is =

If Q is the charge on the plates, the potential difference

is

= = ………….. (1)

Let d' be the new separation between the plates. When a

slab of thickness t and dielectric constant K is

introduced, the new capacitance is ′ = ′− −

Since charge Q remains the same, the new potential

difference is ′ = = [ ′− − ] ………….. (2) ′ = . Equating Eqs. (1) and (2), we get = ′ − ( − ) ′ − = ( − )

Given d′ = = = mm. Thus = − which gives K=3.

Hence the correct choice is (b).

23. A capacitor of capacitance 4µF is charged to 80 V and

another capacitor of capacitance 6µF is charged to 30 V.

When they are connected together, the energy lost by the

4 µF capacitor is [AIEEE]

(a) 7.8 mJ (b) 4.6 mJ (c) 3.2 mJ (d) 2.5 mJ

Ans: (a)

Solution: Common potential is = + + = × − × + × − ×× − + × − =

Energy lost by 4 capacitor = − = − = × × − × − = . × − J = . mJ

Hence the correct choice is (a)

24. Condenser A has a capacity of 15µF when it is filled with

a medium of dielectric constant 15. Another condenser B

has a capacity 1µF with air between the plates. Both are

charged separately by a battery of 100 V. After charging,

both are connected in parallel without the battery and the

dielectric material being removed. The common potential

now is

(a) 400 V (b) 800 V (c) 1200 V (d) 1600

Ans: (b)

Solution: Charge on capacitor A is given by = × = × − = × −

Charge on capacitor B is given by = × = × − = −



Capacity of condensers A after removing dielectric = = × − = µ

Now when both condensers are connected in parallel,

their capacity will be 1µF + µ = µ .

Common potential =

= ( × − ) + ( × − ) × − = 800 V.

25. A long cylindrical shell carries positive surface charge

in the upper half and negative surface charge - in the

lower half. The electric field lines around the cylinder will

look like figure given in: (figures are schematic and not

drawn to scale)

Ans: 4

Sol: It originates from +Ve charge and terminates at Ve

charge. It cannot form close loop.

26. In the circuit shown, the current in the 1Ω resistor is :

(a) 0 A

(b) 0.13 A, from Q to P

(c) 0.13 A, from P to Q

(d) 1.3 A, from P to Q

Ans: b

Sol: Taking the

potential at Q to be 0

and at P to be V, we apply Kirchhoff’s current law at Q:

The current will flow from Q to P.

27. A uniformly charged solid sphere of radius R has

potential V0 (measured with respect to ∞ ) on its surface.

For this sphere the equipotential surfaces with potentials

3 V0/2, 5 V0/4 , 3 V0/4 , V0/4 have radius R1,R2 ,R3 and

R4 respectively. Then

(a) R1 ≠ 0 and (R2 - R1) > (R4 - R3 )

(b) R1 = 0 and R2 < (R4 - R3 )

(c) 2R < R4

(d) R1 = 0 and R2 > (R4 - R3 )

Sol: The potential at the centre is

So R1 = 0

Potential at surface

Potential at

Potential at

Potential at

So Both options (b) and (c) are correct.

28. In the given circuit, charge Q2 on the

2 F capacitor changes as C is varied from 1 F to 3 F. Q2 as a function of

‘C’ is given properly by μ (figures

are drawn schematically and are not

to scale)

Ans: 1

Sol: Let the potential at P be V,

Then, C (E–V) = 1×V+2×V (we take C in F)

29. When 5V potential difference is applied across a wire of

length 0.1 m, the drift speed of electrons is 2.5×10–4

ms–1

.

If the electron density in the wire is 8 × 1028

m–3

, the

resistivity of the material is close to:

(a) 1.6 × 10–7

Ωm (b) 1.6 × 10–6

Ωm

(c) 1.6 × 10–5

Ωm (d) 1.6 × 10–8

Ωm

Sol: Ans: c



30. A conductor lies along the z-axis at –1.5 ≤ z < 1.5 m and

carries a fixed current of 10.0 A in –az direction (see

figure). For a field B = 3.00x10-4

e-2x

aY T, find the power

required to move the conductor at constant speed to x =

2.0 m, y = 0 m in 5 × 10-3 s. Assume parallel

(a) 14.85 W (b) 29.7 W (c) 1.57 W (d) 2.97 W

Sol: Ans : d

31. A parallel plate capacitor is made of two circular plates

separated by a distance of 5 mm and with a dielectric of

dielectric constant 2.2 between them. When the electric

field in the dielectric is 3 × 104 V/m, the charge density of

the positive plate will be close to :

(1) 3×104 C/m

2 (2) 6×10

4 C/m

2

(3) 6×10–7

C/m2 (4) 3×10

–7 C/m

2

Ans: c

Sol: By formula of electric field between the plates of a

capacitor

32. The current voltage relation of diode is given by I =

(e1000V/T

– 1) mA, where the applied voltage V is in volts

and the temperature T is in degree Kelvin. If a student

makes an error measuring ± 0.01 V while measuring the

current of 5 mA at 300 K, what will be the error in the

value of current in mA ?

(a) 0.5 mA (b) 0.05 mA (c) 0.2 mA (d) 0.02 mA

Sol: Ans: c

33. In a large building, there are 15 bulbs of 40 W, 5 bulbs of

100 W, 5 fans of 80 W and 1 heater of 1kW. The voltage

of the electric mains is 220 V. The minimum capacity of

the main fuse of the building will be :

(a) 12 A (b) 14 A (c) 8 A (d) 10 A

Sol: Ans: a

So current capacity

34. Supposing that the earth has a charge surface density of 1

electron/metre2, calculate (i) earth’s potential, (ii) electric

field just outside earths surface. The electronic charge is -

1.6 × − coulomb and earth’s radius is 6.4 ×

metre = . × − / . (a) -0.155 V, -1.8x10

-8 N/C (b) 0.155 V, 2.8x10

-8 N/C

(c) -1.155 V, -1.8x10-8

N/C (d) 2.155 V, 5.6x10-8

N/C

Solution: Let R and be the radius and charge surface

density of earth respectively. The total charge, q on the

earth surface is given by q=4

(i) The potential V at a point on earth’s surface is same as if the entire charge q were concentrated at its centre.

Thus,

V= - .

= . = .

Substituting the given values

V = . × − ×(− . × − 9 / )( . × − −)



= - 0.115 − = − . = - 0.115 volt.

(ii) E =

= . =

= − . × − 9 /. × − / −

= - 1.8× − /

The negative sign shows that E is radially inward.

35. A spherical drop of water carrying a charge of 3 ×− C has potential of 500 volt at its surface. What is the

radius of the drop? If two such drops of the same charge

and radius combine to form a single spherical drop, what

is the potential at the surface of the new drop?

a) 794V b)924V c) 1kV d) 324V

Solution: The potential V of a sphere having charge q

and radius r given by

V=

Here, V=500 and q=3× −

500=× − × × lving, we get r = . m = . cm.

Volume of new drops=

Volume of two drops = Let r' be the radius of the new drop then

′ = or r'= / Change n new drs = = × × −

= 6× − C

Now potential = ′ = ( × − )( × 9)/ × .

= 794 Volt.

36. A particle having charge of 1.6× − C enters midway

between the plates of a parallel plate condenser. The

initial velocity of particle is parallel to the plates. A

potential difference of 300 volt is applied to the capacitor

plates. If the length of the capacitor plate is 10 cm and

they are separated by 2 cm, calculate the greatest initial

velocity for which the particle will not be able to come

out the plates. The mass of the particle is 12× − kg.

a) 10x105 m/s b)1x10

4 m/s c) 10

5 m/s d) 10

3 m/s

Solution:

Here E=

= / =

As the particle does not come out, its maximum

deflection y=1cm= − m

We know that

y= .

, or = . .

= . × − 9× − − =

u= / . 37. A capacitor has square

plates with sides of

length l as shown in

fig. The plates are

inclined at a small

angle . Find the

capacitance of plate.

a) [ − ] b)

[ − ] c) [ − ] d) [ − ]

Solution: Imagine that the capacitor is divided into

differential strips which are parallel. Consider one such

strip at a distance x as shown in fig. The length of strip is l

(perpendicular to the plane of paper) and width dx (in the

plane of paper). The area dA of the strip is l dx

The separation of the plate at a distance x is given by

d= +

The capacitance dC, due to this strip is given by

dC=

= +



C=∫ + = ∫ +

= + θ − dx

= [ − ] = [ − ]

= [ − ]

38. Assume that an electric field E = 30x2i exists in space.

Then the potential difference VA – VO, where VO is the

potential at the origin and VA the potential at x = 2 m is :

(a) -80 V (b) 80 V (c) 120 V (d) – 120V

Sol: Ans : a

39. In the arrangement shown in fig, a dielectric slab of

dielectric constant K is partially inside a parallel plate

capacitor.

Assuming gravity to be absent, calculate the expansion

in the spring if the whole system is in equilibrium. If the

slab is slightly displaced will it perform S.H.M? If the

battery is disconnected and then the slab is slightly

displaced, will it perform S.H.M.? Given that l=length of

the plates, b= breadth of plate and d=distance between

the plates.

a) yes, no b) data insufficient c) no, yes d) yes , yes : = −

When battery remains connected, it will perform S.H.M.

When the battery is disconnected, it will not perform

S.H.M. = , or − =

X= − , Force on slab=

−

C= − +

When battery remains connected, the force on slab is

constant but when battery is disconnected, the force is no

longer a constant but depends on x (length of slab inside

plate).

40. The effective capacitance of two capacitors of

capacitances C1 and C2 (with C2 >C1) connected in

parallel is times the effective capacitance when they are

connected in series. The ratio C1/C2 is

(a) (b) (c) (d)

Ans: (a)

Solution: Given + = + ×

or + =

or + + =

or + – =

Let = . Then, we have + − =

or − + =

Which gives = . Since > , = is not

possible.



SUBJECTIVES

1. A particle of mass 40 mg and carrying a charge 5 10 –9

C is moving directly towards a fixed positive point charge

of magnitude 10 –8

C. When it is at a distance of 10 cm

from the fixed positive point charge it has a velocity of 50

cm sec –1

. At what distance from the fixed point charge

will the particle come momentarily to rest? Is the

acceleration constant during motion?

Solution : Let r be the distance of q from Q when comes

to rest as shown in figure.

The loss of kinetic energy given by

262

622 1010402

1

2

11040.

2

1mv

2

1mu

2

1

= 5 10 –6

Joule

This loss of kinetic energy is used as workdone against

the force of repulsion. In its path, let us consider the

force acting on the mass when it is at a distance x i.e.,

2

989

20 x

10510109

x

Qq

4

1F

= 2

8

x

1045

dW = - F dx

Negative sign is used because work is done against the

force.

The total workdone in moving the mass from (1/10)

metre to a distance x = r is given by

W =

r

10/12

8

dxx

1045dW

or

r

10/1

8

x

11045W

10r

110105 86

10

r

19100

Solving we get r = 4.737 10 –2

m

The force obeys inverse square law i.e., it increases as

distance decreases. Hence acceleration also increases.

2. Two small balls A and B with charges –q and +q

respectively are fixed on a horizontal plane at a distance d

from each other. A third ball C with charge +Q is

suspended from a string. The string makes an angle of 300

with the vertical when the ball C is in equilibrium at a

height d vertically above the ball A. When ball C is in an

identical situation above ball B, find the angle which the

string now makes with the vertical. [Roorkee 2001]

Solution: The first situation is shown in Fig.(a)

Fig (a)

The ball C is in equilibrium. Therefore, the sum of the

torques of all the forces about the point of suspension

should be zero.

Hence sin = + sin

(where K = ⁄ )

Solving the above equation for Q, we get = − …….. (1)

The second situation is shown in Fig

Fig. (b)

In this situation, − sin = sin − = cs − ……… (2) Substituting the value of Q from eq. (1) in eq. (2), [ − − ] sin = − − – cs −cs − = cs cs + sin sin cs − sin – cs = cs ct + sin

Solving we get ct = √ [ − cs ] −

3. a)Two similar balls of mass m are hung from silk threads

of length l and carry similar charges. Prove that

separation x =

/q l

mg

, when is small,

keeps the balls in equilibrum.

(b) Find the rate ( dq/dt ) with which the charge



leaks off each sphere if their approach velocity varies

as ϑ = a/√ where a is a constant.

Solution :

a) The different forces on ball are shown in fig

The electrostatic repulsive force between balls is given

by

F = q q

.x

………………. (1)

Where x is the separation between the two

The restoring force = mg sin

In equilibrium,

m g sin = q

.x

………….. (2)

From Figure , sin = (x / ) x

l l

x q

mg. .l x

………… (3)

or x3 =

q l.

mg

………… (4)

or x =

/q l

.mg

……………. (5)

(b) From equ . (3) ,

q2 =

mg x.

l

………… (6)

Differentiating this equation with respect to time, we get

2 q dq

dt

mg.

l

3x

2

dx

dt

or

/mg x dq

.l dt

= mg a

x .l x

dq

adt

mg

.l

4. A thin fixed ring of radius 1 metre has a positive charge

1 10 –5

coulomb uniformly distributed over it. A

particle of mass 0.9 gm and having a negative charge of

1 10 –6

coulomb is placed on the axis at a distance of 1

cm from the centre of the ring. Show that the motion of

the negatively charged particle is approximately simple

harmonic. Calculate the time period of oscillations.

Solution :

The force dF between dq and qı is given by

20 PA

'qdq.

4

1dF

from P to A

Force on q by the charged ring

F = dF. sin

=

sin

PA

'qdq

4

1

20

= PA

x

PA

'qdq

4

1

20

PA

xsin

where OP = x

=

xPA

'qq.

4

1

30

where dq = q

or F = kx where 30 PA

'qq.

4

1k

F x

i.e. force F is proportional to displacement and is

directed towards the centre, hence the motion of

negatively charged particle is simple harmonic along the

axis of the ring .

q = 1 10 –5

C, q1 = 1 10

–6 C

(PA)2 = (x

2 + r

2) = (0.01)

2 + (1)

2 ( x = 0.01 metre)

2/32

659

101.0

101101.109k



In simple harmonic motion, we know that

Time period

k

m2T

=

659

2/323

1010109

101.0109.02

= 5

seconds

5. Three particles, each of mass 1 gm and carrying a charge

q, are suspended from a common point by insulated

massless strings, each 100 cm long. If the particles are

in equilibrium and are located at the corners of an

equilateral triangle of side length 3 cm, calculate the

charge q on each particle. (Take g = 10 m/s2). [ IIT

JEE 1988]

Solution : Let A, B, C be the three particles suspended

by strings of equal lengths from a common support O.

The particles lie in a horizontal plane at the corner of an

equilateral triangle as shown in figure.

When the particle at A is in equilibrium, the following

forces act on it.

(i) Weight mg vertically downwards

(ii) Tension t in the string

(iii) Electrical force F, which is the resultant of the

repulsive forces FBA and FCA acting on A due to B and C

respectively. Thus

)say(fx

q

4

1FF

2

2

0CABA

where x = 3 cm = 3 x 10 –2

m.

The resultant force F is given by

f360cosf.f2ffF 022

2

2

0 x

q

4

3F

Taking the moment of these forces about O, we have

F . OG = (mg) . AG Also cos 300 = AN/AG

3

x

2/3

2/AB

cos30

ANAG

0

or m103AG 2 ( x = 3 10 –2

m)

Also m1AGOAOG 22

2103mg

OG

AG.mgF

But 2

200

2

103mgx4

q3F

q2 = 4 0 (mg)x

2 10 –2

or

2/1

9

243

109

1010910101q

coulomb q = 3.162 10 –9

6. A small ball of mass × − kg having a charge of 1µC

is suspended by a string of length 0.8 m. Another

identical ball having the same charge is kept at the point

of suspension. Determine the minimum horizontal

velocity which should be imparted to the lower ball, so

that it can make complete revolution

[IIT JEE2001]

Solution: Given : = = − , = × − kg and l=0.8m

Let u be the speed of the particle at its lowest point and v

its speed at highest point.

At highest point three forces are acting on the particle.

(a) Electrostatic repulsion

= (outwards)

(b) Weight w = mg (inwards)

(c) Tension T (inwards)

T=0, if the particle has just to complete the circle and the

circle and the necessary centripetal force is provided by

w – Fe i.e., = − , = − = . × × − ×− . × × −. ⁄



or = . ⁄ ...(i)

Now, the electrostatic potential energy at the lowest and

highest points are equal. Hence, from conservation of

mechanical energy.

Increase in gravitational potential energy = Decrease in

kinetic energy.

or = −

or = +

Substituting the values of from Eq. (i), we get = . + . = . ⁄

u = 5.86 m/s

Therefore, minimum horizontal velocity imparted to the

lower ball, so that it can make complete revolution, is

5.86m/s.

7. Two circular wire loops of radii 0.05 m and 0.09 m are

placed such that their axes coincide and their centres are

0.12 m apart. Charge of 10 –6

C is distributed uniformly

on each loop. Find the potential difference between the

centres of loops.

Solution : Figure shows two wire loops X and Y of radii

rx and ry with their centres at A and B respectively. Let

Q be the charge on each loop. The potential at the centre

A of coil X is

VA = potential due to charge Q on loop X + potential due

to charge Q on loop Y

=

1x0 d

Q

r

Q

4

1 ………. (1)

Now d21 = r

2y + d

2 = (0.09)

2 + (0.12)

2 = 0.0225

Therefore, d1 = 0.15 m

Also rx = 0.05 m and Q = 10 –6

C. Using these values in

(1), we have

15.0

1

05.0

110109V 69

A

Similarly, the potential at the centre B of loop Y is

2y0B

d

1

r

1

4

QV …….. (2)

Now d22 = r

2x + d

2= (0.05)

2 + (0.12)

2 = 0.0169. Thus

d2 = 0.13 m.

13.0

1

09.0

110109V 69

B

= 1.7 105 V

Potential difference between A and B = VA – VB

= 2.4 105 – 1.7 10

5 = 7 10

4 V

8. Two fixed charges – 2Q and Q are located at the points

with co ordinates ( – 3 a,0) and (+3 a,0) respectively in

the x-y plane.

(a) Show that all the points in the x-y plane where

the electric potential due to the two charges is zero, lie

on a circle. Find its radius and the location of its centre.

(b) Give the expression for the potential V(x) at a

general point on the x-axis and sketch the function V(x)

on the whole x-axis.

(c) If a particle of charge +q starts from rest at the

centre of the circle, show by a short qualitative argument

that the particle eventually crosses the circle. Find its

speed when it does so. [ IIT JEE1991]

Solution: (a) the situation is shown in Fig.

Let electric potential be zero at point P with

coordinates (x,y). The electric potential at P is give

by = [– + ] =

Where r1 and r2 are the distance of P from

– 2Q and Q respectively. Then = √[ + + ] and = √[ − + ]

= × [− √[ + + ] + √[ − + ]] =

or √[ + + ] = √ − + Solving this equation, we get + − + =

or − + =

This is an equation of a circle in XY plane having a

radius 4a and coordinates of centre at (5a,0).

Solution:(b)The potential at any general point on X-axis

is given by, = [ − − + ] << = [ − − + ] >

Sketch of potential V(x) versus distance x is shown in

fig.



We now sketch the potential function V(x) on the whole

X-axis. It is obvious from part first that the circle of zero

potential cuts the x axis at (a,0) and (9a,0) respectively.

Hence

(i) V(x) =0 for x = a and x = 9a

(ii) From the expression of V(x), it is clear that → ∞ → → −∞ ℎ → − . (iii) → → ±∞.

(iv) In general V(x) varies as 1/x. The sketch is

shown in Fig.

Solution: (c) When the particle of charge q is at the

centre of the circle (5a,0), the force on the particle is

given by = [ − ] = ( – ) = along + X axis.

Thus the particle moves along X-axis. When it reaches at

the point on the periphery of the circle i.e., at = += , the force on the particle is given by = [ − ] = ∙ ( − ) = ∙ ∙ = along + X axis.

This expression shows that sill there is a force on the

particle along X-axis, so the particle crosses the circle.

Let v be the velocity of particle when it crosses the

circle. From the law of conservation of energy, we have

(K.E.+P.E.) at the centre = (K.E.+P.E.) at periphery + ( − ) = + [ − ] + ∙ ( − ) = +

Solving, we get = √[ ]

9. Two square metallic plates of 1m side are kept 0.01 m

apart, like a parallel plate capacitor, in air in such a way

that one of their edges is perpendicular to an oil surface

in a tank filled with an insulating oil. The plates are

connected to a battery of em.f. 500 V. The plates are

then lowered vertically into the oil at a speed of 0..001

m/s. Calculate the current drawn from the battery during

the process. ( Dielectric constant of oil = 11 , = 8.85 x

10-12

C2 N

-1 m

-2 ) [IIT JEE 1994]

Solution : The situation is shown in figure.

The capacitance of air capacitor is given by

C0 = (0 A/d) = (0 a2/d) = 0/d ( a

2 = 1)

When the capacitor is lowered with a speed of v, then

after a time t, the length of plates within dielectric is v t

and

above dielectric is (a – v t)

Now, the capacitance of air capacitor

d

)vt1(

d

)vt1(aC 00

1

The capacitance of dielectric capacitor

d

vtK

d

avtKC 00

2

Net capacitance after time t

d

vtK

d

)vt1(CCC 00

21

Let q be the charge on the plates of the capacitor, then

q = C V

or dt

dCV)CV(

dt

d

dt

dq (as V = constant)

or ( vt) K vtd

i Vdt d d

= d

v)1K(V

d

vK

d

vV 000

= 001.0

01.0

1085.8111500 12

= 500 8.85 10 –12

= 4.425 10 –9

i = 4.425 10 –9

amp

10. In diagram find the potential difference between

the points A and B and between the points B and C in

the steady state.

Solution : The circuit is redrawn in figure. (a), (b),

(c)From figure (c)

Potential difference between P and q

= Potential difference between R and s = 100 volts

Q = capacity volt = 3/2 10 –6

100 = 150 10 –6

C



Potential difference between A and B = Potential

difference across the two ends of condenser of capacity 6

F.

volts25106

10150

capacity

QV

6

6

1

Again potential difference between C and D = Potential

difference across the two ends of condenser of capacity

2 F.

volts75102

10150V

6

6

2

11. The capacitance of a parallel plate capacitor with plate

area A and separation d is C. The space between the

plates is filled with two wedges of dielectric constants

K1 and K2 respectively. Find the capacitance of resulting

capacitor [IIT JEE 1996]

Solution :

Let the length and breadth of the plates be l and b each.

Let the entire capacitor is broken into small capacitors as

shown in fig below. Consider one such capacitor of

length dx at a distance x from one end p.

The small capacitance dC of this capacitor is given by

dC

= ( b dx) ( b dx)

( x) tan d ( x) tanRS ST

K KK K

All such capacitors are connected in parallel and

hence, the total capacitance C of the condenser is

given by

C = x

x

( b dx)dC

( x) tan d ( x) tan

K K

= ∫ − tan + [ − − tan ] =

b K Klog (l x) K tan

( K K ) tan

+ K1[ d - (l - x ) tan ) ]

From figure tan = d

l

e

b K K l KC log

(K K ) d K

e

A K K Klog

(K K ) d K

12. (a)Two dielectric slabs of dielectric constants K1 and K2

have been filled in between the plates of a capacitor as

shown in fig. What will be the capacitance in each case?

(b) A capacitor is filled with the dielectric of same

dimensions but of dielectric constants 2 and 3

respectively. Find the ratio of the capacitances in the two

possible arrangements.

Solution

(a) Let A be the area of each plate of the capacitor and D

be the distance between the two plates. If the capacitance

be and respectively, then

= / and = /

Let C be the equivalent capacitance, then

C = + = +

[ condensers are in parallel]

C= +

The arrangement shown in fig is equivalent to



two capacitors joined in series. Let their capacitances be

and respectively. Then

= / and = /

Now = +

= + = [ + ]

C = [ + ]

(b)The ratio of capacitances in the possible arrangements

is given by

=

+ + = + =

+× × = 25 : 24

13. A capacitor consists of two stationary plates shaped as a

semi-circle of radius R and a movable plate made of

dielectric with permittivity K and capable of rotating

about an axis O between the stationary plates. The

thickness of movable plate is equal to d which is

practically the separation between the stationary plates.

A potential difference V is applied to the capacitor. Find

the magnitude of the moment of forces relative to the

axis O acting on the movable plate in the position shown

in the figure below

Solution : Let be the capacity of outside condenser,

then = …… …

In the rotated condition, let C1 be the new capacity of

inner condenser. Then

= [ − ϕ] … … . .

where outside plate area = ϕ, because the

circumference is R and area is × × ϕ i. e., ϕ

If C2 be the capacity of outside condenser, then

=

……. (3) In rotated position, the arrangement is equivalent to two

capacitors [of capacity and ] connected in parallel.

Hence

C = +

C = ( )

-

+

Or c = [ + − ] , Initial energy

= = ( ), Final energy

= . ( )

[ + − ]

∆U = − = −

If M be the moment of force, then

M ϕ = −

or M = −

14. A system consists of a thin charged wire ring of radius R

cm and a very long uniformly similarly charged straight

wire, oriented along the axis of the ring, with one of its

ends coinciding with the centre of the ring. The total

charge on the ring is Q coulomb and the charge per unit

length on the wire is . What will be the interaction force between the ring and the straight wire?

Solution : The situations shown in Fig. The electric field

on an axial point of a charged ring of radius R at a

distance x is given by

= ∙ + ⁄

Charge on a small element dx of the wire dq=λ dx

= = − + ⁄

(-ve sign is used to indicate that dF/dx is negative) or = − ∫ + ⁄∞

Substitute = tan θ, then = θ + =

= − ∫ ⁄ = − ∫ sin ⁄

=

15. A thin conducting ring of radius r has an electric charge

+Q. What would be the increase in the tension of the

wire, if a point charge +q is placed at the centre of the

ring?

Solution : Let c be the point where a test charge q0 is

placed. Now the force F exerted on this test charge by

either charge Q is given by = + The



resultant force on the test charge will be 2F cos θ

So = cs , = ∙ + ∙ cs = ∙ + ∙ + ⁄ = + ⁄ ……. (1)

(a) For the force to be maximum ⁄ = =

[ + ⁄ ∙ − + ⁄ ∙ ∙+ ] = + ⁄ = + ⁄ ∙ + = = = /√

(b) The force is radial and away from the centre of the

circle.

16. A point charge q is situated at a distance d from one

end of a thin non-conducting rod of length L having a

charge Q uniformly distributed along its length as

shown in Fig.

What is the magnitude of electric force between the two?

Solution :Consider an element of length dx at a distance

x from point charge q. The force between charge q and

charge on element dx is given by = ∙ ×

Where = × = ∙

The total force F between q and Q is given by = ∙ ∫ ( )+

= ∙ [− ] + = ∙ [ − + ] = ∙ +

17. Two identical oppositely- charged metallic spheres

placed 0.5m apart attract each other with force of 0.108

newton. When they are connected to each other by a

copper wire for a short while, they begin to repel each

other with a force of 0.036 newton. What were the initial

charges on them?

Solution : Let q1and q2 be the charges on the spheres.

Before connection, = or . = . × . = . × .. × 9 = . × − ……..(1) After connection: After connecting through a wire, the

charge flows from one sphere to another till both acquire

the same charge (spheres are identical). Since q1 and q2

are of opposite signs hence after connection each sphere

will have a charge −

. Due to same sign of charge,

the two spheres will now repel each other. Hence,

. = . × − − .

Or . = ( . × 9) − .

− = . × . ×. × ⁄

Or − = . × − ….. (2) Solving eqs. (1) and (2), we find that the charges on two

spheres were + . × − coulomb and − − coulomb.

18. A bob of mass m carrying a positive charge q is

suspended from a light inextensible string of length l

inside a parallel plate condenser with its plates making

an angle with the horizontal. The upper plate of the

condenser is negatively charged and the intensity of

electric field inside the condenser is E. Find the period of

vibration of the pendulum and the angle between the

thread in equilibrium position and the vertical.

Solution : Different forces are shown in fig

From the figure

T sin α = q Esin …… (1)

T cs α = q Ecs = mg

T cs α = mg – q Ecs … (2)

From eqs. (1) and (2), we have

tan α = E− E

Or α = tan− E− E …..(3)

The net acceleration on the bob due to m g and q E is



a=√[ + + cs − ] =√[ + + cs ]

, = √ = [ √ + − ]

19. A thin fixed ring of radius 1 metre has a positive

charge 1 10 –5

coulomb uniformly distributed over

it. A particle of mass 0.9 gm and having a negative

charge of 1 10 –6

coulomb is placed on the axis at a

distance of 1 cm from the centre of the ring. Show

that the motion of the negatively charged particle is

approximately simple harmonic. Calculate the time

period of oscillations.

Solution : Fig. shows a ring AB of radius one metre

having a positive charge × − C uniformly

distributed over it with centre O. Let OX be the axis of

the ring. P is a point on the axis of the ring one cm from

the centre where a charge q′= × − is placed. Let

AC be a small element of ring having a charge dq on it.

The force dF between dq and ′ is given by

20 PA

'qdq.

4

1dF

from P to A

Now we resolve dF along and perpendicular to OP. The

components are dF sin θ and dF cosθ respectively. Now we take another small element opposite to A, i.e., at

B and calculate the force between q′ and the charge on this element. Resolving along X axis and Y axis, we

observe that the perpendicular components are cancelled

out by an equal but opposite components established by

the charge element on the opposite side of the ring.

Therefore, the resulting force lies along the axis of the

ring.

Force on q by the charged ring

F = dF. sin

=

sin

PA

'qdq

4

1

20

= PA

x

PA

'qdq

4

1

20

PA

xsin

where OP = x

=

xPA

'qq.

4

1

30

where dq = q

or F = kx where 30 PA

'qq.

4

1k

F x

i.e., force F is proportional to displacement and is

directed towards the centre, hence the motion of

negatively charged particle is simple harmonic along the

axis of the ring

q = 1 10 –5

C, q1 = 1 10

–6 C

(PA)2 = (x

2 + r

2) = (0.01)

2 + (1)

2

( x = 0.01 metre)

2/32

659

101.0

101101.109k

In simple harmonic motion, we know that

Time period

k

m2T

=

659

2/323

1010109

101.0109.02

=5

seconds

20. A small conducting sphere of mass m is suspended

form a string of length l between two parallel charged



plates where a vertical electric field of strength E is

established. Find the period of this pendulum, when

sphere is given a charge +q and lower plate is

charged (i) positively and (ii) negatively.

Solution : In case of simple pendulum, the time period T

is given by = √

(i) When the lower plate is charged positively, the

sphere experiences a vertical upward pull qE. Now the

net force on the sphere = –

Acceleration = = –

The time period in this case is given by = √( − ⁄ )

= √([ − ⁄ ])

(ii) When the lower plate is charged negatively, the

sphere experiences a vertical downward pull qE. The net

force on the sphere = +

Acceleration + = [ + ⁄ ]

The time period in this case is given by = √( + ⁄ )

21. An inclined plane making an angle 30o with the horizontal

is placed in a uniform horizontal electric field E of 100

V/m. A Particle of mass 1 Kg and charge 0.01 C is

allowed to slide down from rest from a height of 1m. If

the coefficient of friction is 0.2, find the time it will take

the particle to reach the bottom.

Solution : The different forces on the particle are shown

in Fig.

Form Fig (b), = cs + cs = cs + cs Now the total force F working along the inclined plane = sin − – cs = sin − cs

− cs − cs

= = sin − cs − cs − cs = . × . − . × . × (√ ⁄ ) − . × . × × . − . × × √

) = . − . × . − . − . × . = . − . − . − . = .

Now = + = √ ⁄ = √ × [ = / sin = ] = √ . = . sec.

22. A thin straight rod of length 2a carrying a uniform

distributed charge q is located in vacuum. Find the

magnitude of the electronic field strength as a function of

the distance r from the rod’s centre along the straight line

(A)Perpendicular to the rod and passing through its

centre.

(B) Coinciding with the rod’s direction (at the point lying

outside the rod).

Solution : a) Consider an element dl of the rod. The

charge on this element is (q/2a) x dl.

Electric field strength dE due to this charge = ( ) × The component of dE in the direction of r is given by = × cs

From figure, 0 and r0

Substituting these values, we get = × cs 0 = ∫ cs = sin The value of sin will be sin = √ +

= √ +

or = √ +



It should be remembered that the perpendicular

components of dE will be cancelled due to the upper and

lower parts of the rod. Thus the resultant electric field

strength will be in perpendicular direction to the rod.

(b) See fig.

Here, = × −

If the element lies on outer side, then

= ( ) × + = ∫ = ∫ ( ) × − +∫ ( ) × +

Integrating this expression, we get = × −

23. A very long straight uniform charged thread carries

a charge per unit length. Find the magnitude and direction of the electric field strength at a point which

is at a distance y from the thread and lies on the

perpendicular passing through one of the thread’s ends.

Solution : The situation is shown in Fig.

Consider a small element dx of the rod at a distance x

from one end. = ∙ + = cs = × + × cs From figure, = tan or =

= × + × cs

or = cs + =

or = ∫ cs ⁄ = [sin ] ⁄

or =

Similarly, dEy=dE sinθ

= × × ∫ sin ⁄ = × ( ) × [− cs ] ⁄

= × ( ) =

= √ + ( ) = √ , tan = = or = ] 24. A thin half-ring of radius R=20 cm is uniformly

charged with a total charge q=0.70 nC. Find the

magnitude of the electric field strength at the

curvature centre of this half-ring.

Solution : A thin wire ring of radi The situation is shown

in Fig.

Consider a small element dl of the wire at an angle θ and (θ=dθ) with centre O. The charge on this element. = × = ×

= From the symmetry of the problem = ∫

where dEx is the projection of field strength dE due to dl

on x-axis. Here, = cs = cs

= ∫ cs + ⁄− ⁄

= [sin ]− ⁄+ ⁄ = =

Substituting the values, we get = × −× . × . × − × . = ⁄ .

25. A thin wire ring of radius r carries a charge q. Find

the magnitude of the electric field strength on the axis

of the ring as a function of distance l from the centre.

Investigate the obtained function at l> > r. Find the

maximum strength magnitude and the corresponding

distance l.

Solution : See Fig.



Consider a small element of length dl of the ring. Charge

dq on the element is given by = ×

Electric field strength due to this charge = × × +

Electric field strength due to this charge along x-axis = cs = × × cs+ = × + ⁄

[ cs = √ + ] = × + ⁄ ∫ = × + ⁄ × = × + ⁄ …… (1)

For l > > r, we have = ×

Thus the ring behaves like a point charge. For Emax,

dE/dl=0. From eq.(1), we get = [ + ⁄ ∙ − ⁄ + ⁄ ×+ ] =

+ ⁄ = ⁄ + ⁄ ×

Solving, we get = √ …. (2) Substituting the value of l in eq. (1), we get = × ( √⁄ ) + / ⁄ = √

26. A ball of radius R carries a positive charge whose volume

density depends on a separation r from the ball’s centre as = − ⁄ , where ρ0 is a constant. Assuming the

permittivity of the ball and the environment to be equal

to unity, find:

(a) the magnitude of the electric field strength as a

function of the distance r both inside and outside the

ball,

(b) The maximum intensity and the corresponding

distance rm.

Solution : (a) Le the ball be divided into a number of

spherical shells. Consider one shell at a distance r and

of thickness dr. Volume (dV) of this shell is given by

dV=4 =

Given that = − ⁄

= − ×

Now total charge enclosed in the sphere of radius r is

given by ∫ = ∫ −

= ∫ −

or = [ − ] = [ − ] So, E= , = × × [ − ] = −

when r> > R, = ∫ − r r dr =

and =

(b) For maximum electric field

= − or = − =

− = or =

Now = − × =

27. ABCD is a square of 4 cm side. Charges of ×− , − × − × − coulomb are

placed at points A,C and D respectively. Find the

intensity of the field at point B.

Solution : The Electric field at B due to a charge + × − at A given by = ∙ = × × −. = × along AB

The electric field at B due to a charge − × − at C

is given by = × × −9. = × along BC

Similarly, = × × −9× − = × along DB.

As E is vector quantity, the total intensity at B is the

vector sum of three intensities due to charges at A,C and

D. Hence, = + +

Now, = ( + ) ⁄ = [ + cs + − sin ] ⁄

= [ + ⁄ + cs + + ⁄ − sin ] ⁄

=[ + + + cs − sin ] ⁄ = √ × newton/coulomb and tan = ⁄ = − sin+ cs = × [ − √⁄ ]× [ + √⁄ ]



= . θ= 44′.

28. A charge of 4 × − coulomb is uniformly

distributed over the surface of a ring shaped

conductor of radius 0.3 m. Calculate the intensity of

the electric field at point on the axis of the ring at a

distance of 0.4 m from the plane of the conductor and

specify its direction. What is the intensity at the

centre of the ring?

Solution: = = ∙ ∙ cs [See Fig.]

= ∙ + ⁄ cs = = + ⁄ = × [ × − .[ . + . ] ⁄ ] = .

Along the axis and away from centre O.

29. A very long straight uniformly charged thread carries a

charge per unit length. Find the magnitude and direction of the electric field strength at a point which is at a

distance y from the thread and lies on the perpendicular

passing through one of the thread’s ends. Solution : The electric field at a point P at a distance y

from the end A due to an element dx at a distance x from

A: =

It’s component along x = sin

Also from the figure =

And = or = ∙ = = ∙ = sin

or = ∫ = ∫ sin ⁄ = , Similarly = cs

That is, = and = √ + = √

This field is directed at an angle of − = with the thread.

30. A non-conducting disc of radius a and uniform positive

surface charge density is placed on the ground with its

axis vertical. A particle of mass m and positive charge q

is dropped, along the axis of the disc from a height H

with zero initial velocity. The particle has q/m=4 .

[1998, 8M]

(a) Find the value of H if the particle just reaches the

disc.

(b) Sketch the potential energy of the particle as a

function of its height and find its equilibrium position.

Solution : Potential at a height H on the axis of the disc

V (P).The charge dq contained in the ring shown in

figure

dq=

Potential at P due to this ring

dV= . where x=√ +

dV= . √ +

= √ +

Potential due to the complete disc = ∫ ==

= ∫ √ +== = [√ + − ]

Potential at centre, (O) will be

= H=0

(a) Particle is released form P and it just reaches

point O. Therefore, form conservation of mechanical



energy Decrease in gravitational potential energy =

Increase in electrostatic potential energy (∆ = because = = )

mgH=q[ − ] = ( ) [ − √ + + ] …

= , =

Substituting in Eq.(i), we get

gH=2g [ + − √ + ] or

= + − √ +

or √ + =a+

or + = + +

or =

H= /

(b)Potential energy of the particle at height

H=Electrostatic potential energy + gravitational potential

energy

U =qV+mgH Here V=Potential at height H

U= [√ + − + ]..(ii)

At equilibrium position

F=− =

Differentiating Eq.(ii) w.r.t.H

Or mg + [ √ + − ] =

( = )

mg+2mg[ √ + − ] =

or 1+√ + − =

or √ + =

or + =

or = or H=√

From Eq. (ii), we can write U-H equation as

U=mg( √ + − ) (Parabolic Variation)

U=2mga at H=0

and U= = √ mga

at H=√

Therefore, U-H graph will be as shown

Note that at H=√ , U is minimum.

Therefore, H=√ is stable equilibrium position.

31. Four point charges +8 µC,-1µC,-1µC and + 8µC are fixed

at the points -√ / m,-√ / m,+√ / m and +√ /

m respectively on the y-axis. A particle of mass 6×− kg and charge +0.1µC moves along the x-direction.

Its speed at x=+∞ is . Find the least value of for

which the particle will cross the origin. Find also the

kinetic energy of the particle at the origin. Assume that

space is gravity free. / = × Nm / .

[ IIT JEE]

Solution:

In the figure q=1µC= − , = + . µ = −

and m=6× − kg and Q=8µC=8× −

Let P be any point at a distance x from origin O.

Then

AP=CP=√ +

BP=DP=√ +

Electric potential at point P will be

V= − where K= = × Nm /C

V=2× × [ × −√ + − −√ + ]



V=1.8× [√ + − √ + ] …(i)

Electric field at P is

= − = . × [ (− ) ( + )− (− ) ( + )− / ]

E=0 on x-axis where x=0 or

+ / = + /

⇒ /+ / = + /

⇒ + = +

This equation gives x=±√ m. The least value of kinetic energy of the particle at

infinity should be enough to take the particle upto

x=+√ m, Because at x=+√ m, E=0. ⇒ Electrostatic force on charge q is zero or = . For at x> √ m, is repulsive (towards positive x-axis)

and for x < √ m, E is attractive (towards negative x-

axis)

Now, from Eq.(i), potential at x=√ m

V=1.8× [√ + − √ + ]

V=2.7× volt

Applying energy conservation at x=∞ and x=√ m

m = …(ii)

= √

Substituting the values

= √ × − × . ×× − .

Minimum value of rm . i , tential at rgin = is

= . × [ √ − √ ]

= . × V

Let K be the kinetic energy of the particle at origin.

Applying energy conservation at x=’0 and at x=∞

K+ = m

But = [from Eq.(ii)]

K= −

K= − × − ×

K=3× −

Note E=0 or on is zero at x=0 and x=±√ m of

these x=0 is stable equilibrium position and x=±√ is

Unstable equilibrium position.

32. Three concentric metallic shells A,B,C of radius a, b and

c (a<b<c) have surface charge densities and

respectively.

(i) Find the potential of three shells A,B and C

(ii) If the shells A and C are at the same

Potential, obtain the relation between the

radii a, b and c [IITJEE]

Solution: The three shells are shown in fig.

Potential of A

=(Potential of A due to + on A)

+ (Potential of A due to- on B)

+(Potential of A due to + on C)

= [ − + ] = [ − + ] Potential of B

= (Potential due to + on A)

+(Potential due to – on B)

+(Potential due to + on C)

= [ − + ] =

[ − + ] Potential of C

=(Potential due to + on A)

+(Potential due to – on B)



+(Potential due to + on C)

= − +

= [ − + ]

(ii) Given that =

[ − + ] = [ − + ]

a-b-c= − +

Solving we get c= + .

33. A particle of mass 9× − kg and a negative charge of

1.6 × − coulomb projected horizontally with a

velocity of m/s into a region between two infinite

horizontal parallel plates of metal. The distance between

the plates is 0.3 cm and the particle enters 0.1 cm below

the top plate. The top and bottom plates are connected

respectively to the positive and negative terminals of a

30 volt battery. Find the component of the velocity of

the particle just before it hits one on the plates.

Solution: We know that

E= and .

E=

Here V=30 volt and r=0.3 cm =3× − m

E= × − = rce n the article f negative charge mving

between the plates

F=e× = . × − × = . × − newtn

The direction of force will be towards the positive plate.

Now acceleration of the particle

f=F/m, = . × − / × −

= 1.77× m/sec

As the electric intensity E is acting in the vertical

direction the horizontal velocity v of the particle remains

same. If y is the displacement of the particle, then

y=

Here= y=0.1cm= − m. f = . × m/sec

− = × . × −

Solving we get t=1.063× − second

Component of velocity in the direction of field is given

by = acceleration × time

= . × . × −

= 1.881× / . 34. In fig. , an electric dipole is placed at a distance x from

an infinitely long rod of linear charge density .

(a) Find the net force acting on the dipole.

(b) What is the work done in rotating the dipole through

?

(c) If the dipole is slightly rotated about its equilibrium

position, find the time period of oscillation. Assume that

the dipole is linearly restrained.

Solution: Let us first calculate the electric field due to

infinitely long rod of linear charge density . For this

purpose see fig.

Consider an element of length dx at a distance x from O.

The charge on this element is dq= dx. The field dE at a

point P due to this charge dq

dE = × =

d = sin and d = cs

Further consider an element of length dx at a distance x

from O towards left. In this case, the components of dE

along X and Y axis will be dE sin and dE cos

respectively. Due to symmetry, the horizontal

components of the field intensity will cancel each other

and only vertical components remain. Hence the total

intensity of the electric field will be 2 dE cos in Y

direction.

E = ∫ cs∞

E = ∫ × × cs∞

= ∫ ∞

From figure = tan or = θ dθ

dx = r sec θ dθ and = + +

when x=0, θ=0 and when x=∞, = /



E= ∫ +/

= ∫ + /

= ∫ /

+ =

= ∫ = [sin ] //

= [ − ] = . (a) The electric field at a distance x from the rod is =

The net force in the dipole is given by

F=p = [− ] = − …(1)

Negative sign shows that the force is attractive.

(b) We know that work done is equal to the

difference of potential energy in the two positions of

dipole.

Hence W=∆ = − . Here = −. = − cs = −

and = −. = − cs =

W= pE-(-pE)=2pE

= =

(c) The restoring torque = − sin = − =− / .

Further = =

+ =

+ =

or + = where =

This represents the equation of S.H.M. Time period T is

T=2 √ =√

35. A conducting sphere S1 of radius r is attached to an

insulating handle. Another conducting sphere S2 of

radius R is mounted on an insulating stand. S2 is

initially uncharged.S1 is given a charge Q, brought into

contact with S2 and removed. S1 is recharged such that

the charge on it is again Q and it is again brought into

contact with S2 and removed. This procedure is repeated

n times. [IIT JEE]

(a) Find the electrostatic energy of S2 after n such

contacts with S1.

(b) What is the limiting value of this energy as n → ∞?

Solution:-

Capacities of conducting spheres are in the ratio of their

radii. Let C1 and C2 be the capacities of S1 and S2, then =

(a) Charges are distributed in the ratio of their

capacities. Let in the first contact, charge acquired by S2

is q1. Therefore, charge on S1 will be Q-q1. Say it is q1

′ = − = =

= + …(i)

In the second contact, S1 again acquired the same charge

Q.

Therefore, total charge in S1 and S2 will be

Q+q1 = Q + +

This charge is again distributed in the same ratio. The

charge on S2 in second contact,

q2 =Q + + + = [ + + + ] Similarly,

q3=Q[ + + + + + ] and qn+Q[ + + + + ⋯+ + ] or qn= Q [ − + ] … (ii)

[ = −− ] Therefore, electrostatic energy of S2 after n such contacts

Un= =

or Un =

Where qn can be written from Eq.(ii).

(b) As n → ∞

∞ = , ∞ = = /

∞=

36. Two parallel plate capacitors A and B have the same

separation d=8.85× − between the plates. The

plate areas of A and B are 0.04m2 and 0.02m

2

respectively. A slab of dielectric constant (relative

permittivity) K = 9 has dimensions such that it can

exactly fill the space between the plates of capacitor

B. [ IIT JEE]



(a) the dielectric slab is placed inside A as shown in

figure (i) A is then charged to a potential difference

of 110V. Calculate the capacitance of A and the

energy stored in it.

(b) The battery is disconnected and then the dielectric

slab is removed from A. Find the work done by the

external agency in removing the slab from A.

(c) The same dielectric slab is now placed inside B,

filling it completely. The two capacitors A and B are

then connected as shown in figure (iii). Calculate the

energy stored in the system. Capacitors before and after

the introduction of the dielectric.

Solution:

(a) Capacitor A is a combination of two capacitors

and in parallel. Hence, = + = + = +

Here, A=0.02m2. Substituting the values, we have = + . × − .. × − = . × −

Energy stored in capacitor A, when connected with a 110

V battery is

= = × −

= . × −

(b) Charge stored in the capacitor = = . × − = . × −

Now, this charge remains constant even after battery is

disconnected. But when the slab is removed,

capacitance of A will get reduced. Let it be ′

′ = = . × − .. × −

′ = . × −

Energy stored in this case would be

′ = = . × −. × −

′ = . × − >

Therefore, work done to remove the slab would be

W= − = . − . × −

or W= 4.84× −

(c) Capacity of B when filled with dielectric is = = . × − .. × −

= . × −

These two capacitors are in parallel. Therefore, net

capacitance of the system is

C= ′ + = . + . × −

C=2.2× −

Charge stored in the system is

q= = . × −

Therefore, energy stored, U=

U=( . × − ). × −9 or U=1.1× −

37. The figure shows two identical parallel plate

capacitors connected to a battery with the switch S

closed. The switch is now opened and the free space

between the plates of the capacitors is filled with a

dielectric of dielectric constant (or relative

permittivity)3. Find the ratio of the total electrostatic

energy stored in both capacitors before and after the

introduction of the dielectric.

Solution: Before opening the switch potential difference

across both the capacitors is V, as they are in parallel.

Hence, energy stored in them is, U= =

= = …(i)

After opening the switch, potential difference across it is

V and its capacity is 3C

= =

In case of capacitor B, charge stored in it is q=CV and its

capacity is also 3C.



Therefore, = =

= +

= = = …(ii)

From Eqs.(i) and(ii) =

38. A rectangular parallel-plate capacitor has a dielectric

slab that partially fills the space between the plates as

shown in fig.

Show that the capacitance is

C= [ − − ]

(b) The capacitor shown in above figure is connected to

a battery with terminal voltage V. This connection is

maintained while the dielectric slab is held in place by

the application of a force F. Determine the required

force F.

Solution: F = − .

(a) The system shown in figure is equivalent to

capacitors in parallel-one with free space and plate area

b x and the other with dielectric and plate area b ( a-x).

The resultant capacity is given by

C= + −

= [ + − ] = + [ − − ] (b)The energy of the capacitor is given by

U= = × [ − − ] This is a function of x.

If F is the force acting on the slab, then F dx would be

the work done by the field on the slab in increasing x by

dx. This is equal to the loss of energy dU.

dU = -F dx or F = -

F = × [ − − ]

= − = −

39. What amount of heat will be generated in the circuit

shown in the fig, after the switch Sw is shifted form

position 1 to position 2?

Solution: When the switch is in position 1. + − = ⇒ = −

⇒ = −

When in position

2, new charge = +

∆ = − = − + =

This must flow through the battery .. Besides it also

provides the energy C for the capacitor. Total

energy increment when the circuit is changed form

position 1 to 2:

-∆ + − −

=- + - −

=- + − − +

=- . Heat produced = . 40. Determine the potential difference VA -VB between

points A and B of the circuit shown in fig. Under

what condition is it equal to zero?

Solution: Let a charge Q1 flows in the upper branch and

a charge Q2, in the lower one. Then obviously.

+ = = +

Also, + = ⇒ = +

Now, − = − − − = −



= + − + = [ − + + ] − will be zer, when

= or =

41. What amount of heat will be generated in the circuit

shown in fig.(a) when the switch S is thrown from

position a to b.

Solution: When the switch is on a, the effective

capacitance can be calculated with the help of equivalent

circuit shown in fig.

Effective capacitance =++

Charge on capacitor 2= ++

Charge on capacitor 1= ++ +

= + … (i)

Now consider the case when switch is on b. In this case

the capacitor 1 will be in series with capacitors 2 and 3

in parallel.

Now charge on capacitor 1= ++ … (2)

The charge flown can be calculated by subtracting eq.(2)

from eq.(1). Hence

Charge flown= + , Heat generated = +

42. Two parallel capacitors each of capacitance C were

connected in series to a battery of e.m.f. E. Then one of

the capacitor was filled with dielectric of permittivity . (a) How many times did the electric field strength in

that capacitor decrease? ( b) What amount of charge

flows through the battery?

Solution: Here the two capacitors are in series and hence

p.d. across each = . The electric field strength between

the plates = (where r= distance between the plates).

When one capacitor is filled with dielectric, its capacity

= C

Effective capacity of two condensers

= . + = +

charge on condensers = . +

Potential = ℎ = . + ×

= + , Field = +

The electric field decreases from to + ,i.e., + times.

Charge flows through the battery

= original charge – final charge

= − + = −+



BOARD EXAM QUESTIONS

I. VERY SHORT ANSWER QUESTIONS:

1. What do you mean by electrostatics?

2. Define charge. What is the SI unit of electric charge?

3. Name the methods by which you can charge a neutral

object.

4. What is the basic difference between electric force and

gravitational force?

5. Name a simple device used to detect whether a body is

charged or neutral.

6. When we rub a glass rod with silk cloth, it acquires

positive charge. What does this signify?

7. What is charging by induction?

8. What do you mean by a point charge?

9. Give the value of electric permittivity of free space along

with its units.

10. Define Coulomb’s law of electrostatics. 11. Define test charge

12. Define field.

13. What do you mean by electric field lines?

14. Draw electric field lines due to a dipole

15. Why is electric field intensity inside a charged conductor

zero?

16. The electric lines of force do not pass through a closed

conductor. Why?

17. Define electric flux.

18. Define electric dipole.

19. Define electric dipole moment.

20. What is the net force on a dipole in a uniform electric

field?

21. What do you mean by Gaussian surface?

22. What do you mean by continuous charge distribution?

23. Name the different types of charge distributions.

24. State Gauss’s law of electrostatics. 25. Define linear charge density.

26. Define surface charge density.

27. Define volume charge density.

28. Does the charge outside the Gaussian surface contribute

to total electric flux?

29. What is the strength of an electric field inside a charged

spherical shell?

30. A sphere S1 of radius r1 encloses a total charge Q. If there

is another concentric sphere S2 of radius r2 (> r1) and

there be no additional charges between S1 and S2, find the

ratio of electric flux through S1 and S2.

31. What is an ideal dipole?

32. What do you mean by 1 eV?

33. Does the electric potential increase or decease along the

electric line of force?

34. For what position of an electric dipole inside a uniform

electric field its potential energy is 9a0 minimum and (b)

maximum?

35. Define equipotential surface?

36. Draw an equipotential surface for a uniform electric field.

37. Electric potential of Earth is taken to be zero. Why?

38. A hallow metal sphere of radius 10 cm is charged such

that potential on its surface in 5 V. what is the potential at

the center if the sphere?

39. In which orientation, a dipole placed in a uniform electric

field is in (a) stable and (b) unstable equilibrium?

40. Can two equipotential surfaces intersect each other? Give

reasons to justify your answer.

41. What meaning would you give to the capacitance of a

single conductor?

42. What is dielectric constant of a medium in terms of force

between electric charges?

43. What is electrostatic shielding?

44. Define dielectrics.

45. Why does a given capacitor store more charge at a given

potential difference when a dielectric is filled in between

the plates?

46. If the plates of a charged capacitor are suddenly

connected to each other by a wire, what will happen?

47. Define dielectric strength.

48. We have two metal spheres of same radius r but one is

solid and other is hollow. Which sphere has higher

electric capacitance and why?

49. Define capacitance of a capacitor.

50. What is the net charge on a charged capacitor?

II. SHORT ANSWER QUESTONS:

1. Distinguish between conductors and insulators. Give

examples.

2. Two charges attract each other with a force of 1.5 N.

What will be the force if the distance between then is

reduced to one-ninth of its original value?

3. What is the force between two small charged spheres

having charges 2 x 10-7

C and 3 x 10-7

C, respectively,

placed 30 cm apart in air?

4. Can a charged body attract another uncharged body?

Explain.

5. How can you charge a metal sphere positively without

touching it?

6. Four point charges qA = 2 μC, qB = -5 μC, qC = 2μC, and qD = -5 μC are located at the corners of a square ABCD of slide 10 cm. What is the force on a charge of 1μC placed at the center of the square?

7. Four charges of same magnitude and same sign are

placed at the corners of a square, each of side 0.1 m.

What is electric field intensity at he centre of the square?

8. Explain what is meant by an electric line of force. Give

its two important properties.

9. Define the term electric field intensity. Electric field

inside a conductor is zero, explain.

10. Derive an expression for the electric field intensity at a

distance from a point charge q.



11. Two point electric charges of unknown magnitude and

sign are placed at a distance of apart. The electric field

intensity is zero at a point, not between the charges but

on the line joining them. Write two essential conditions

for this to happen.

12. A 3.0 μC point charge is placed in an external uniform electric field of 1.6 x 10

4 N/C. At what distance from the

charge is the net electric field zero?

13. A point charge of 2.0 μC is at the center of a cubic

Gaussian surface 9.0 cm on edge. What is the net electric

flux through the surface?

14. A point charge causes an electric flux of -1.0 x

103 Nm

2/C to pass through a spherical Gaussian surface

of 10.0 cm radius centered on the charge. (a) if the radius

of the Gaussian surface were doubled, how much flux

would pass through the surface? (b) What is the value of

the point charge?

15. A system has two charges qA=+2.5 x 10-7

C and qB = -2.5

x 10-7

C located at points A: 0, 0, -15cm) and B:

(0,0,+15 cm), respectively what are the total charges and

electric dipole moment of the system?

16. Define electric flux. Write its SI units. A spherical rubber

balloon caries a charge that is uniformly distributed over

its surface. As the balloon is blown up and increases in

size, how does the total electric flux coming out of the

surface change? Give reason.

17. An electric dipole of dipole moment 20 x 10-6

cm is

enclosed by a closed surface. What is the net flux coming

out of the surface?

18. Figure shows three charges, labeled qq, q2, and q3. A

Gaussian surface is drawn around q1 and q2. (a) Which

charges determine the electric flux through the Gaussian

surface? (b) Which charges produce the electric field at

the point P ? Justify your answers.

19. Derive Coulomb’s law from Gauss’s law.

20. Use Gauss’s law to derive an expression for electric field at a point due to a uniformly charged spherical shell.

21. Using Gauss’s law, derive an expression for infinite plane sheet of charge.

22. In which orientation, a dipole placed in a uniform electric

field is in (a) stable, (b) unstable equilibrium?

23. A charge, +q, is placed inside a spherical Gaussian

surface. The charge is not located at the center of the

sphere. (a) Can Gauss’s law tell us exactly where the charge is located inside the sphere? Justify your answer.

(b) Can Gauss’s law tell us about the magnitude of the

electric flux through the Gaussian surface? Why?

24. Two large, thin metal plates are parallel and close to each

other. On their inner faces, the plates have surface charge

densities of opposite signs and of magnitude 17.0 x 10-22

C/m2. What is the electric field DE (a) in the outer region

of the first plate, (b) in the outer region of the second

plate, and (c) between the plates?

25. During lightning, while driving a car, a person should

stop the car and stay inside the closed car instead of

coming out in the open. Comment.

26. A sensitive instrument is to be shifted away from a strong

electrostatic field in its vicinity. Suggest a possible way.

27. Distinguish between electric potential and potential

energy

28. Define electric potential at a point in an electric field.

What is its unit?

29. Define the dipole moment of an electric dipole. How

does the electric potential due to a dipole vary on the

dipole axis as a function of r (distance of the field point

from the mid-point of the dipole) at large distances?

30. Establish a relation between electric field and electric

potential.

31. In fig, how much work must we do to bring a particle of

charge Q = +16e that is initially at rest, along the dashed

line from infinity to the indicated point near two fixed

particles of charges q1 = +4e and q2 = -q1/2? Distance d =

1.40, θ1 = 430, and θ2 = 60

0.



32. Consider a point charge q = 1.0 μC, point A at distance d1

=2.0 m from q , and point B at distance d2 = 1.0 m. (a) if

A and B are diametrically opposite each other, as in

Fig(a), what is the electric potential difference VA – VB?

(b) What is that electric potential difference if A and B

are located as in Fig b?

33. Two particles, of charges q1 and q2, are separated by

distance d as in fig. The net electric field due to the

particles is zero at x = d/4. With V = 0 at infinity, locate

(in terms of d) any point on the x-axis (other that at

infinity) at which the electric potential due to the two p

articles is zero.

34. A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the center

of the hexagon.

35. Show mathematically that the potential at a point on the

equatorial line of an electric dipole is zero.

36. Three point charges have identical magnitudes, but two

of them are positive and one is negative. These charges

are fixed to three corners of a square (Fig). No matter

how the charges are arranged, the potential at the empty

corner is positive. Explain why.

37. Two conductors of identical shape and size, but one of

copper and the other of aluminum (which is less

conducting), are both placed in an identical electric field.

In which metal more charges will be induced?

38. (a) Why does the electric field inside a dielectric decrease

when it is placed in an external electric field?

(b) a parallel-plate capacitor with air between the plates

has a capacitance of 8 pF. What will be the capacitance if

the distance between the plates is reduced by half and the

space between them is filled with a substance of

dielectric constant K = 6?

39. What will be the effect on capacitance of parallel-plate

condenser when the area of each plate is doubled and

distance between them is also doubled?

40. In a parallel-plate capacitor shown in Fig. the potential

difference of 102 V is maintained between the plates.

What is the electric field at points A and B?

41. Write an expression for potential energy of two charges

q1 and q2 at and in uniform electric field .

42. Explain electrostatic shielding. Give one example.

43. Two charged conducting spheres of radii a and b are

connected to each other by a wire. What is the ratio of

electric fields at the surfaces of the two spheres? Use the

result obtained to explain why charge density on the

sharp and pointed ends of a conductor is higher than on

its flatter portions.

44. What is a capacitor? Define capacitance. Discuss its

units.

45. A parallel-plate capacitor is charged to a potential

difference V by a dc source. The capacitor is then

disconnected from the source. If the distance between

the plates is doubled, state with reason how the

following change: (a) electric field between the plates,

(b) capacitance, and (c) energy stored in the capacitor.

III. LONG ANSWER QUESTIONS:

1. How will you show that there are two kinds of charges?

2. Explain the phenomenon of charging by induction.

3. Explain what is meant by quantization of charge and

conservation of charge.

4. State and prove Coulomb’s law in vector form. Does it obey Newton’s third law of motion?

5. A charge q is placed at the centre of the line joining two

equal charges Q. Show that the system of three charges

will be in equilibrium if = −

6. Explain the concept of electric field. Deduce a relation

between electric field strength and force.

7. Derive an expression for electric field intensity at a point

due to (a) a point charge and (b) a group of charges.

8. What is meant by electric lines of force? Give their

important properties.

9. Derive an expression for electric field at the axial line of

electric dipole.

10. Briefly explain the meaning of electric dipole and dipole

moment. Give some examples of electric dipole.

11. State Gauss’s law and using this law derive an expression for electric field due to a charged spherical shell for a

point (a0 outside, (b0 on the surface, and © inside the



spherical shell? Plot the variation of electric field with

the distance from the center of the shell

12. State and prove Gauss’s law. Using this law derive an expression of electric field for an infinite long straight

wire carrying charge q.

13. Using Gauss’s theorem, deduce an expression for the electric field at a point due to a uniformly charged

infinite plane sheet.

14. (a) Define electric pole moment. Is it a scalar or a

vector? Derive the expression for the electric field of a

dipole at a point on the equatorial plane of the dipole.

(b) Draw the equipotential surfaces due to an electric

dipole. Locate the points where the potential due to the

dipole is zero .

15. Use Gauss’s law to derive the expression for the electric field between two uniformly charged large parallel sheets

with surface charge densities – respectively.

16. Figure a, b shows the field lines of a positive and

negative point charge, respectively. (a) What is the sigh

of the potential difference (i) Vp – VQ and (ii) VB – VA?

(b) What is the sign of the potential energy difference of

a small negative charge between points (i) Q and P and

(ii) A and B? (c) What is the sign of the work done by the

field in moving a small positive charge from Q to P? (d)

What is the sign of the work done by the external agency

in moving a small negative charge from B to A? (e) Does

the kinetic energy of a small negative charge increase of

decrease in going from B to A?

17. The electric field at a single location is zero. Does this

fact necessarily mean that the electric potential at the

same place is zero? Use a spot on the line between two

identical point charges as an example to support your

reasoning.

18. (a) Define electric flux. Write its S.I. unit. (b) Using

Gauss’s law, prove that the electric field at a point due to

a uniformly charged infinite plane sheet is independent of

the distance from it .

19. The two graphs drawn below, show the variation of

electrostatic potential (V) with , ( r being distance of the

field point from the point charge) for two point charges

(i) what are the signs of the two charges?

(ii) Which of the two charges has a larger magnitude and

why ?

20. Find the P.E. associated with a charge ‘q’ if it were

present at the point P with respect to the ‘set-up’ of two charged spheres, arranged as shown. Here O is the mid-

point of the line .

21. Derive an expression for the potential energy of (a) a

single charge in an electric field and (b) a system of two

charges in an external field.

22. Derive an expression for the capacitance of a parallel-

plate capacitor.

23. Explain the principle of a capacitor.

24. Explain the concept of electrostatic potential from

electrostatic potential energy

25. Define ‘dielectric constant’ of a medium. Briefly explain why the capacitance of a parallel plate capacitor

increases, on introducing a dielectric medium between

the plates.

26. Deduce the effect of introducing (a) a conducting slab

and (b) a dielectric slab in between the plates of a

parallel-plate condenser on the capacitance of the

condenser.

27. Three condensers C1, C2, and C3 are connected in series.

Derive an expression for the equivalent capacitance.

28. Discuss briefly the principle, construction, and working

of a van de Graff electrostatic generator. How is the

leakage of charge minimized from the generator?

29. Derive expression for energy stored between he plates of

a capacitor (using integration method). Also extend the

result to obtain expression for energy stored per unit

volume.



30. Prove that the force on each plate of a parallel plate

capacitor has a magnitude equal to (1/2) qE, where q is

the charge on the capacitor, and E is the magnitude of

electric field between the plates .

IV. NUMERICALS:

1. Consider three charges q1, q2, and q3 each equal to q at

the vertices

of an equilateral triangle of side I. What is the force on a

charge Q (with the same sign as q) placed at the centroid

of the triangle, as shown in Fig

2. Consider the charges q1, q1, and -q placed at the vertices

of an equilateral triangle, as shown in fig. What is the

force on each charge?

3. Three point charges of 2 μC, -3 μC, and -3 μC are kept at the vertices, A, B, C, respectively, of an equilateral

triangle of side 20 cm as shown in Fig. What should be

the sign and magnitude of the charge to be placed at the

midpoint (M) of side BC so that the charge at A remains

in equilibrium?

4. A particle of mass − kg and charges 5 enters into a

uniform electric field of 2 x − , moving with a

velocity of 20 − in a direction opposite to that of the

field. Calculate the distance it would travel before

coming to rest

5. (i) Can two equipotential surfaces intersect each other?

Give reasons.

(ii) Two charges – + are located at points A , , − , . + respectively. How much

work is done in moving a test charge from point 7, , to − , , ?

6. Two charges are located on the x-axis: q1=+6.0 μC at x1

= +4.0 cm and q2 = +6.0 μC at x2= -4.0 cm. To other

charges are located on the y-axis: q3 = +3.0 μC at y3 =

+5.0 cm and q4 = -8.0 μC at u4 = +7.0 cm. Find the net

electric field (magnitude and direction) at origin.

7. Two point charges q1 and q2 of magnitude + 10-8

C and -

10-8

C, respectively, are placed 0.1 m apart. Calculate the

electric fields at points A,B, and C shown in Fig.

8. Two point charges qA = 3 μC and qB = -3 μC are located 20 cm apart in vacuum, (a) What is the electric field at

the mid point O of the line AB joining the two charges?

(b) if a negative test charge of magnitude 1.5 x 10-9

C is

placed at this point, what is the force experienced by the

test charge?

9. Two point charges 4 and -2 are separated by a

distance of 1 m in air, calculate at what point on the line

joining two charges the electric potential is zero.

10. Two point electric charges of values q and 2q are kept at

a distance d apart form each other in air. A third charge

Q is to be kept along the same line in such a way that the

net force acting on q and 2q is zero. Calculate the

position of charge Q in terms of q and d.

11. Figure shows a closed Gaussian surface in the shape of a

cube of edge length 2.00 m, with one corner at x1 = 5.00

m and y1= 4.00 m. The cube lies in a region where the

electric field vector is

given by = − . − . + . N/C,

with y in meters. What is the net charge contained by

the cube?

12. A thin conducting spherical shell of radius R has charge

Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point

outside the shell. Draw graph of electric field E® with

distance r from the center of the shell for 0 ≤ ≤ ∞ ?

13. A small metallic sphere carrying a charge +Q is located

at the center of a spherical cavity in a large uncharged

metal sphere as show in figure. Using Gauss’s theorem, find the electric field at points P1 and P2



14. A cubical Gaussian surface encloses a charge of 8.85 x − C in vacuum at the centre. Calculate the electric

flux passing through one of its faces.

15. Find the amount of work done in rotating an electric

dipole, of dipole moment 3 x −8 cm, from its position

of stable equilibrium to the position of unstable

equilibrium, in a uniform electric field of intensity

N/C.

16. In Fig, a thin conducting spherical shell of radius R =

2.00 cm has a uniformly spread surface charge of q =

5.00 x 10-15

C. What are the electric potential and

magnitude of the electric field on (a) the surface and at

(b) r = 2.00 R and (c) r = 5.00 R?

17. Two charges 3 x 10-8

C and -2 x 10-8

C are located 15 cm

apart. At what point, on the line joining the two charges,

is the electric potential zero? Take the potential at infinity

to be zero.

18. Four charges are arranged at the corners of a square

ABCD of side d, as shown in Fig. (a) Find the work

required to put together this arrangement. (b) A charge q0

is brought to the center e of the square, the four charges

being held fixed at its corners. How much extra work is

needed to do this?

19. A cube of side b has a charge q at each of its vertices.

Determine the potential and electric field due to this

charge array at the center of the cube.

20. (a) Can two equipotential surfaces intersect each other?

Give reasons. (b) Two charges –q and +q are located at

points A (0,0, -a) and B (0,0,+a), respectively. How much

work is done in moving a test charge from point P (7,0,0)

to point Q (-3,0,0)?

21. A metal sphere of radius 15 cm has a net charge of 3 x

10-8. (a) What is the electric field at the sphere’s surface?

9b0 if V = 0 at infinity, what is the electric potential at

the sphere’s surface? (c) At what distance from the sphere’s surface has the electric potential decreased by

500 V?

22. A spherical conductor of radius 12 cm has a charge of 1.6

x 10-7

C distributed uniformly on its surface. What is the

electric field (a) inside the sphere, (b) just outside the

sphere, and (c0 at a point 18 cm from the center of the

sphere?

23. A capacitor of unknown capacitance is connected across

a battery of V volts. The charge stored in it is 360 .

When potential across the capacitor is reduced by 120 V,

the charge stored in it becomes 120 .

Calculate :

(i) The potential V and the unknown capacitance C.

(ii) What will be the charge stored in the capacitor, if the

voltage applied had increased by 120 V ?

24. A parallel plate capacitor, each with plate area A and

separation d, is charged to a potential difference V. The

battery used to charge it is then disconnected. A

dielectric slab of thickness t and dielectric constant K is

now placed between the plates. What change, if any,

will take place in :

(i) charge on the plates.

(ii) electric field intensity between the plates.

(iii) capacitance of the capacitor.

Justify your answer in each case.

25. Two capacitors with capacity are charged to

potential respectively and then connected in

parallel. Calculate the common potential across the

combination, the charge on each capacitor, the

electrostatic energy stored in the system and the change

in the electrostatic energy from its initial value.

26. A network of four 10 μF capacitors is connected to a 500 V supply, as shown in Fig. Determine (a) the equivalent

capacitance of the network and (b) the charge on each



capacitor. (Note that the charge on a capacitor is the

charge on the plate with higher potential, which is equal

and opposite to the charge on the plate with lower

potential.)

27. (a) A 900 pF capacitor is charged by 100 V battery. How

much electrostatic energy is stored by the capacitor? (b)

The capacitor is disconnected from the battery and

connected to another 900 pF capacitor. What is the

electrostatic energy stored by the system?

28. A parallel-plate capacitor, each with plate area A and

separation d , is charged to a potential difference V. The

battery used to charge it is then disconnected. A dielectric

slab of thickness d and dielectric constant K is now

placed between the plates. What change if any , will take

place in (a)charge on the plates; (b) electric field intensity

between the plates; and (c)capacitance of the capacitor.

Justify your answer in each case.

29. Two capacitors, of capacitances 3 and 6 are

charged to potentials of 2 V and 5 V respectively. These

two charged capacitors are connected in parallel. Find

the charge across each of the two capacitors now.

30. Two capacitors of capacitances 3 and 6 are

charged to potentials of 2 V and 5 V respectively. These

two charged capacitors are connected in series. Find the

potential across each of the two capacitors now.


CURRENT ELECTRICITY 49

CURRENT ELCETRICITY Ohm’s law, Series and parallel arrangements of resistances and cells; Kirchhoff’s laws and simple applications; Heating effect of current

Electric Current : The time rate of flow of charge through

the conductor ( i.e. through any cross section of the

conductor ) is defined as electric current. Fig shows how

electric charges (electrons) move in a conductor under the

influence of a driving force ( battery ).

Thus, dt

dqi (Definition of current)

The SI unit for current is the coulomb per second, also

called ampere (A).

Current is scalar.

Fig shows a loop of

copper in electrostatic

equilibrium. The entire

loop is at a single

potential, and the

electric field is zero at

all points inside the

copper. (b) Adding a

battery imposes an

electric potential

difference between the

ends of the loop that

are connected to the

terminals of the

battery. The battery thus produces an electric field within

the loop, from terminal to terminal, and the field causes

charges to move around the loop. This movement of charges

is a current i.

Resistance and Resistivity :

Fig shows an assortment of resistors. The circular bands are

color coding marks that identify the value of the resistance.

The resistance of a conductor is defined as the ratio

of the potential difference applied across the conductor to

the current flowing through it.

Ohm’s Law: The current flowing through a conductor is

always directly proportional to the potential difference

across its two ends.

The Resistance ‘R’ is then,i

VR

,SI unit

1 ohm = 1 = 1 volt per ampere

In fig A potential difference V is applied between the ends

of a wire of length L and cross section A. establishing

current i.

The resistance of a conductor is directly proportional to its

length and inversely proportional to the area of cross

section A, i.e

R ∝ L

and RA

or R = ρ L / A

Where, ρ is a constant of proportionality and is known as

resistivity or specific resistance of the material of the

conductor.

Resistivity of the material of a conductor is equal

to the resistance offered by a wire to the resistance of unit –

length and unit – area of cross – section of the material. Its

SI unit is ohm – meter

Following fig (a) A device to whose terminals a potential

difference V is applied, establishing a current i. (b) A plot of

current i versus applied potential difference V when the

device is a 1000 resistor.

Power in Electric Circuits:

Fig: A battery B sets up

a current i in a circuit

containing an

unspecified conducting

device.

In Fig the charge dq

moves through a

decrease in potential of

magnitude V, and thus

its electric potential

energy decreases in

magnitude dU = dq V

= i dt V

The power P associated

with that transfer is the

rate of transfer dU/dt, which is P = iV (rate of electric



energy transfer). The unit of power that follows from above

equation is the volt-ampere (V.A). We can write it as

W1s

J1

s

C1

C

J1A.V1

.

Current in a single – Loop Circuit Energy Method:

Fig: A single-loop

circuit in which a

resistance R is

connected across an

ideal battery B with

emf E. The

resulting current i is

the same

throughout the

circuit.

From Fig, in a time interval dt, dq = I dt will have moved

through battery B, and the work that the battery will have

done on this charge is dW = dq = i dt.

From the principle of conservation of energy, the work

done by the (ideal ) battery must be the thermal energy that

appears in the resistor. i dt = i2 R dt.

This gives = i R. So R

i

Internal Resistance:

Fig (a) A single-loop circuit containing a real battery having

internal resistance r and emf E. (b) The same circuit, now

spread out in a line.

The potentials encountered in traversing the circuit

clockwise from a are shown in the graph. The potential Va is

arbitrarily assigned a value of zero, and other potentials in

the circuit are graphed relative

to Va.

The internal resistance ‘r’ of the battery is the electrical resistance of the conducting materials of the battery.

If we apply the loop rule clockwise beginning at point a, the

changes in potential gives us - ir – iR = 0. So rR

i

Combinations of Resistors:-

Resistances in Series:

Fig. (a) Three resistors are connected in series between

points a and b. (b) An equivalent circuit, with the three.

Resistance connected in series can be replaced with an

equivalent

resistances Req

that has the same

current I and the

same potential

difference ‘V’ as the actual

Resistances.

For Req , we apply

the loop rules. For

fig a, starting at

terminal ‘a’ and going clockwise

around the circuit

we find

- iR1 – iR2 – iR3 = 0,

Or 321 RRR

i

For Fig b, - iReq = 0 ,

Or eqR

i

Above eqs shows that

Req = R1 + R2 + R3

Resistances in Parallel:

Fig (a) Three resistors connected in parallel across points a

and b. (b) An equivalent circuit, with the three resistors

replaced with their

equivalent resistance

Req.

Resistance connected

in parallel can be

replaced with an

equivalent resistance

requirement that has

same potential

difference V and the

same total current i as

the actual resistances.

For Requ in Fig b, the

current in each actual resistance form fig a

321321

R

1

R

1

R

1Viiii

For fig b,eqR

Vi

From above Eqs 321eq R

1

R

1

R

1

R

1 ,



Kirchoff’s Rules:

i. Junction rule:- The sum of the magnitude of the currents

directed into a junction equals the sum of the magnitudes of

the currents directed out of the function

ii. Loop rule :- Around any closed circuit loop (eg :-

Wheatstone bridge),the sum of the potential drops equals the

sum of potential rises.

Multi Loop

Circuits:

Fig A multi-loop

circuit consisting of

three branches left-

hand branch bad,

right-hand branch

bcd, and central

branch bd. The

circuit also consists of three loops: left-hand loop badb

right-hand loop bcdb, and big loop badcb.

There is no variation in the charge at the junction d, the total

incoming current must equal the outgoing current.

i1 + i3 = i2 …………… (1) This rule is often called Kirchoff’s Junction rule ( or kirchoff’s Current law ). Let’s choose the left hand loop in a anti-clockwise

direction from point b, then 1 – i1R1 + i3R3 = 0 ………. (2) If we traverse the right-hand loop in a counterclockwise

direction from point b, then i3R3 – i2R2 - 2= 0……….. (3)

We now have three equations in the three unknown currents,

and they can be solved by a variety of techniques.

Ammeter and Voltmeter:

Fig shows A

single-loop

circuit, showing

how to connect

an ammeter (A)

and a voltmeter

(V).

An instrument

used to measure

currents is called

an ammeter. It is

essential that the

resistance RA of

the ammeter be

very small

compared to other resistances in circuit.

A meter used to measure potential difference is called a

voltmeter. It is essential that resistance RV of the voltmeter

be very large compared to the resistance of any circuit

element across which the voltmeter is connected.

RC Circuits:

Fig When switch S is

closed on a, the

capacitor C is

charged through the

resistor R. When the

switch is closed on b,

the capacitor

discharges through R.

Charging Capacitor

Fig (a) shows a plot of the buildup of charge on the

capacitor. Fig (b) shows a plot of decline of the charging

current in the above circuit. The curves are plotted for R =

2000 . C = 1 µF, and E = 10 V;

The small triangles

represent successive

intervals of one time

constant.

For examining the

charging process, apply

the loop rule to the

circuit, traversing it

clockwise from the

negative terminal of the

battery.

Then 0C

qiR ………….. (i)

Substituting dt

dqi

we find C

q

dt

dqR (Charging equation) …….. (ii)

The solution is q = t / RC

C ( e )

------- (iii)

The derivative of q (E) is the current i ( t) charging the

capacitor RC/te

Rdt

dqi

…….. (iv)

We find that the potential difference VC (t) across the

capacitor during the

charging process is

)e1(C

qV

RC/tC

The product RC that

appears in the above

eqn has the

dimensions of time

and it is called the

capacitive time

constant (Q)

Discharging Capacitor: With no battery in the discharge

loop, , Thus 0C

q

dt

dqR (Discharging equation)

The Solution is q = q0e –t/RC

[ q = q0 at t = 0 )

RC/t0 eRC

q

dt

dqi

( i =

q

RC

at t = 0 )

[ - sign indicates that q decreases with ‘t’ ]



JEE Corner

OBJECTIVES

1. In the circuit shown, some potential difference is applied

between A and B. Find the equivalent resistance

between A and B.

(a) 15 (b) 12.5 (c) 5

18 (d) 2

Ans(c)

Solution:

Rearranging the circuit, it will be of the form as shown in

below

It represents a

balanced Wheatstone

bridge and hence no

current follows

through 5 ohms.

Equivalent resistance between A and B

= x = ohms.

2. For the circuit shown in figure the current I2 will be

(a) 1 ampere (b) 0.5 ampere

(c) 0.6 ampere (d) 1.2 ampere

Ans(c)

Solution:

Beginning from right end the circuit is gradually reduced in

stages as shown in figure.

I = 25.4

17 = 4 amperes

From (b)

I1 =

39

9 I = 3 amperes.

From the circuit given in the problem,

I2 =

82

2 I1

= 10

2 3 = 0.6 amperes

3. A copper wire is stretched to make it 0.1 percent longer.

What is the percentage change in resistance?

(a) 0.1% (b) 0.2% (c) 0.5% (d) 0.4%

Ans(b)

Solution: R = A

l

If d is the density of wire and m the mass of wire Ald = m or

A = ld

m

R = ld

m

Since m

d is constant.

Log R = log m

d + 2 log l

Differentiating l

dl2

R

dR

R

dR 100 =

l

dl2 100

percentage change in resistance

= 2 0.1 = 0.2 %

4. P and Q are two points on a uniform ring of resistance R. The

equivalent resistance between and Q is :



(a)

)2(4

R2

2

(b) R

2

1

(c) R

2 (c) R

4

2

Ans(a)

Solution:

Resistance of section PSQ, R1 =

2

Rr

r2

R

Resistance of section PTQ, R2 = r2

)2(Rr

R2 =

2

)2(R

As R1 and R2 are in parallel

So, R eq = 21

21

RR

RR

=

2

2

4

R

(2 )

5. Two conductors have the same resistance at 0o C by their

temperature coefficients of resistance are α1 and α2 the

respective temperature coefficients of their series and

parallel combinations are nearly

(a ) + , +

(b) + , +

(c) + , + (d) + , +

Ans: (a)

Solution: The resistance of a conductor at temperature t0C

is given by = +

where R0 is the resistance at 00C

For series combination = +

At 00C, = + =

+ = + + +

= +

For parallel combination = +

At00C, = + = ⇒ =

( + = + + +

( + − = + − + + − ⇒ ( − = − + − <<

= +

Directions for 6 and 7 : Consider a block of conducting

material of resistively ‘ρ’ shown in Fig. Current ‘I’ enters at

‘A’ and leaves from ‘D’. We apply superposition principle to find voltage ‘ΔV’ developed between ‘B’ and ‘C’. The calculation is done in the following steps:

(i) Take current ‘I’ entering from ‘A’ and assume it to spread over a hemispherical surface in the block.

(ii) Calculate field E(r) at distance ‘r’ from A by using

Ohm’s law E=ρJ. Where J is the current per unit area at

‘r’. (iii) From the ‘r’ dependence of E(r). Obtain the

potential V(r) at r.

(iv) Repeat (i),(ii) and (iii) for current ‘I’ leaving ‘D’ and superpose results for ‘A’ and ‘D’.

6. ΔV measure between B and D is [AIEEE]

(a) − (b) − +

(c) − + (d) − +

Ans: (b)

Solution: = = .

Also = − ⟹ = −

Therefore, potential difference between B and C due to

current at A is = − ∫ = − ∫+

= − |− | + = − +

Similarly, potential difference between B and C due to

current at D is

′ = − +

From superposition principle, the potential difference

between B and C due to current at A and at D is = + ′ = − + .

7. For current entering at A, the electric field at a distance ‘r’ from A [AIEEE]



(a) (b) (c) (d)

Ans: (d)

Solution: = = ,

Hence the correct choice is (d).

8. A meter bridge is set-up as shown in figure, to determine an

unknown resistance X using a standard 10Ω resistor. The galvanometer shows null point when tapping –key is at 52

cm mark. The end-corrections are 1 cm and 2 cm

respectively for the ends A and B. The determined value of

X is [IIT JEE]

(a) 10.2 Ω (b) 10.6 Ω (c) 10.8 Ω (d) 11.1 Ω

Ans: (b)

Solution: Using the concept of balanced Wheat stone

bridge, we have, =

+ = +

= × = . Ω

9. A capacitor is charged using an external battery with a

resistance x in series. The dashed line shows the variation of

lnI with respect to time. If the resistance is changed to 2x,

the new graph will be [IIT JEE]

(a) P (b) Q (c) R (d) S

Ans: (b)

Solution: Charging current,

= − −

Taking log both sides, log = log ( ) − When R is doubled, slope of curve decreases. Also at t=0,

the current will be less. Graph Q represents the best. Hence,

the correct option is (b)

10. Find the time constant for the given RC circuit in correct

order (in µs). [IIT JEE]

= Ω, = Ω C = , = . (a) 18,4, 8/9 (b) 18,8/9, 4

(c) 4,18, 8/9 (d) 4, 8/9, 18

Ans: (b)

Solution: = = + + = = ( + ) ( + ) = × = = + ( + ) = ( ) =

11. A moving coil galvanometer has 150 equal divisions. Its

current sensitivity is 10 divisions per milliampere and voltage

sensitivity is 2 division per mill volt. In order that each

division reads 1V, the resistance (in Ω) needed to be connected in series with the coil will be[AIEEE]

(a) 103 (b) 10

5 ( c) 99995 (d) 9995

Ans: (d)

Solution: Current for full scale deflection of 150 divisions

is

g = × = = × −

Voltage for full scale deflection of 150 divisions is

g = × = = × −

Resistance of galvanometer coil is = gg = × − × − = Ω

If each division = 1 V, the required voltage for full scale

deflection is = × =

Let R be the required resistance to be connected in series

with G, then = g + ⇒ = × − + ⇒ R+5 = 10000 ⇒ R = 9995Ω



12. If each of the resistances in the network shown in the figure

is R, what is the resistance between the terminals A and B?

[IITJEE]

a) 2R (b) 4R (c) R (d) R/2

Ans : (c)

Solution: The given circuit makes a balanced Wheatstone’s bridge. The resistance between P and Q can be removed. All

resistors have value R.

= ( =

= = .. =

When R1 and R2 are inter changed, then + = × .. = Ω

Now potential difference across RL will be = [ ⁄+ ⁄ ] =

Earlier it was 9V

Since, = or P ∝

In new situation potential difference has been decreased

three times. Therefore, power dissipated will decrease by a

factor of 9.

13. Two bars of radius r and 2r are kept in contact as shown. An

electric current I is passed through the bars. Which one of

following is correct? [IIT JEE]

(a) Heat produced in bar BC is 4 times the heat produced in

bar AB

(b) Electric field in both halves is equal

(c) Current density across AB is double that of across BC

(d) Potential difference across AB is 4 times that of across

BC

Ans : (a)

Solution: Current flowing through both the bars is equal.

Now, the heat produced is given by

H = I2 Rt

or ∝

or =

= ⁄⁄ ( ∝ ∝ ) = =

14. In the circuit shown in the figure, the current through

[IIT JEE]

(a) the 3Ω resistor is 0.50 A

(b) the 3Ω resistor is 0.25 A

(c) the 4Ω resistor is 0.50 A

(d) the 4Ω resistor is 0.25 A

Ans: (d)

Solution: Net resistance of the circuit through 9 Ω. Current drawn from the battery, = = =Current through 3Ω resistor

Potential difference between A and B is − = − + = = = . = − = .

Similarly, potential difference between C and D − = − − + = − = − . = = = .

Therefore, = − = . − . = .

15. Three resistors and three batteries are connected as shown in

the figure. Then

(a) potential difference between points a and b is 5 volt

(b) the current through the 7 V battery is 2.4 A

(c) the potential difference between the terminals of 15 V

battery is 2.8 volt

(d) the potential difference between the terminals of 8 V

battery is 10 volt

Ans(a)



Solution:

By Kirchoff’s law, For loop 1, 2x + 5 (x – y) = 15 + 8

7x – 5y = 23

for loop 2,

(1 + 4)y + (3 + 2) (y – x) = 7 – 8

10y – 5x = – 1

Solving, x = 5 A, y = 2.4 A

Potential across ab = – 8 + 2.6 (5) = 5 volt A correct

Current through the 7 V battery = 2.4 A

Potential across 15 V battery = 15 – (1 5) = 10 volt

Potential across 8 V battery = 8 – (2.6 2) = 2.8 volt

16. In order to increase the resistance of a given wire of uniform

cross section to four times its value, a fraction of its length

is stretched uniformly till the full length of the wire

becomes 2

3 times the original length what is the value of

this fraction?

(a) 4

1 (b)

8

1 (c)

16

1 (d)

6

1

Ans (b)

Solution:

If L is the full length, let x out of L be stretched so that new

length becomes 2

3 L. If x’ is the stretched length, (L – x) +

xı = 2

L3

Or xı – x = 2

L …….(1)

If a is the original cross-section let a’ be the cross – section

of stretched portion.

xı aı = xa

aı = 'x

xa

……..(2)

New resistance = 'a

'x

a

)xL(

Where is specific resistance.

Original resistance = a

L

4 a

L =

'a

'x

a

)xL(

= 'xa

'x

a

)xL( 2

a

L4

xa

2

Lx

a

)xL(

2

x(L – x ) +

2

2

Lx

= 4Lx

Solving for x,

x = 8

L

17. A coil of emf 3.4 V and internal resistance 3 is connected

to an ammeter of resistance 2 and to an external

resistance of 100 . A voltmeter is connected across the

100 resistor. The ammeter reads 0.04 A. What are the

voltmeter reading and its resistance and what would be its

reading, if the voltmeter is ideal?

(a) 3.2 V, 400 and 3.24 V

(b) 4.8V, 600 and 4.86 V

(c) 6.4 V, 800 and 6.48 V

(d) 8 V, 900 and 8 V

Ans: (a)

Solution:

Voltmeters have usually high resistances while ammeters

have low resistances. Let RV be the resistance of voltmeter.

The parallel combination of RV and

100 = V

V

R100

R100

= Rı say.

Hence, the current in the circuit = . VRı+ + = 0.04 A.

Solving, we get Rı = 80 and RV = 400 .

The voltmeter resistance = 400

Voltmeter reading = (0.04A) 400100

400100

= 3.2 V

Voltmeter reading = 3.2 V

If the voltmeter is ideal.

RV = i.e., no

resistance across 100

. Then current

through the circuit =

3.4/(3 + 2 + 100) = (3.4

/ 105) A.

The voltage across voltmeter

= (3.4 / 105) 100 = 340 / 105 = 3.24 V

18. A d.c. generator with internal resistance R0 is loaded with

three identical resistances R interconnected as shown in



figure. At what value of R will the power generated in the

circuit is maximum?

(a) R = 3R0 (b) R = 2R0

(c) 3

RR 0 (d) 0R

2

3R

1. Ans: (a)

Solution:

The circuit can be redrawn as follows:

The three resistances are in parallel.

Equivalent resistance = R/3

By circuit equation i = 3/RR

E

0

Power delivered = i2

3

R

= 2

0

2

)3/RR(

E

R/3

P = 2

0

2

)RR3(

RE3

P is maximum, when dR

dP = 0

20 )RR3(

R

dR

d = 0

(3R0 + R)2 – R 2 (3R0 + R) = 0

R = 3R0

19. In the given circuit the current flowing through the

resistance 20 ohms is 0.3 ampere while the ammeter reads

0.8 ampere. What is the value of R1?

(a) 30 . (b) 40 .

(c) 50 . (d) 60 .

Ans(d)

Solution:

If I is the main current, the current divides in the

branches in the ratio

15

1:

20

1:

R

1

1

It is given the current through 20 =0.3 A

ratio 300 : 15 R1 : 20 R1

0.8

1

1

R35300

R15 = 0.3

120 R1 = 900 + 105 R1

R1 = 60

20. In the circuit diagram shown in the figure, which of the

following statements is correct?

(a) the current through 10 ohm is 2 A

(b) the current through 5 ohm is 2.5 A

(c) the current through 25 V battery is 6 A

(d) the potential difference between the terminals of 10 V

battery is 20 V

Ans(b)

Solution:

Using Kirchhoff’s law

25 = 2i + 10 i1 …….(1) 10 i1 = 5 i2

i2 = 2i1 ……..(2) 10 = 1(i1 – i + i2) + 5i2 = i1 – i + 6i2

= i1 – i + 12i1 = 13 i1 – i ……..(3) from equations (1) and (3),

i1 = A4

5 ; i2 = 2.5 A ; i = 6.25 A

Potential difference between the terminals of 10 V

battery

= 10 V – 1 (i1 – i + i2)

= 10 V – 1

2

5

4

25

4

5 = 12.5 V

21. A 10 F capacitor is connected in series with resistor 16 ohm

and 9 ohm to a source of emf 10 volt. Which of the

following statements is wrong at steady state of circuit?

The energy stored in capacitor is 500 J

(a) The energy dissipated in 16 is 320 J



(b) The energy dissipated in 9 is 180 J

(c) Energy supplied by the battery during charging the

capacitor is equal to that stored in the capacitor.

(d) Ans(d)

Solution:

During charging

q = q0 (1 – e – t/CR

) where R = R1 + R2, q0 = CE

Given C = 10 F, R1 = 16 , R2 = 9

Energy stored in capacitor,

Uc = C

q

2

12

= C2

)e1(q CR/t20

= 2

CE2

(1 – e – t/CR

)

At steady condition. T

Uc = 2

CE2

= 2

1010 2 = 500 J

Half of the energy supplied by the batteryis stored in the

capacitor and rest half islost as heat. Current through resistor

C = dt

dq =

CR

q 0 e-t/CR

= )RR(

E

21 e

–t/C(R 1 + R 2 )

Energy dissipated in R1 = 16 = i2 R1

=

0

2

21 RR

Ee )RR(c

t2

21

R1 dt

=

2

21 RR

E

R1

0

)RR(C

t2

21e dt

= 2

)RR(CR

RR

E 211

2

21

0

)RR(C

t2

21e

= )RR(2

E

21

2

R1 =

252

1016102

= 320 J

Similarly, energy dissipated in R2 = 9 = 180 J

22. V-I graph of a conductor at temperatures T1 and T2 are shown

in the figure. (T2 – T1) is proportional to :

(a) cos 2 (b) sin 2 (c) cot 2 (d) tan 2

Ans(c)

Solution:

Slope of line gives resistance

So, R1 = tan = R0(1 + T1)

R2 = tan (90 ) = cot = R0(1+ T2)

cot tan = R0 (T2 – T1)

cos

sin

sin

cos = R0 (T2 – T1)

or

cossin

sincos 22

= R0 (T2 – T1)

R0 (T2 – T1) =

2

)2(sin

2cos

Or T2 – T1 cot 2

23. The charges flowing through a resistance R varies with time

as Q = 2t – 8 t2. The total heat produced in the resistance is

(a) 6

R

joules (b)

3

R joules

(c) joules2

R (d) R joules

Ans(a)

Solution:

i = dt

d

dt

dQ (2t – 8t

2)

i = 2 – 16t

i = 0 for t = 8

1

16

2 seconds

The current flows for 8

1 seconds.

If the current flows for dt seconds, the heat produced = i2R

dt.

total heat produced

= 8

1

0

2Ri dt

= 8

1

0

2 R)t162( dt

= 8

1

0

2 R)t256t644( dt

= R 8/1

0

32

3

t256

2

t64t4

= 6

R jouls

24. Three 60W, 120V light bulbs are connected across 120V

power line as shown in fig.

The total power dissipated in the three bulbs is

(a) 60 W (b) 180 W (c) 90 W (d) 40 W

Ans: (d)

Solution: The resistance of each bulb is given by



= = = Ω

Total resistance of the circuit = + ×+ = + ×+ = Ω

The current in the circuit = = = . The power produced is = = × = W

25. A Heater is designed to operate with a power of 1000 W in a

100 V line. It is connected in combination with a resistance

of 10Ω and a resistance R, to a 100V mains as shown in the figure. What will be the value of R so that the heater

operates with a power of 62.5W? [IIT JEE]

(a) 5 (b) 7 (c) 8 (d) 10

Ans: (a)

Solution: From = ∙ Resistance of heater, = = = Ω =

Current required across heater for power of 62.5W = √ = √ . = .

Main current in the circuit, = + +

= ++ = ++

This current will distribute in inverse ratio of resistance

between heater and R. = ( + ) . = ( + ) [ ++ ] = +

Solving this equation, we get R = 5Ω

26. Capacitor of capacitance and capacitor of

capacitance are separately charged fully by a common

battery. The two capacitors are then separately allowed to

discharge through equal resistors at time t=0. (1989, 2M)

(a) The current in each of the two discharging circuits at t=0

are equal but not zero

(b) The currents in the two discharging circuits at t=0 are

equal but not zero

(c) The currents in the two discharging circuits at t=0 are

unequal

(d) Capacitor , loses 50% of its initial charge sooner than

loses 50% of its initial charge

Sol: The discharging current in the circuit is, = − /

Here, = initial current =

Here, V is the potential with which capacitor was charged.

Since, V and R for both the capacitors are same, initial

discharging current will be same but non-zero. Correct

option is (b)

Further, = < or <

or loses its 50% of initial charge sooner than . Option d is also correct.

27. A microammeter has a resistance of 100 and full scale

range of 50 . It can be used as a voltmeter or as a higher

range ammeter provided a resistance is added to it. Pick the

correct range and resistance combination (s) (1991, 2M)

(a) 50 V range with 10 k resistance in series

(b) 10 V range with 200 k resistance in series

(c) 5 mA range with 1 resistance in parallel

(d) 10 mA range with 1 resistance in parallel

Ans: (b,c)

Sol: To increase the range of ammeter a parallel resistance

(called shunt) is required which is given by = igi − ig

For option (c) = x −6x − − x −6 ≈ Ω

To change it in voltmeter, a high resistance R is put in series,

where R is given by = −

For option (b) = x −6 − ≈ 200k

Therefore, options (b) and (c) are correct.

28. For the circuit shown in the figure



(a) the current / through the batter is 7.5 mA

(b) the potential difference across is 18

(c) ratio of powers dissipated in and is

(d) if and are interchanged, magnitude of the power

dissipated in will decrease by a factor of 9

Sol: = + x .+ . = .

= . = . = = ( + ) = .. x . = .

=

29. In the circuit shown in figure. The point F is grounded.

Which of the following is wrong statement?

(a) D is at 5 V

(b) E is at zero potential

(c) The current in the circuit will be 0.5 A

(d) None of the above

Ans: (b)

Sol: Effective emf of circuit = 10-3=7V

Total resistance of circuit=2+5+3+4=14

Current I=7/14=0.5A

Potential difference between A and D=0.5x10=5A

Potential at D=10-5=5V

Potential at E=5-3=2V

Hence, E cannot be zero potential, as there is potential drop

at E.

30. In the circuit in figure the potential difference across P and Q

will be nearest to

(a) 9.6 V (b) 6.6 V (c) 4 V (d) 3.2 V

Ans: (d)

Sol: Total resistance of circuit=100+100+80+20=300

Current I= = .

Potential difference across P

and Q=20 x 0.16=3.2 V

31. What is the potential difference between points A and D of

circuit shown in figure?

(a) 5 V (b) 9 V (c) 10.4 V (d) 11.4 V

Ans: ( c)

Sol: Let be the currents drawn from cells of emf 6

V and 4 V in the circuits. Then, = + + = and = + = . V − V = x = V; V − V = V and V − V = . x = . V − = − + − + − = + + . = . V

32. The V-I graph for a conductor at temperatures T1 and T2 are

as shown in the figure, The term T2-T1 is proportional to

(a) cos 2 (b) sin 2 (c) cot 2 (d) tan 2

Ans: (c)

Sol: = tan = + = cot = + cot − tan = + − + = − − = cot − tan = (cossin − sincos ) = cos sin = cot

33. A 6 V battery is connected to the terminals of a three metre

long wire of uniform thickness and resistance of 100 . The

difference of potential between two points on the wire

separated by a distance of 50 cm will be

(a) 2 V (b) 3 V (c) 1 V (d) 15 V

Ans: (c)

Sol: Potential gradient along the wire, = V/cm

Potential difference across 50 cm length is



V = kx = x = volt Alternative Here, = = = or =

The resistance of 50 cm wire is = ′ = . =

Current in the wire =

Potential difference across the given portion of wire is = = = volt 34. Find the current supplied by battery just after switch is

closed.

(a) zero (b) 8 A (c) 2 A (d) 17 A

35. Find the current supplied by battery after a long time.

(a) zero (b) 8A (c) 4 A (d) 2 A

36. Find the heat dissipated in circuit in a long time.

(a) 56 μJ (b) 40 μJ

(c) 10 μJ (d) None of the above

37. Two resistances 300 Ω and 400 are connected to a 60 V

power supply as shown in figure. A voltmeter connected

across the 400 resistor reads 30 V.

Sol: Just after the switch is closed, the capacitor would be

treated as short circuit and the circuit can be redrawn as if P,

Q, R are at the same potential.

Therefore, all resistors connected between these points are

uneffective. After redrawing the circuit, I=8 A. = +. = . Ω

When steady state is reached, capacitor would be treated as

an open circuit.

In this case, solve the circuit, I comes out to be 4 A.

Potential difference across =

So, charge on , =

Potential difference across CD=0

So, charge on , =

Charge on , =

Charge on , =

So, H=work done by battery-ΣU = x μJ − μJ = μJ 38. A letter A is constructed of a uniform wire with resistance

1.0 per cm. The sides of the letter are 20 cm and the cross

piece in the middle is 10 cm long. The apex angle is °. The resistance between the ends of the legs is close to [JEE

Main Online 2013]

(a) 50.0 (b) 10 (c) 36.7 (d) 26.7

Ans: (d)

Sol: We have, in series, required + = + =

and in parallel, = = + = ⇒ = ℎ , = + + = + + = = . = .

39. Figure shows a circuit in which three identical diodes are

used. Each diode has forward resistance of 20 and infinite

backward resistance. Resistors = = = . Battery voltage is 6 V. The current through is [JEE

Main Online 2013]

(a) 50 mA (b) 100 mA (c) 60 mA (d) 25 mA

Sol: As diode is conducting in forward bias condition and

now conducting in reverse bias condition. Diode D1 is in



forward bias, and diode D2 is in forward bias but D3 is

reverse bias. So the figure can be drawn as

Here, 20 , R and R3 are in series.

Equivalent resistance=50+50+20=120 = = = ⇒ =

40. To find the resistance of a galvanometer by the half

deflection method the following circuit is used with

resistances R1=9970 , R2=30 and R3=0. The deflection in

the galvanometer is d. With R3=107 the deflection changed

to . The galvanometer resistance is approximately [JEE

Main Online 2013]

(a) 107 (b) 137 (c) 107/2 (d) 77

Ans: (d)

Sol: As at initial condition the deflection is d while R3=0,

then equivalent resistance of R2 and R3

=R2+R3=R2=30

Now, when R3=107 and R2=30

Then, equivalent resistance should be =

It is only when equivalent resistant resistance and R3 and Rg

will be parallel to R2 giving resistance 15

Let R3 – Rg=equivalent=30 =R + = + = =

Thus, Rg must will be 77 in order to maintain

R3-Rg=30 ⇒ 107-Rg=30=Rg=77

41. Which of the four resistances P,Q,R and S generate the

greatest amount of heat when a current flows from A to B?

[JEE Main Online 2013]

(a) Q (b) S (c) P (d) R

Sol: We know that i ∝ iR , i = + i = i

or i = + i = i Power rate in 2 of upper series = x ( i) = i

Power rate in 4 of upper series = x ( i) = i

power rate of 1 in lower series = x ( i) = i

Power rate of 2 in lower series = x ( i) = i → / → / , → / , → /

42. Two conductors have the same resistance at 0°C but their

temperature coefficients of resistance are and . The


parallel combinations are nearly [AIEEE 2010]

(a) + , + (b) + , +

(c) + , + (d) + , +

Ans: (c) Sol: Let, be the initial resistance of both conductors

At temperature their resistances will be, = + θ = + θ

For series combination, = + + = + + +

Where, = + = + = + + or = +

For parallel combination, = + ( + = + + + + + ℎ , = + = ( + = + + + + + as and are small quantities. , is negligible. So, neglect , , = + + + = + [ − ( + ) ]

[Binomial expansion] + is negligible = +

43. The resistance of a wire is 5 at 50°C and 6 at 100°C.

The resistance of the wire at 0°C will be [AIEEE 2007]

(a) 2 (b) 1 (c) 4 (d) 3

Ans: (c)

Sol: From = + = + = +



= + + or =

Putting value of α in Eq. i , we get = R ( + x ) or =

44. A material B has twice the specific resistance of A. A circular

wire made of B has twice the diameter of a wire made of A.

Then, for the two wires to have the same resistance, the ratio

IB/IA of their respective lengths must be [AIEEE 2006]

(a) 1 (b) 1/2 (c) 1/4 (d) 2/1

Sol: Resistance of = = / = /

Resistance of = = / = /

From given information, = = and = / = / or / = or = = :

45. The resistance of the series combination of two resistances is

S. When they are joined in parallel, the total resistance is P. If

S=nP, then the minimum possible value of n is [AIEEE-

2004]

(a) 4 (b) 3 (c) 2 (d) 1

Ans: (a) Sol: Let resistances are R1 and R2. Then,

S=R1+R2 = + + = +

[from S=nP] or + = ⇒ = [ + + ] = [ + + ]

We know,

Arithmetic Mean ≥ Geometric Mean + ≥ √

⇒ + ≥

So, n (minimum value)=2+2=4

46. Time taken by a 836 W heater to heat one litter of water from

10°C to 40°C is [AIEEE 2004]

(a) 50 s (b) 100 s (c) 150 s (d) 200 s

Sol: Let time taken in boiling the water by the heater is t

second. ℎ . = ⇒ = . = − or . = or = . =



SUBJECTIVES

1. An electrical circuit is shown in figure. Calculate the

potential difference across the resistor of 400 Ω as will be measured by the voltmeter V of resistance 400 Ω either by applying Kirchhoff’s rules or otherwise. [IIT JEE]

Solution: The given circuit actually forms a balanced

Wheatstone’s bridge (including the voltmeter) as shown below

= Ω

Here, we see that =

Therefore, resistance between A and B can be ignored and

equivalent simple circuit can be drawn as follow

The voltmeter will read the potential difference across

resistance Q.

Currents = = + =

Potential difference across voltmeter = = ( ) =

Therefore, reading of voltmeter will be

2. An infinite ladder network of resistances is constructed with

1Ω and 2Ω resistances, as shown in figure. The 6V battery

between A and B has negligible internal resistance.

[IIT JEE]

(a) Shown that the effective resistance between A and B is

2Ω.

(b) What is the current that passes through the 2Ω resistance nearest to the battery?

Solution:

(a) Let = . Then, we can break one chain and

connect a resistance of magnitude x in place of it.

Thus, the circuit remains as shown in figure.

Now, 2Ω and x are in parallel. So, their combined resistance is

+

Or = + +

But is assumed as x. Therefore, = + +

Solving this equation, we get = Ω

(b) Net resistance of circuit = + ×+ = Ω

Current through battery = =

This current is equally distributed un 2Ω and 2Ω resistances. Therefore, the desired current is or 1.5A.

3. Two resistor 400Ω and 800Ω are connected in series with a 6 V battery. It is desired to measure the current in the

circuit. An ammeter of 10Ω resistance is used for this purpose. What will be the reading in the ammeter?

Similarly, if a voltmeter of 1000Ω resistance is used to measure the potential difference across the 400Ω resistor, what will be the reading in voltmeter? [IIT JEE]

Solution:

Refer figure (a) Current through ammeter,



= = + + = . × − = .

Refer figure (b) Combined resistance of 1000Ω voltmeter and 400Ω resistance is, = ×+ = . Ω = . + = . × −

Reading of voltmeter = = ′ = . × − . = . V

4. In the circuit shown in figure E1 = 3V,

E2 = 2 V, E3 = 1V and R = r1 = r2 = r3 = 1Ω. (a) Find the potential difference between the points A and B

and the currents through each branch.

(b) If r2 is short circuited and the point A is connected to

point B, find the currents through E1, E2, E3 and the resistor R

[IIT JEE]

Solution:

(a) Equivalent emf of three batteries would be: = Σ( ⁄Σ( ⁄

= ( ⁄ + ⁄ + ⁄( ⁄ + ⁄ + ⁄ =

Further r1, r2 and r3 each are of 1 Ω. Therefore, internal resistance of the equivalent battery will be Ω as all three

are in parallel.

The equivalent circuit is therefore shown in the figure.

Since, no current is taken from the batter. = (From V= E – i r )

Further, = − = −

= − + = − + =

Similarly, = − + = − + =

And = − + = − + = −

(b)r2 is short circuited means resistance of this branch

becomes zero. Making a closed circuit with a battery and

resistance R. Applying Kirchhoff’s second law in three loops so formed. − − + + = ….. (i) − + + = …… (ii) 1- i3 – (i1 + i2 +i3 )= 0 ……(iii) Form Eq. (ii) + + =

Substituting in Eq. (i), we get, =

Substituting in Eq. (iii) we get, = −

=

5. At the temperature 00C, the electric resistance of conductor

B is n times that of conductor A. Their temperature

coefficients of resistances of A and B 1 2

respectively. Find the resistance and temperature

coefficients of resistance of a circuit segment consisting of

these two conductors when they are connected in series.

Solution: Let R0 be the resistance of conductor A at 0 .

Then resistance of conductor B at 0 . Now at

temperature t, we have

and

Let Rs be the series resistance of the conductors at

temperature t , then

If R0 be the resistance of combination of two conductors at 0

and

in series, then

Comparing eqs. (i) and (ii), we get

and

6. A cylindrical tube of length l has inner radius a while outer

b as shown in fig. Show that the resistance of the tube

between its ends is R = (l/(b2-a

2) Where, ρ is the

resistivity of the material.

Solution: The electric field at a point inside the conductor is

given by

J=σ E or = = = …… (i)



Let the tube be divided into a large number of coaxial

annular discs. Consider one such disc of thickness dx at a

distance x from left end. We have = = − ……. (ii) Substituting these values in eq. (i), we get − = [ − ]

Integrating this expression within proper limits, we get − ∫ = ∫ −

or = − = − or

= = −

7. A cylindrical tune of length l has inner radius a while outer

b as shown in fig (a).

Show that the resistance between its inner and outer surface

is =

Where ρ is the resistivity of its material.

Solution: We know that, E=ρ j Here the field is radial as shown in fig (b). Let the tube be

divided into a large number of concentric cylindrical shells

as shown in fig. (a). Consider one such shell of radius r and

thickness dr. Here = =

− =

Integrating within proper limits, we get − ∫ = ∫ or − ∫ = ∫

or = − = [ g ] = [ g − g ] = g ( )

8. A metal ball of radius a is surrounded by a thin concentric

metal shell of radius b. The space between these electrodes

is filled up with a poorly conducting homogeneous medium

of resistivity ρ. Find the resistance of the inter-electrode

gap. Analyze the obtained solution as b → ∞.

(b) The space between two conducting concentric spheres of

radii a and b(a<b) is filled up with homogeneous poorly

conducting medium. The capacitance of such a system

equals C.

Find the resistivity of the medium if the potential difference

between the spheres, when they are disconnected from an

external voltage, decreases -fold during the time interval

Δt. Solution:

(a) Consider a thin spherical layer of inner and outer

radii r and r + dr. The lines of current at all points of this

layer will be perpendicular to it. Hence the layer may be

regarded as a spherical conductor of thickness dr. Its cross

sectional area will be . For the definition of specific

resistance ρ, we have ... (1)

Integrating this expression, we get

For b → ∞, …. (3)

(b) Current ….. (4)

Further …. (5) From eqs. (4) and (5), we have

or



Hence, the resistivity ρ of the medium is given

9. In the circuit a voltmeter reads 30 V when it is connected

across 400 ohm resistance. Calculate what the same

voltmeter will read when it is connected across the 300 Ω resistance?

Solution:

Potential difference across 400 ohm = 30V

Potential difference across 300 ohm

=(60 – 30) =30 V

This shows that the potential difference is equally shared.

Let R be the voltmeter resistance. The resistance 400 and

voltmeter resistance R are in parallel. Their equivalent

resistance R′ is given by ′ = + = + +

But R′ should be equl to 300 ohm. Hence + = = ℎ

Thus, voltmeter resistance is 1200 ohm.

When the voltmeter is connected across 300 ohm, the

effective resistance R′′ is given by ′′ = + = + +

′′ = = ohm.

Now the potential difference is shared between 240 ohm and

400 ohm.

Potential difference across 240 ohm : Potential difference

across 400 ohm. =240 : 400 = 3:5

As total potential is 60 V, hence potential difference across

240 ohm, i.e., across resistance 300 ohm will be × =. V

10. A potential difference of 220 volt is maintained across a

12000 Ω rheostat as shown in fig. The voltmeter V has a

resistance of 6000Ω and point C is at one-fourth of the

distance from a to b. What is the reading of voltmeter?

Solution: The resistance of part a c = × = Ω resistance of part b c = 9000 ohm.

Resistance of voltmeter = 6000 ohm.

Voltmeter resistance and rheostat’s part a c resistance are in parallel. Hence equivalent resistance R′ is given by ′ = + =

R′ = 2000 ohm. Resistance R′ and are in series. Hence total resistance

=9000 ohm + 2000 ohm = 11000 ohm.

The current in the circuit = . amp.

Now potential difference across b c is given by

current resistance =0.02 9000 = 180 V

Potential difference across a c

=220 – 180 volt=40 volt

Reading of voltmeter = 40 volt.

11. In the circuit shown in fig, V1 and V2 are two voltmeter

having resistances 6000 ohm and 4000 ohm respectively.

E.M.F of battery is 250 volt, having negligible internal

resistance. Two resistance R1 and R2 are 4000 ohm and

6000 ohm respectively. Find the reading of the voltmeters

V1 and V2 when

(i) switch S is open

(ii) switch S is closed.

Solution: (i) When switch S is open.

R1 and R2 are in series. Let their resistance be R′. Then R′ = 4000 + 6000 = 10,000 ohm

The voltmeters are also in series. Let their resistance be R′′. Then

R′′ = 4000 + 6000 = 10,000 ohm

The resistances R′ and R′′ are connected in parallel. Their equivalent resistance is given by.

current form battery

Current i1 in the voltmeter branch

Potential difference across V1



Potential difference across V2

(ii) When switch S is closed.

The circuit redrawn in this case is shown in fig. In this case

V1 and R1 are in parallel.

Similarly V2 and R2 are in parallel. Let R′ be equivalent resistance of V1 and R1

Similarly, for R2 and V2 combination

So, the two resistances are equal and joined in series. Hence

potential difference across V1 is equal to potential difference

across V2. This is equal to volt

Reading of voltmeter V1 = 125 volt

And reading of voltmeter V2 = 125 volt.

12. A battery of e.m.f 1.4 V and internal resistance 2 Ω is connected to a resistance of 100 ohm through an ammeter.

The resistance of ammeter is 4/3 ohm. A voltmeter has also

been connected to find the potential difference across the

resistor.

(i) Draw the circuit diagram.

(ii) The ammeter reads 0.02 A. What is the

resistance of voltmeter.

(iii) The voltmeter reads 1.10 V. What is the

error in the reading?

Solution: (i) The circuit diagram is shown in fig.

(ii) Let R be resistance of the voltmeter.

The voltmeter is connected in parallel with 100 ohm

resistance. Hence effective resistance = +

The resistance of the circuit = + + +

= + + = + +

Current in the circuit = . ++

. = . + +

Solving, we get R= 200 ohm.

(iii) Equivalent resistance R′, of voltmeter (R=200 ohm) and 100 ohm resistor is given by ′ = + = + =

R′= ℎ

Potential difference across voltmeter = ′ = . × = = .

Error in voltmeter reading

= 1.33 – 1.10 = 0.23 V.

13. For the fig, calculate the current through 3 ohm resistor and

power dissipated in the entire circuit.

The e.m.f. of battery is 2 volt and its internal resistance is

2/3 ohm.

Solution: The diagram can be redrawn as shown in fig.

The effective resistance RAC between A and C = + = = ℎ

The effective resistance RCB between C and B = + = = ℎ , = + = + = ℎ

Corresponding to points X and Y, the resistances 3 ohm, 4

ohm and 6 ohm are in parallel, hence effective resistance

RXY is given by



= + + = + + = = = ℎ . Total resistance R of the circuit is (4/3)+(2/3) = 2 ohm

Current in the circuit = 2/3 = 1.5 amp.

Power dissipated in the circuit =i2R=4.5 W.

Potential difference between X and Y is,

i . RXY = 4.5 X (4/3) = 6 V

Potential difference across 3 ohm resistor is, 6 V

Current in the 3 ohm resistor is,

6/3=2 amp.

14. A battery is made by joining m rows of identical cells in

parallel. Each row consists of n cells joined in series. The

total number of cells is equal to N. This battery sends a

maximum current i in a given external circuit of resistance

R. Now the cells are so arranged in the battery that instead

of m rows, n rows are joined in parallel and each row

consists of m cells joined in series. What would be the

current in the external circuit now interiors of i.

Solution: (i) Let E be the e.m.f. of each cell. The e.m.f. of

each row will be n E and internal resistance of each row will

be n r (where r is the internal resistance of each cell).

Hence m rows are joined in parallel. So, the total internal

resistance of the circuit. = = =

Total resistance of the circuit = +

current in the circuit = +( )

For current to be maximum, = . Hence

[ + ( ) ] =

[ ∙ + − ] =

+ − − + − =

+ − [ + − ] = − = = =

Hence maximum current in the circuit will flow when the

external resistance is equal to the internal resistance of the

battery = + ( ) = + … .

(ii) In this case e.m.f. of each row = m E and internal

resistance of each row = m r

The rows are joined in parallel, hence total internal

resistance of battery = mr/n

Total resistance of the circuit

= + = +

Current in the circuit ′ = [ + ] = + …

Dividing eq. (2) by eq. (1) and solving, we get ′ = + ×

15. 12 cells each having the same e.m.f. are connected in series

and are kept in a closed box. Some of the cells are wrongly

connected. This battery is connected in series with an

ammeter and two cells identical with the others. The current

is 3 A when the cells and battery aid each other and is 2 A

when the cells and battery oppose each other. How many

cells in the battery are wrongly connected?

[Roorkee 93]

Solution: Let x cells be connected correctly and y cells are

connected wrongly. According to the given problem

x + y =12 …. (1) If E be the e.m.f. of one cell, then net e.m.f will be −

Let R be the resistance of the circuit which remains

constant.

(i) When the cells aid the battery, the net e.m.f. = − + curent = net e. m. fResistance = − + = …

(ii) When the cells oppose the battery, net e.m.f = − − curent = net e. m. fResistance = − − = …

Dividing eq. (2) by (3), we have − +− − = − +− − =

Solving, we get x – y = 10 ….. (4) On solving (1) and (4) we get x=11, y=1.

Hence only one cell s wrongly connected.

16. A source of e.m.f. volts is connected between the points A

and B in the circuit. The resistance R1 = 20 s and

resistance R2 = 30 . For what value of RX will the thermal

power generated in it is independent of small variations of

that resistance?

Solution:

In the circuit, 30 ohms and Rx are in parallel.

Effective resistance = x

x

R30

R30



20 and x

x

R30

R30

are in series.

Total resistance = x

xx

R30

R30R20600

= x

x

R30

R50600

= x

x

R30

)R12(50

So the current in the main circuit

i = )R12(50

)R30(V

x

x

The current through Rx is given by,

ix = )R12(50

)R30(V

x

x

)R30(

30

x

= )R12(50

V30

x

= xR12

V6.0

Thermal power dissipated

P =

2

xR12

V6.0

Rx

If P is to be independent of Rx

xRd

Pd= 0

0.36 V2

4x

xx2

x

)R12(

R)R12(21)R12(

= 0

(12 + Rx)2 = 2 (12 + Rx) Rx

12 + Rx = 2Rx

Rx = 12 ohms

17. In a Wheatstone’s bridge a battery of 2 volt and internal resistance 2 ohm is used. Find the value of the current

through the galvanometer in that unbalanced condition of the

bridge when P = 1 ohm, Q = 2 ohm, R = 2 ohm, S = 3

ohm and resistance of galvanometer is 4 ohm.

Solution :

The wheatstone’s bridge is shown in figure.

The current in the different branches are also shown in

figure.

Applying Kirchoff’s second law to loop ABDA, we have i1 1 ig 4 – (i – i1) 2 = 0

or 3 i1 + 4 ig – 2 i = 0 ………. (1) Applying Kirchoff’s second law to loop BCDB, we get (i1 – ig) 2 – (i – i1 + ig) 3 – ig 4 = 0

or 5 i1 – 9 ig – 3 I = 0 ………….. (2) Applying Kirchoff’s second law to loop ADCEA, we get (i – i1) 2 + (i – i1 + ig) 3 + i 2 = 0

or - 5 i1 + 3 ig + 7 i = 2 …….. (3) Multiplying eq. (1) by 3 and eq. (2) by 2 and then

subtracting we get

i1 = 30 ig ……… (4) Adding eqs. (2) and (3), we have

2

1i

2

3i g ……… (5)

Substituting the values of i1 and i from eqs. (4) and (5) in eq.

(2), we get

02

1i

2

33i9i305 ggg

02

1i

2

33i9i150 ggg

2

3

2

i273 g or amp

91

1ig

18. At time t = 0, a battery of 10 V is connected across points A

and B in the given circuit. If the capacitors have no charge

initially, at what time (in second) does the voltage across

them become 4 V? [IIT JEE]

Solution: Voltage across the capacitors will increase from 0

to 10 V exponentially. The voltage at time t will be given by = − (− ⁄

Here = = × × − = = − − ⁄

Substituting V = 4 volt, we have = ( − − ⁄

Or − ⁄ = . =

Taking log both sides we have, − = −

or = − =

Hence, the answer is 2.

19. When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat

produced in R is J1. When the same batteries are connected

in parallel across R, the rate is J2. If J1 = 2.25 J2 then the

value of R in Ω is [IIT JEE] Solution: In series, = +



= = ( + ) ∙

In parallel, = . + = = ( . + ) ∙ = . = . ++

or . = . ++

Solving we get, R= 4Ω

20. At t= 0, switch S is closed. The charge on the capacitor is

varying with time as = − − . Obtain the value

of Q0 and in terms of circuit parameters. [IIT JEE]

Solution:

Q0 is the steady state charge stored in the capacitor.

Q0=C [PD across capacitor in steady state]

=C [steady state current through R2] (R2) = ( + ) ∙ = + =

Here, is equivalent resistance across capacitor after

short circuiting the battery. Thus, = + (As R1 and R2 are in parallel) = +

= +

21. In the circuit shown the battery is an ideal one, with emf V.

The capacitor is initially uncharged. The switch S is closed

at time, t=0. [IIT JEE]

(a) Find the charge Q on the capacitor at time t.

(b) Find the current in AB at time t. What is its limiting

value as t→ ∞ ?

Solution: Let at any time t charge on capacitor C be Q and

currents are as shown. Since, charge Q will increase with

time t. Therefore,

(a) Applying Kirchhoff’s second law in loop MNABM = − +

or = − …… (i) Similarly, applying Kirchhoff’s send law in loop MNSTM, we have = + + …… (ii) Eliminating I from Eqs. (i) and (ii), we get = +

or = −

or = −

or = −

or − =

or ∫ − = ∫

This equation gives = − − ⁄

(b) = = − ⁄

From eq. (i) = + = + − ⁄

Current through AB = − = + − ⁄ − − ⁄

= − − ⁄ = → ∞

22. A leaky parallel plate capacitor is filled completely with a

material having dielectric constant K = 5 and electrical

conductivity = . × − Ω− m-1

. If the charge on the

capacitor at instant t = 0 is q = 8.85 µC, then calculate the

leakage current at the instant t = 12s. [IITJEE]



Solution: The problem is basically of discharging of CR

circuit, because between the plates of the capacitor, there is

capacitor as well as resistance,

= ( = )

Or =

Time constant, = =

Substituting the values, we have = × . × −. × − = .

Charge at any time decreases exponentially as = − ⁄

Here, = . × − (charge at time t = 0)

Therefore, discharging (leakage) current at time t will be

given by = − = (– / or current at t = 12s is = . × −. − / . = . × − = . = .

23. Find the emf (V) and internal resistance (r) of a single battery

which is equivalent to a parallel combination of two batteries

of emfs V1 and V2 and internal resistances r1 and r2

respectively, with polarities as shown in figure

[IITJEE]

Solution: (a) PD across the terminals of the battery is equal

to its emf when current drawn from the battery is zero, in

the given circuit:

Current in the internal circuit

= = ++

Therefore, potential difference between A and B would be − = − − = − ( ++ ) = −+

So, the equivalent emf of the battery is = −+

Note that if V1r2 = V2r1 : V = 0

If V1r2 > V2r1 : VA – VB = Positive i.e., A side of the

equivalent battery will become the positive terminal and

vice-versa. (b) r1 and r2 are in parallel. r = (r1 r2/ r1+ r2)

24. In the given circuit:

E1 = 3E2 = 2E3 = 6 V and R1 = 2R4 = 6Ω, R3 = 2R2 = 4Ω, C=5 F.

Find the current in R3 and the energy stored in the capacitor.

[IITJEE]

Solution: In steady state no current will flow through R1 =

6Ω. Potential difference across R3 or 4Ω is E1 or 6V

Current through it will be = . from right to left.

Because left hand side of this resistance is at higher

potential.

Now, suppose this 1.5 a distributes in i1 and i2 as shown.

Applying Kirchhoff’s second law in loop dghfed − − × . − + = = − = − .

To find energy stored in capacitor we will have to find

potential difference across it or Vad.

Now, − + =

or − = − = − .

or − = . =

Energy stored in capacitor:



= = × − . = . × −

25. A part of circuit in steady state along with the currents

flowing in the branches, the values of resistances etc, is

shown in the figure, Calculate the energy stored in the

capacitor C(4µF)

[IIT JEE]

Solution: Using Kirchhoff’s first law at junctions a and b, we have found the current in other wires of the circuit on

which currents were not shown.

Now, to calculate the energy stored in the capacitor we will

have to first find the potential difference Vab across it. − × − × + × = − = = =

= × − J = . mJ

26. In the circuit shown in figure E,F,G,H are cells of emf 2,1,3

and 1V respectively, and their internal resistances are 2,1,3

and 1 Ω respectively.Calculate

(a) the potential difference between B and D and

(b) the potential difference across the terminal of each cells

G and H. [IIT JEE]

Solution: Applying Kirchhoff’s second law in loop BADB − − − − − = … . .

Similarly applying Kirchhoff’s second law in loop BDCB − + − − − = … .

Solving Eqs. (i) and (ii), we get = , =

And − = −

(a) Potential difference between B and D. + − = − = − − =

(b) = − = − × = = − = + × =

27. An electric circuit is shown in figure. Calculate the

potential difference across the resistor of 400 ohm, as will

be measured by the voltmeter V of resistance 400 ohm,

either by applying Kirchoff’s rules or otherwise.

Solution :

The current distribution is also shown in figure.

Applying Kirchoff’s second law to mesh A, we have 100 I1 + 200 (I1 + i3) = 10

or 300 I1 + 200 I3= 10

or 30 I1+ 20 I3 = 1 ……… (1) Applying Kirchoff’s voltage law to mesh B, we have 100 I2 + 100 I3 – 100 I1 = 0

or I1 = I2 + I3 ……… (2) Applying KVL to mesh C, we have

0I100II2002

II400 331

32

or 2 I1 = 2 I2- 5 I3 ……….. (3)



Solving these equations, we get

I3 = 0, I1 = I2 = (1/3)) amp

Required potential difference

=

67.660

1400

2

II400 32

Volts

Alternate method :

The resistance 400 ohm of voltmeter and another 400

resistance in parallel with it gives a resistance [(400

400)/(400 + 400)] = 200

The equivalent circuit is now drawn. This is shown in

figure.

This is a balanced wheat stone bridge. Obviously no current

flows between B and D.

Now

Amp30

1

200100

10I2

.

Half of I2 passes through voltmeter and rest half passes

through other resistance of 400

Hence potential difference across voltmeter

67.660

400400

30

1

2

1400

2

IV 2 Amp.

28. In the circuit shown in figure. E, F, G and H are cells of emf

2, 1, 3 and 1 volts and their internal resistances are 2, 1, 3 and

1 ohm respectively. Calculate

(i) the potential difference between B and D and

(ii) the potential difference across the terminals of each of

the cells G and H.

Solution :

Figure shows the current distribution.

Applying Kirchoff’s first law at point D, we have i = i1 + i2 …………. (1) Applying Kirchoff’s second law to mesh ADBA, we have 2i + 1 i + 2 i1 = 2 – 1 = 1

or 3 i + 2 i1 = 1 …………… (2) Applying Kirchoff’s second law to mesh DCBD, we get

3 i2 + 1 i2 – 2 i1 = 3 – 1

4 i2 – 2 i1 = 2 ………….. (3) Solving eqs. (1), (2) and (3) we get

.amp13

1i1 , amp

13

6i2 and amp

13

5i

(i) Potential difference between B and D

= volt13

2

13

12i2 1

(ii) Potential difference across G

= V61.113

363RiE 2

Potential difference across H

= V46.1)1(13

61

29. Two bulbs rated at 25 watt, 110 volt and 100 watt, 110 volt

are connected in series to 220 volt electric supply. Perform

the necessary calculations to find out which of the two

bulbs, if any, will fuse. What would happen if the two bulbs

were connected in parallel to the same supply?

Solution: Let i1 and i2 be the currents which can flow

through the two lamps safely, then = = . . = = . . The resistances of two bulbs are given by = = . ℎ = . ℎ

When the two bulbs are connected in series, their total

resistance = + = . + . = ℎ

When these two lamps are connected in series to 220 volt,

the current passing through them is given by = = . . Thus, the first bulb will fuse because the current passing

through it, i.e., 0.363 is more than i1(0.227).

When the two bulbs are connected in parallel, the effective

resistance R′ is given by ′ = + = . + . = .

′ = . ℎ . Current flowing through circuit ′ = ′ = × . .



be the current passing through the two bulbs

as shown in fig.

Now the potential difference across the two bulbs is the

same.

Hence, = ∙ . = ∙ .

Or = ….. (1) Again + = ′ = × .

….. (2) Solving eqs. (1) and (2), we get = . . = . . Thus, both the bulb will fuse.

30. Find the current flowing through the resistance R in the

circuit shown in fig (a). The internal resistances of the

batteries are negligible.

Solution: The current distribution is shown in fig.. At

junction (2) and (3) = + … … = + … . .

Applying Kirchhoff’s law to meshes 12781, 1236781 and 34563 respectively, we have = … … + = … … . − + = … … . .

Substituting the value of i1 from eq. (2) in eq. (4), we get + + = + + = = −+ … . .

Substituting the value of i3 from eq. (6) in eq. (5), we get − ( −+ ) + = − + + ( + ) + = [ + ( + )] = + +

= + ++ + = + ++ +

31. Two squares ABCD and BEFC have the side BC in

common. The sides are of conducting wires with resistances

as follows:

AB, BE, FC, CD each 2 ohm, AD, BC, EF each 1 ohm. A

cell of E.M.F. 2 volt and internal resistance 2 ohm is joined

across AD. Find the currents in all various parts of the

circuit.

Solution: The current distribution is shown in fig.

Applying Kirchhoff’s second law to the loop containing the cell and AD, we have × + × − = − = … …

For loop ABCDA, we have × + × − + × − × − = − + − = … . .

Similarly, for loop BEFCB, we have × + × + × − × − = − + = … . Solving eqs. (1), (2) and (3), we get = , = = . As i, i1 and i2 are known and hence the currents in all other

branches can be calculated.

32. Twelve equal wires, each of resistance 6 ohm are joined up

to form a skeleton cube. A current enters at one corner and

leaves at the diagonally opposite corner. Find the joint

resistance between the corners.

Solution: The skeleton ABCDEFGH, is shown in fig.

This skeleton consists of twelve wires. Let the resistance of

each wire be r. Hence the currents i enters at corner A and

leaves at corner G. The current i at corner A is divided into

three equal parts (i/3) because the resistance of each wire is

the same. At B, D and E, the current i/3 is divided into two

equal parts each having magnitude i/6. At the corners C, F

and H, the currents again combine to give currents, each of

magnitude i/3 along CG, FG and HG respectively. At corner



G, all these currents combine so that the current leaving at G

is i.

Let R be the equivalent resistance between the corners A and

G. Taking any one of the paths say ABCG, we have = + + = + + =

According to given problem r = 6 ohm. = × = ℎ .

33. A wire forms a regular hexagon and the angular points are

jointed to the centre by wires each of which has a resistance

(1/n) of the side of the hexagon.

Find the resistance to current entering at one angular point

and leaving it by opposite point if r is the resistance of any

one side of hexagon.

Solution:

The current distribution is also shown in fig. Applying

Kirchhoff’s law at E and F, we get = + … = + … .

ℎ , + − = + = … . ℎ , + − = = … .

From eqs. (1) and (4), we get = ( + ) = ( + ) From eq. (3) + + = = ( ++ ) = ++

Putting the value of i1 in eq. (2), we get = ++ + = + ++ = ++ +

Let Req. be the equivalent resistance, then = ( ) = ++ +

34. In a Wheatstone’s bridge a battery of 2 volt and internal resistance 2 ohm is used. Find the value of the current

through the galvanometer in that unbalanced condition of

the bridge when P = 1 ohm, Q = 2 ohm, R = 2 ohm, S = 3

ohm and resistance of galvanometer is 4 ohm.

Solution:

The Wheatstone bridge is shown in fig.

The current in the different branches are also shown in fig.

Applying Kirchhoff’s second law to loop ABDA, we have × + g × − − = + g − = …… (1) Applying Kirchhoff’s second law to loop BCDB, we get ( − g × − ( − + g × − g × = − g − = …… (2) Applying Kirchhoff’s second law to loop ADCEA, we get − × + ( − + g × + × = − + g + = …… (3) Multiplying eq. (1) by 3 and eq.(2) by 2 and then

subtracting, we get = g … … ..

Adding eqs. (2) and (3), we have = g + … … …

Substituting the value of i1 and i from eqs. (4) and (5) in eq.

(2), we get ( g − g − ( g + ) =

g − g − g − = g = g = .

35. A galvanometer together with an unknown resistance in

series is connected across two identical batteries each of

1.5V. When the batteries are connected in series the

galvanometer records a current of 1 ampere and when the

batteries are in parallel the current is 0.6 ampere. What is

the internal resistance of the battery [IITJEE]

Solution:



Emf of each cell,ε = 1.5V

The internal resistance of each cell be r. Let the resistance of

the galvanometer be G.

Le the unknown resistance in series with the galvanometer

be R.

(i) Le the cells be in series.

The emf of the circuit = 2ε = 3V

The resistance of the circuit = (R+G+2r)Ω

The current in the circuit + + = … …

(ii) When the cells are in parallel the emf of the circuit

ε = 1.5V

The resistance of the circuit = + + ∙+ = + + Ω

The current in the circuit .+ + = . … .

From equations (1) and (2), we get + + = + + = .. = .

Subtracting these two equations, we get = . ⟹ = Ω

Internal resistance of each cell = Ω

36. In the circuit V1 and V2 are two voltmeters of resistances

3000 ohm and 2000 ohm respectively. The resistances

R1=2000 ohm and R2=3000 ohm and the emf of the battery

ε = 200 V. The battery has negligible internal resistance.

Find the readings of the voltmeters V1 and V2 when

(i) the switch S is open and

(ii) the switch S is closed.

Solution:

(i) When S is open

V1 and V2 in series have a resistance

= 3000 + 2000 = 5000Ω

R1 and R2 in series have a resistance

= 2000 + 3000 = 5000Ω

5000 ohm and 5000 ohm in parallel are equivalent to 2500Ω

Cuircuit current = =

Current in the branch of V1 and V2 = ( ) = . . = ℎ = . . = ℎ =80V

the voltmeters V1 and V2 read 120V and 80V

respectively.

(ii)When S is closed

For V1 and R1 in parallel, the equivalent

resistance

= ×+ × = ℎ

Similarly V2 and R2 in parallel have an equivalent resistance

of 1200 ohm.

As these two equivalent resistances are same

p.d across AS= p.d across SB

p.d. across AS = 100V

This is registered by V1

Similarly p.d. across SB = 100V

This is registered by V2.

37. (a) Find the emfs E1 and E2 in the circuit of the following

diagram and the potential difference between the points a

and b

(b) If in the above circuit, the polarity of the battery E1, be

reversed, what will be the potential difference between a

and b?

Solution:



(a) 1 A current flow in the circuit from b to a.

Applying Kirchhoff’s law to the loop PabP.

20 – E1 = 6 + 1- 4 – 1 = 2

Hence E1 = 18V

Also applying Kirchhoff’s law to the loop PaQbP, − = + + × + × =

Hence E2 = 7V

Thus the potential difference between the points a and b is

Vab = 18 – 1 – 4 = 13V

(b) On reversing the polarity of the battery E1,the current

distributions will be changed.

Le the currents be I1 and I2 as shown.

Applying Kirchhoff’s law for the loop PabP. + = + − + = − … . .

Similarly for the loop abQa, + + + + + + + = or 3I1 + 8I2 = - 25 …… (2) Solving (1) and (2) for I1 and I2, we get

I1 = 2.52 A and I2 = - 4.07A

Hence = − × . +

= - 20.35 + 18 = - 2.35V

38. A convention if often employed in circuit diagrams where

the battery (or other power source) is not shown explicitly

but the points connected to the source are indicated by

voltage and ground respectively. The following two circuit

diagrams are drawn on this convention. Assume the battery

resistance is negligible.

(a) In figure (a), what is the potential difference Vab when

the switch S is open?

(b) What is the current through switch S when it is closed?

(c) In figure (b), what is the potential difference Vab when

switch S is open?

(d) What is the current through switch S when it is closed?

(e) What is the equivalent resistance in the circuit (b),

when (i) switch S is open and (ii) switch is closed?

Solution:

The given circuit is equivalent to

(a) Potential at the point a. = − ( × ) =

Potential at the point b, = − ( × ) =

Hence Vab = Potential difference between a and b

= Va – Vb = 12 – 24 = 12V

(b) When the switch S is closed, the currents and

potentials will readjust to new values.[It is important to

remember that there will be a smart (even if negligible)

resistance in the switch]

The equivalent circuit is now

Let the current distributions be I1 and I2 as shown. Let the

current in the switch be Iab from a to b and x the resistance

of switch. Then for the loop QabQ, + − = ….. (1) For the loop PQaRP, = + ….. (2) = − ….. (3) From (1) and (3), we get = + + … … .

Proceeding to the limit ×→



Without → ∞, from (4), we get = =

Substituting in (2), we get

I1 = 3 A and I2 = 6A.

Hence the current through the switch,

Iab=I1 – I2= - 3A

The current flows in the switch from b to a.

(c) In figure (b) we have a resistance of 3Ω added to the switch circuit. However this will NOT affect the current and

potential distributions when the switch S is open.

Hence the potential difference Vab = -12 V (as in the case

(a) above).

(d) When the switch S is closed, the currents and

potentials will redistribute to new values. Let the currents be

I1, I2 and I3 shown.

For the loop QROQ, − +× − = … …

For the loop PQOSP, = + + = + + = … … .

For The loop QRSOQ, + − − + − = − + = − + = … … .

Solving (1),(2) and (3) for I1, I2 and I3, we get = = . = = . = = .

Hence the current that flows through the switch when it is

closed =1.71 A (flowing from b to a).

(e) When the switch S is open in circuit diagram (b) the

total current in the circuit = the same total current as in

circuit diagram (a) with switch S open. Hence equivalent

resistance is the same as in case (a) above, and = ×+ = = . Ω

When the switch S is closed in diagram (b), the total current

drawn from the battery is = + = + =

Hence the equivalent resistance in the circuit is = = ⁄ = = . Ω

39. A circuit shown below has resistances R1 = 20Ω and R2 =

30Ω. What value of the resistance RX will the thermal

power generated in it be practically independent of small

variations of that resistance? The Voltage between the point

A and B is supposed to be constant in this case.

Solution:

The equivalent resistance in the circuit is

= + ( ×+ ×)

Hence the current × through the resistance × is given by = ( + ) = ( + ) [ + + ]

= ∙ + + … . .

The thermal power generated in the resistance RX is given

by = = [ + + ] … = ( + +

For the power to be independent of small variations of RX, it

should be an extremum (in this case, a maximum). For this,

the variable part of the function (denominator) must be an

extremum.

Hence we must have [ + + ] = + [ + + ] − [ + + ] = + = + +

This gives = ×+ = ×+ =



40. Two cells of emfs 1.5V and 2V and internal resistances 2

ohm and 2 ohm respectively have their negative terminals

joined by a wire of 6 ohm and positive terminals by another

wire of 4 ohm. A third resistance of 8 ohm connects the

midpoints of these wires. Find the potential difference

between the ends of this third wire. Give the circuit diagram.

Solution:

The circuit diagram is drawn as follow:

Applying Kirchhoff’s law for the mesh CE1DC + + + + = . + = . … … . For the mesh CDE2C + + + + = + = … … . × ⟹ + = × ⟹ + =

Subtracting = = =

Now from (1) + ( ) = . = − = − = = + =

P.d. across 8 ohm = 8(i1 + i2) = × = .

41. (a) Find the potential difference between the points a and b,

in the following circuit. (b) If a and b are connected , find

the current in the 12V cell.

Solution: (a) Since when a and b are not connected, no

current flow across the 10V battery, let I be the current in

the loop PQRS.

Applying Kirchhoff’s law. 12 - 8= (2 + 2 + 1 + 2 + 1 + 1)I

or 4 = 9 I =

The potential difference

Vab = 5I + 8 – 10 = × − = = . (b)If a and b are connected there will be a redistribution of

current in the circuits.

Let the current be I1, I2 and I3 as shown in the Figure.

For the loop PQRSP,

12 – 10 = (1 + 1 + 2) I1 + ( 1 + 3 ) I2 ….. (1) Or 2I1 + 2I2 = 1

For the loop PQRSP,

12 – 8 = (2 + 1 + 1)I2 + ( 2 + 1 + 2 )I3

Or 4I1 + 5I3 = 4 ….. (2) Also we have I1 = I2 + I3 ….. (3) Solving (1), (2) and (3), we get,

I1 = 0.464 A, I2 = 0.036 A, I3 = 0.428 A

Hence the current through the 12 V battery = 0.464 A.

42. N cells, each of emf E and internal resistance r are

connected in a closed ring so that the positive terminal of

each cell is joined to the negative terminal of the next as

shown in the adjoining diagram. Any two points of this ring

are connected through an external resistance R. Find the

current in R1.



Solution:

If n be the number of cells connected on one side of resistor

R then the number of cells connected on the other side of R

is N-n. Current distribution in the equivalent circuit is

shown in the adjoining circuit dragram.

Here n cells are shown as equivalent to a single cell of emf

nE and inernal resistance nr. Similarly (N-n) cells are shown

equivalent to a single cell of emf (N-n) E and internal

resistance (N-n)r

Applying Kirchhoff’s second law to closed a b c d a × [ − ] = − ×= − −− ×= − − … … . i

Applying Kirchhoff’s second law to closed mesh c d a f b c × − + − = − + × [ − + ] = × = + ×= + ×= + … … ii

Combining equation (1) and equation (2) we get − − = + [ + − ] =

y = 0

i.e., current through resistor R is zero.

43. Eight identical resistors each of resistance ‘r’ are connected along edges of a pyramid having square base ABCD as

shown in figure 1.

(i) Between A and D

(ii) Between A and O

Solution:

(i) When a battery is connected across terminals A and D,

circuit becomes symmetric about a plane passing through

mid points of edges BC and AD vertex O. Therefore,

current through different resistors will be a shown in fig.2.

A

Since current in AO and Od is equal, therefore, they may be

assumed in series with each other. Similarly, BO and OC

may be assumed in series. Hence, the circuit may be

assumed as shown in Fig. c. Therefore, equivalent resistance

of each series combination (BO and OC, AO and OD) is

equal to 2r. Series combination of BO and OC is parallel

with BC

Equivalent resistance of this combination = ×+ =

Now this combination is in series with resistors AB and CD.

Therefore, equivalent resistance of this series combination. = + =

This series combination, resistor AD and series combination

of resistors AO and OD are the parallel with each other.

Hence equivalent resistance R of the circuit is given by. = + + =

(ii) Now the battery is connected across the terminals A so

O so the circuit becomes symmetric about a plane passing

through edges AO and OC, and vertex ‘O’ current through

different resistors will be as shown in figure.

Applying Kirchhoff’s law on mesh OCDO + − = = … .

For mesh ODAO



+ ( + ) − − − = + = …

From equation (1) and (2) = =

For mesh OBA – V – O + ( + ) − =

Substituting values of i1 and i2 i r =

But equivalent resistance of the circuit = Vi

Equivalent resistance = r

44. An electric circuit is shown in fig. Calculate the potential

difference across the resistor of 400 ohm, as will be measured

by the voltmeter V of resistance 400 ohm, either by applying

Kirchhoff’s rules or otherwise. Solution: The current distribution is also shown in fig.

Applying Kirchhoff’ second law to mesh A, we have + + = + = + = … .

Applying Kirchhoff’s voltage law to mesh B, we have + − = = + … …

Applying KVL to mesh C, we have ( − ) − + − = = − … …

Solving these equations, we get = , = = ( ) . Required potential difference = × − = × = .

Alternative method: The resistance 400 ohm of voltmeter

and another 400 Ω resistance in parallel with it gives a

resistance [ ×+ ] = Ω

The equivalent circuit is now drawn. This is shown in fig.

This is a balanced Wheatstone bridge. Obviously no current

flows between B and D. Now = + = .

Half of I2 passes through voltmeter and rest half passes

through other resistance of 400Ω

Hence potential difference across voltmeter = × = × × = = . .

Consider an infinite ladder network shown if fig.

A voltage is applied between points A and B. If the voltage

is halved after each section, find the ratio R1/R2. Suggest a

method to terminate it after a few sections without

introducing much error with attenuation.

[Roorkee]

45. A Nichrome wire of uniform cross-sectional area is bent to

form a rectangular loop A B C D. Another Nicrome wire of

the same cross-section is connected to form the diagonal a c.

Find out the ratio of the resistance across B D and A C if A

B = 0.4 and B C = 0.3m. [Roorkee]

Solution:

(see fig.). We know that = = , ℎ =

is the specific resistance of nichrome and A is area of

cross – section of nichrome wire. The current distribution is

shown in fig. Where i = 2 i1 + i2.



Let R1 = RAB = 4K, R2 = RBC = 3K and R3 is resistance of

wire AC = 5K. The equivalent resistance between AC is

given by = + + = + = = … .

Let the potential between B and D be VBC.

For path B A D, i1 R1 (i1 + i2) R2 = VBD … (2) For path BCAD,

2(i1 + i2) R2 + i2 R3 = VBD ….. (3) From eqs. (2) and (3), we get

i1(R1 + R2) + i2 R2 = VBD

Or i1(7 K) + i2 3K = VBD ….. (4) and 2i1 R2 + i2 (2R2 + R3) = VBD

or 6i K + i2 (11 K) + VBD ….. (5) Solving eqs. (4) and (5), we get = = … …

The equivalent resistance between B and D is given by = = + = + = … . .

The ratio of resistance across BD and AC is given by = =

46. In the given circuit. (see fig), E1=6, E2=2, E3=3 volt, R1=6

ohm, R2=2 ohm, R3=4 ohm, R4=3 ohm and C= 5 μ F.

Find the current in R3 and energy stored in the capacitor.

[ IIT 88]

Solution;

The distribution of current is shown in fig.

Applying Kirchhoff’s second law to mesh bcdeb, we have = = = . . Current in resistor R3 = 1.5 amp.

Applying Kirchhoff’s second law of mesh dhgfed, we have + − = − − − = − − × . = −

Solving, we get i2 = 0.2 amp.

To find out the potential difference between b f, we consider

the path b e f

Vb + 2i2 + 2 = Vf − = + = × . + = . . It is obvious that there is no current in resistor R1 hence

three will be 2.4 volt potential difference across the

condenser. The energy stored in capacitor c is given by = = . = . × − .

47. In the following R-C circuit, the capacitor is in the steady

state. The initial separation of the capacitor plates is x0. If at

t = 0, the separation between the plates starts changing so

that a constant current flows through R. Find the velocity of

the moving plates as a function of time. The pate area is A.

Solution:

Let q be the instantaneous charge on the capacitor when a

steady current i flow through the circuit. From the circuit,

we have = + = + … . .

When separation between the plates of the capacitor is x,

0 A/x

Differentiating eq. (1), with repsect to time, we get = ( ) + = = − ( ℎ = ) … .

From eq. (2), substituting the value of q in eq. (1), we have = − +

= = −− )

− = − ) … . .

Integrating the above expression with respect to time, we

get



− = [ − ] … . .

Again differentiating with respect to time

− = −

= ( ) = − ) … …

From eqs. (4) and (5), we get

= − ) [( ⁄− ) + ]

48. A circuit shown below has resistances R1 = 20Ω and R2 = Ω. At what value of the resistance Rx will the thermal power

generated in it be practically independent of small variations

of that resistance? The voltage between the points A and B is

supposed to be constant in this case.

Solution:

The equivalent resistance in the circuit is

= + ( x+ x)

Hence the current ix through the resistance Rx is given by

x = ( + ×) = ( + x) [ + x+ x ]

x = + × + … …

The thermal power generated in the resistance Rx is given

by = x x = x[ + x + ] … … … = xx [ + x + ]

For the power to be independent of small variations of Rx, it

should be an extremum (in this case, a maximum). For this

the variable part of the function (denominator) must be an

extremum.

Hence we must have

x [ + x + ]x = + [ + × + ]x

− [ + × + ]x = x + = + x +

This gives

x = ×+ = ×+ =

49. In the circuit shown below, resistances R1 and R2 as well as

the e.m.fs E1 and E2 are known. The internal resistances of

the sources are negligible. At what value of the resistance R

will the thermal power generated in it be the highest? Also

what is its value?

Solution:

Let the currents be as shown. Then by Kirchhoff’s laws.

For junction A. i = i1 + i2 …. (1) For the loop ABE1 A. E1 = i1 R1 + iR …. (2) For the loop ABE2 A. E2 = i2 R2 + iR …. (3) Substituting for in (2) and (3) from (1)

E1 = i1 R1 + (i1 + i2) R = i1(R1+R) + i2R …. (4) and E2 = i2 R2 + (i1 + i2 )R

= i2R+i2 (R2 + R) …. (5) Eliminating i1 from (4) and (5),

E1R – E2 (R1 + R) = i2 [R2 – (R1 + R)(R2 + R)]

Which gives = − +[ − + + ] = − +[ − + + ] Hence current i = i1 + i2 = − +[ − + + ] The thermal power generated in R is

P = i2R = [ +[ − + + ] ] = +[ − + + ]

For the power to be maximum, the denominator should be a

minimum. Hence we must have [ [ − + + ] ] =



[ [ + + ] ] = [ + + ] + − [ + + ] = = + + = +

The value of maximum thermal power generated.

= ++ [ + ]

= + +

50. An electric kettle is rated at (1.5 kW – 240 V). It takes 7

minutes for 1.8 kg of water to increase its temperature

from20°C to 100°C. (a) Determine the efficiency of the

kettle and (b) determine the mass of water which may be

boiled by 1 kWh of electrical energy. The specific heat

capacity of water is 4.18 kJ kg-1

K-1

.

Solution:

Energy input to kettle = power × time

= 1.5 kW × (7 × 60s)

=630 kJ

Heat required by water

= m s ∆ T = . . − − − =

(a) Efficiency of kettle = × % = × % = . %

(b) The energy required to ball 1.8 kJ

Of water is 630 kJ

mass of water boiled by 1 kW of energy = . × . ×× = .

51. Calculate the current through the 3 ohm resistor and the

power dissipated in the entire circuit shown in figure. The

emf of the battery is 1.8V and its internal resistance is ohm.

Solution:

The given circuit can be redrawn as follows

(i) The resistors 8 ohm and 2 ohm in parallel between A

and B = ×+ = . Ω

(ii) The resistors 4 ohm and 6 ohm in parallel between B

and C = ×+ = . Ω

(iii) Now 1.6 ohm and 2.4 ohm are in series along ABC

and they are equal to

1.6 + 2.4 = 4Ω

(iv) The circuit now reduces to 3 ohm, 4 ohm and 6 ohm in

parallel = + + = + + = = = Ω

(v) The total resistance of the circuit including the internal

resistance of the battery = + = Ω

Circuit current = . Ω = .

(vi) To find the current through 3 Ω. The circuit current divides itself in the three branches of 3

ohm, 4 ohm and 6 ohm resistors in the inverse ratio of their

resistances. . . , :

the current through the 3 ohm resistor = + + × . = .

(vii) Power dissipated in the entire circuit = VI=(1.8V) (0.9

amp)

= 1.62W



BOARD EXAM QUESTIONS

I. VERY SHORT ANSWER QUESTIONS: -

1. If the current following in a copper wire be allowed to

flow in another copper wire of same length but of double

the radius, what will be the effect on the drift velocity of

the electrons? Also if the same current is allowed to flow

in an iron wire of the same thickness, what will happen?

2. Specific resistances of copper, silver and constantan are

1.18 x 10-6

Ωcm, 1 x 10-6 Ωcm, and 48 x 10-6 Ωcm,

respectively. Which is the best electrical conductor and

why?

3. Define electric current. State its SI unit.

4. Name the carries of current in (a) solids, (b) liquids, and

(c) gases.

5. What is the direction of current flow?

6. Distinguish between static electricity and current

electricity.

7. Is electric current a vector or scalar quantity? Explain.

8. The connecting wires are always made up of copper.

Why?

9. On what factors does the resistance of a conductor

depend? Write a relation between them.

10. Define drift velocity of electrons. What is its order?

11. Write an expression for drift velocity in terms of

relaxation time.

12. Write the mathematical relation between mobility and

drift velocity of charge carriers in a conductor.

13. The standard resistance coils are made of manganin.

Explain.

14. What is the effect of rise in temperature on the electrical

resistivity of semiconductor?

15. What is (a) electric power and (b) electric energy? Define

their units.

16. Define the commercial unit of electric energy

17. What is the color coding of the strips on the carbon

resistor of resistance 200 Ω ± 20%?

18. What is the resistance of carbon resistor on which the

color of rigs in sequence is black, brown, black, and

gold?

19. What is emf of a cell? Write its dimensional formula and

SI unit.

20. State the condition in which terminal voltage across a

secondary cell is equal to its emf.

21. What is the internal resistance of a cell due to?

22. When cells are connected in parallel, what will be the

effect on (a) current capacity and (b) emf of the cells?

23. For what basic purpose the cells are connected (a0 in

series, 9b) in parallel, and 9c) in mixed grouping?

24. A battery has an internal resistance of 0.012 Ω and an emf of 9.00 V. What is the maximum current that can be

drawn from the battery without the terminal voltage

dropping below 8.90 V?

25. State Kirchhoff’s loop rule. 26. State Kirchhoff’s junction rule. 27. Are Kirchhoff’s laws applicable to both alternating

current and direct current?

28. Name the principle on which meter bridge works.

29. Why is a meter bridge so called?

30. Why should the current not be passed through

potentiometer or meter bridge wire for a long time?

II. SHORT ANSWER QUESTIONS :-

1. A current flowing in a copper wire is passed through

another copper wire of same length by double the radius.

How will the drift velocity of free electrons change?

2. The storage battery of a car has an emf of 12 V. If the

internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

3. A certain wire has a resistance R. What is the resistance

of a second wire, made of the same material, that is half

as long and has half the diameter?

4. The voltage-current variations of two metallic wires X

and F at a constant temperature are shown in Fig.

Assuming that the wires have same length and diameter,

explain which wire will have more resistivity and why?

5. Define the terms “resistivity” and “conductivity” and state their SI units. Draw a graph showing the variation

of resistivity with temperature for a typical

semiconductor.

6. The resistivity of semiconductors and insulators

decreases with increase of temperature. Why?

7. Discuss the effect of temperature on the resistance of (a)

metals, (b0 semiconductors, and (c) insulators.

8. Write the mathematical relation between mobility and

drift velocity of charge carriers in a conductor. Name the

mobile charge carriers responsible for conduction of

electric current in (a) an electrolyte and (b) an ionized

gas.

9. Define the term “resistivity” and wire its SI unit. Derive the expression for the resistivity of a conductor in terms

of number density of free electrons and relation time.

10. Deduce Ohm’s law using the concept of drift velocity.

11. A cell of emf ε and an internal resistance r is connected across a variable resistor R. Plot a graph showing the

variation of terminal potential difference V with



resistance R. Predict from the graph the condition under

which V becomes equal to ε

12. One electrical appliance operates with a voltage of 120

V, while another operates with 240 V. bases on this

information alone, is it correct to say that the second

appliance uses more power than the first? Give your

reasoning.

13. Two light bulbs are designed for use at 120 V and are

rated at 75 W and 150 W. Which light bulb has the

greater filament resistance? Why?

14. In order to obtain large current from a low voltage

source, it must have a low internal resistance. Why?

15. Two identical cells, each of emf and internal resistance

r, are put in parallel across an external resistance R. Find

the expression for current flowing through the circuit.

16. Define internal resistance (r) of a cell. State the factors on

which it depends.

17. State Kirchhoff’s laws of electrical network. Give their mathematical forms. Also explain the conservation laws

associated with them.

18. Draw a circuit diagram of a meter bridge arranged to

compare the two resistances. Explain the principle of the

experiment and give the formula used.

19. Fig. shows a 2.0 V potentiometer used for the

determination of internal resistance of a 1.5 V cell. The

balance point of the cell in open circuit is 76.3 cm. when

a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the

potentiometer wire. Determine the internal resistance of

the cell.

20. Wire an expression for the power rating of an electric

lamp. Draw a circuit diagram to show how you would

connect an ammeter and a voltmeter to measure the

electric power consumed by the lamp.

III. LONG ANSWER QUESTIONS :-

1. Derive an expression for drift velocity of free electrons in

a conductor in terms of relaxation time.

2. Define resistivity of a conductor. How does it vary with

temperature? Derive an expression for the resistivity of a

wire in terms of material parameter, number density of

free electrons and collision time.

3. Explain the dependence of resistivity and resistance of

metallic conductor on its temperature.

4. What is a thermistor? How is it made? How does it differ

from a resistance? Give some of its uses.

5. Give reason for the heat produced in a current carrying

conductor. Derive the necessary formula for heat

produced. State the factors on which it depends.

6.

7. Define the term electric power. Derive an expression for

it. Define its SI unit. Name its commercial unit and larger

units. Define the term efficiency of an electrical

appliance. What is its significance?

8. Define the term electrical energy and derive an

expression for it. State and define the commercial unit of

electrical energy. State and explain five applications of

heating effect of current.

9. What do you understand by internal resistance and

terminal potential difference of a cell? On what factors

do they depend?

10. Two cells of emf having internal

resistances respectively are connected in

parallel as shown. Deduce the expressions for the

equivalent emf and equivalent internal resistance of a cell

which can replace the combination between the points

11. A cell of emf (E) and internal resistance (r) is connected

across a variable external resistance (R). Plot graphs to

show variation of:(i) E with R, (ii) Terminal p.d of the

cell (V) with R.

12. State and explain Kirchhoff’s laws. Derive the condition for obtaining balance in a Wheat stone’s bridge.

13. What is the effect of following on the balance condition

of a Wheatstone’s bridge: (a) Current supplied by the cell; (b) resistance of galvanometer; (c) internal

resistance of cell; and (d) resistors connected in series

with cell and the galvanometer?

14. Explain a Wheatstone’s bridge. What for is it used? Using Kirchhoff’s laws of electrical network obtain the balancing condition. Why is this arrangement considered

to be an accurate one?

15. State the principle underlying meter bridge or slide wire

bridge with the help of a circuit diagram. How can it be

used for (a) finding specific resistance of a wire, (b) to



compare two unknown resistances, ad (c) finding an

unknown temperature?

IV. NUMERICALS:

1. (a) Estimate the average drift speed of conduction

electrons in a copper wire of cross-sectional area 1.0 x

10-7

m2 carrying a current of 1.5 A. Assume that each

copper atom contributes roughly one conduction electron.

The density of copper is 9.0 x 103 kg/m

3 and its atomic

mass is 63.5 μ. (b) Compare the drift speed obtained

above with (i) thermal speeds of copper atoms at ordinary

temperatures and (ii) speed of propagation of electric

field along the conductor which causes the drift motion.

2. A wire of unknown composition has a resistance of R0 =

35.00Ω when immersed in water at 20.00C. when the

wire is placed in boiling water, its resistance rises to

47.60 Ωcm,. What is the temperature on a hot summer day when the wire has a resistance of 37.80 Ω?

3. A conductor has a cross-section of 15 cm2 and a specific

resistance of 7.6 μΩcm at 00C. If the temperature

coefficient of resistance of the material is 0.0050C

-1,

estimate its resistance in Ω per 2 km, when its temperature is 50

0 C.

4. It is desired to make a coil of resistance 20 Ω having a zero temperature coefficient of resistance. To achieve

this, a carbon resistor of resistance R1 is placed in series

with an iron resistor of resistance R2. The proportion of

carbon and iron resistance are so chosen that R1 + R2 =

20 Ω for all temperatures near the room temperature. If the temperature coefficient of resistance for carbon and

iron are a1 = -0.5 x 10-3

/0C and a2 = 5 x 10

-3/0C,

respectively, calculate the values of R1 and R2.

5. Two wires of equal length, one of aluminum and the

other of copper have the same resistance. Which of the

two wires is lighter? Hence explain why aluminum wires

are preferred for overhead power cables. (ρA1= 2.63 x 10-

8 Ωgm, ρcu = 1.72 x 10

-8 Ωgm, relative density of Al =

2.7, of Cu = 8.9.)

6. Find the heat developed per minute in each of the three

resistors R1, R2, and R3 shown in Fig.

7. A 20 V battery of internal resistance 1 Ω is connected to three coils of 12 Ω, and 4 Ω in parallel, a resistor of resistance 5 Ω and a reversed battery (emf: 8 V; internal resistance: 2 Ω) as shown in Fig. Calculate (a) current in

the circuit and (b) potential difference across each

battery.

8. A network of resistors is connected to a 16 V battery with

internal resistance of 1 Ω as shown in Fig. (a) Compute the equivalent resistance of the network. (b) Obtain the

current in each resistor. (c) Obtain the voltage drops VAB,

VBC, and VCD.

9. (a) Three resistors 2 Ω, 4 Ω, and 5 Ω are combined in parallel. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 20

V and negligible internal resistance, determine the

current through each resistor and the total current drawn

from the battery.

10. Determine the current in each branch of the network

shown in Fig.

11. The circuit Fig. is known as a Wheatstone bridge circuit.

Find the voltage between points B and D, and state which

point is at the higher potential.



12. A 6 V battery of negligible internal resistance is

connected across a uniform wire AB of length 100 cm.

The positive terminal of another battery of emf 4 V and

internal resistance 1 Ω is joined to the point A as shown in Fig. Take the potential at B to be zero. (a) What are the

potentials at the points A and C? (b) Om the wire AB,

what is the length of AD in which the potential is equal to

the potential at C? (c) If the points C and d are connected

by a wire, what will be the current through it? (d) If the 4

V battery is replaced by 7.5 V battery, what would be the

answers of part (a) and (b)?

13. In the simple Wheatstone bridge circuit in which where

the length AB of bridge wire is 1 m, the resistors X and Y

have values 5 Ω and 2 Ω, respectively. When X is shunted by a length of a wire, the balance point is found

to be 0.625 m from A. What is the resistance of the

shunt? If the shunt wire is 0.75 m long and 0.25 mm in

diameter, what is the resistivity of the material of the

wire?

14. (a) In a meter bridge, the balance point is found to be at

40 cm from the end A when the resistor R is of 15 Ω. Find resistances S. (b) if the cell and galvanometer are

interchanged at the balance point, would it affect the flow

of current through the galvanometer? (c) Calculate the

balance point of the bridge if R and S are interchanged.

15. The length of a potentiometer wire is 1200 cm and it

carries a current of 80 mA. For a cell of emf 4.0 V and

internal resistance 20 Ω, the null point is found to be at 1000 cm. if a voltmeter is connected across the cell, the

balancing length is decreased by 20 cm: Find (a) the

resistance of the whole wire, (b) reading of the voltmeter,

and (c) resistance of voltmeter.

MODEL PAPERS


BOARD EXAM MODEL PAPER 92

Board Exam Model Paper

I. Very Short Answer Questions

1. A point charge + Q is placed at point O as shown in the

figure. Is the potential difference VA – VB positive,

negative or zero ?

1. How does the electric flux due to a point charge enclosed

by a spherical Gaussian surface get affected when its

radius is increased?

2. Why alloys like constantan and manganin are used for

making standard resistors?

3. Define the term drift velocity ?

4. State the working principle of potentiometer .

5. What is the electric flux through a cube of side 1 cm

which encloses an electric dipole?

6. Graph showing the variation of current versus voltage for

a material GaAs is shown in the figure. Identify the region

of (i) negative resistance (ii) Where Ohm’s law is obeyed.

7. A resistance R is connected across a cell of emf E and

internal resistance r. Now, a potentiometer measures the

potential difference between the terminals of the cells as

V. Write the expression for r in terms of E, V and R.

8. In an experiment on meter bridge, if the balancing length

AC is X, what would be its value, when the radius of the

meter bridge wire AB is doubled? Justify your answer.

9. When electrons drift in a metal from lower to higher

potential, does it mean that all the free electrons of the

metal are moving in the same direction?

II. Short Answers Question

10. Two cells of ems 1.5 V and 2.0 V having internal

resistances 0.2 Ω and 0.3 Ω respectively are connected in parallel. Calculate the emf and internal resistance of

equivalent cell.

11. On the basis of electron drift, derive an expression for

resistivity of a conductor in terms of number density of

free electrons and relaxation time. On what factors does

resistivity of a conductor depend?

12. Use Kirchhoff’s rules to obtain conditions for the balance condition in a Wheatstone bridge.

13. Draw a plot showing he variation of (i) electric field (E)

and (ii) electric potential (V) with distance r due to point

charge Q.

14. Two charged conducting spheres of radii r1 and r2

connected to each other by a wire. Find the ratio of

electric fields at the surfaces of the two spheres.

III. Long Answer Questions

15. a) Explain, using suitable diagrams, the difference in the

behavior of a (i) conductor and (ii) dielectric in the

presence of external electric field. Define the terms

polarization of a dielectric and write its relation with

susceptibility.

b) A thin metallic spherical shell of radius R carries a

charge Q on its surface. A point charge Q2 is placed at its

centre C and an other charge + 2 Q is placed ou8tside the

shell at a distance x from the centre as shown in the

figure. Find (i) the force on the charge at the centre of

shell and the point A, (ii) the electric flux through the

shell.

16. In the following potentiometer circuit AB is a uniform

wire of length 1 m an d resistance 10 Ω. Calculate the potential gradient along the wire and balance length AO (

= l)

BOARD EXAM MODEL PAPER

93

17. A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the

terminal voltage V versus (i) r and (ii) the current I .

It is found that when r = 4 Ω , the current is 1 A and when R is increased to 9 Ω, the current reduces to 0.5 A.

find the values of the emf E and internal resistance r

18. a) An electric dipole of dipole moment p consists of

point charges + q and –q separated by a distance 2a apart.

Deduce the expression for the electric field E due to the

dipole at a distance x from the centre of the dipole on its

axial line in terms of the dipole moment p. Hence show

that in the limit x >> a, E ⟶ 3 p / (4πε0x3)

b) Given the electric field in the region E = 2xi , find the

electric flux through the cube and the charge enclosed by

it.

IV. Numerical

19. Two capacitors of unknown capacitances C1 and C2 are

connected first in series and then in parallel across a

battery of 100 V. If the energy stored in the two

combinations is 0.045 J and 0.25 J respectively,

determine the value of C1 and C2 . Also calculate the

charge on each capacitor in parallel combination.

20. State the principle of working of a galvanometer. A

galvanometer of resistance G is converted into a voltmeter

to measure up to V volts by connecting a resistance R1 in

series with the coil. If a resistance R2 is connected in

series with it, then it can measure up to V/2 volts. Find

the resistance, in terms of R1 and R2, required to be

connected to convert it into a voltmeter that can read up to

2 V. Also find the resistance G of the galvanometer in

terms of R1 and R2.

21. Two parallel plate capacitors X and Y have the same area

of plates and same separation between them. X has air

between the plates while Y contains a dielectric of εr = 4 .

(i) Calculate capacitance of each capacitor of equivalent

capacitance of the combination is 4μF. (ii) Calculate the potential difference between the plates

of X and Y.

(iii) Estimate the ratio of electrostatic energy stored in X

and Y

22. A cubical Gaussian surface encloses a charge of 8.85 X

10−

10C in vacuum at centre. Calculate the electric flux

through the one of its faces

23. The electric field components due to a charge inside the

cube of side 0.1 m are shown

Ex = ∝x,

where α = 500 N/C – m, Ey = 0, Ez = 0

Calculate (i) the flux through the cube and (ii) the charge

inside the cube.

24. a) State Kirchhoff’s rules. Apply these rules to the loops

PRSP and PRQP to write the expressions for the currents

I1, I2 and I3 in given circuit

25. The sequence of colored bands in two carbon resistor R1

and R2 is (a) brown, green, blue and

(b) orange, black, green. Find the ratio of their

resistances.


JEE MAINS MODEL PAPER 94

JEE (Main) Model Paper

I Answer the following:

1. Two charges, each equal to q, are kept at x = − a and x = a on the x-axis. A particle of mass m and charge q0= is

placed at the origin. If charge qo is given a small

displacement ( y << a ) along the y- axis, the net force

acting on the particle is proportional to

(a) − y (b) y (c) − y (d) y

2. In a uniformly charged sphere of total charge Q and radius

R, the electric field E is plotted as a function of distance

from the centre. The graph which would correspond to the

above will be

3. Two identical charged spheres suspended from a common

point by two mass less strings of length l are initially a

distance d(d < < l ) apart because of their mutual repulsion .

The charge begins to leak from both the spheres at a

constant rate. As a result the charges approach each other

with a velocity v. Then as a function of distance x between

them

(a) v ∝ x−1/2

(b) v ∝ x−1

(c) v ∝ x1/2

(d) v ∝ x

4. Two identical charged spheres are suspended by strings of

equal lengths. The strings make an angle of 30o with each

other. When suspended in a liquid of density 0.8 gcm−3

, the

angle remains the same. If density of the material of the

sphere is 1.6 gcm-3

, the dielectric constant of the liquid is

(a) 1 (b) 4 (c) 3 (d) 2

5. A thin semi –circular ring of radius r has a positive charge q

distributed uniformly over it . The net field E at the centre

O is

(a)

π εo j (b) π εo j

(c) − π εo j (d) −

π εo j

6. Assume that an electric field exists in space. Then the

potential difference VA − V 0 where V0 is the potential at

the origin and VA the potential at x = 2 m is

(a) 80 J (b) 120 J (c) − 120 J (d) − 80 J

7. A charge Q is uniformly distributed over a long rod AB of

length L as shown in the figure. The electric potential at the

point O lying at a distance L from the end A is

(a) πϵ L (b)

8πϵ L (c) πϵ L (d)

πϵ L

8. This question has Statement 1 and Statement 2. Of the four

choices given after the statements, choose the one that best

describes the two statements.

An insulating solid sphere of radius R has a uniformly

positive charge density ρ. As a result of this uniform charge distribution there is a finite value of electric potential at the

centre of the sphere, at the surface of the sphere and also at

a point outside the sphere. The electric potential at infinity

is zero.

Statement 1 : When a charge q is taken from the centre to

the surface of the sphere, its potential energy changes by ε

.

Statement 2: The electric field at a distance

r (r < R ) from the centre of the sphere is ε

(a) Statement 1 is true, Statement 2 is false.

(b) Statement 1 is false, Statement 2 is true.

(c )Statement 1 is true, Statement 2 is true, Statement 2 is

the correct explanation of Sttement1

(d) Statement 1 is true, Statement 2, is true;

Statement 2 is not the correct explanation of

Statement 1.

9. The electrostatic potential inside a charged spherical ball is

given by Φ = ar2 + b where r is the distance from the centre;

a, b are constants. Then the charge density inside the ball is

(a) − 24πaε0r (b) − 6aε0r

(c) − 24πaε0 (d) − 6aε0

10. Let there be a spherically symmetric charge distribution

with charge density varying as ρ ( r) = ρ0 − up to r =

R, and ρ(r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r ( r < R ) from the

origin is given by

(a) ρ

ε − (b)

πρε

−

(c ) ρ

ε − (d)

ρε

−

11. A parallel plate capacitor is made of two circular plates

separated by a distance of 5mm and with a dielectric of

dielectric constant 2.2 between them. When the electric

field in the dielectric is 3 X 104 V/m , the charge density of

the positive plate will be close to



(a) 6 X 104 C/m

2 (b) 6 X 10

--7 C/m

2

(c) 3 X 10--7

C/m2 (d) 3 X 10

4 C/m

2

12. Two capacitors C1 and C2 are charged to 120 V and 200 V

respectively. It is found that by connecting them together

the potential on each one can be made zero. Then

(a) 9C1 = 4C2 (b) 5C1 = 3C2

( c) 3C1 = 5C2 (d) 3C1 + 5C2 = 0

13.

The figure shows an experimental plot for discharging of a

capacitor in an R – C circuit. The time constant t of this

circuit lies between

(a) 0 and 50 s (b) 50 s and 100 s

( c) 100 s and 150 s (d) 150 s and 200 s

14. In a large building, there are 15 bulbs of 40 W, 5 bulbs of

100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage

of the electric mains is 220 V. The minimum capacity of

the main fuse of the building will be

(a) 14 A (b) 8 A (c) 10 A (d) 12 A

15. The supply voltage to a room is 120 V. the resistance of the

lead wires is 6 Ω . A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240

W heater is switched on in parallel to the bulb?

(a) 10.04 Volt (b) zero Volt

(c) 2.9 Volt (d) 13.33 Volt

16. Two electric bulbs marked 25 W-220 V and 100 W – 220 V

are connected in series to a 440 V supply. Which of the

bulbs will fuse?

(a) 100 W (b) 25 W (c) neither (d) both

17. If a wire is stretched to make it 0.1% longer, its resistance

will

(a) increase by 0.05% (b) increase by 0.2%

(c) decrease by 0.2% (d) decrease by 0.05%

18. Two conductors have the same resistance at 00C but their

temperature coefficients of resistance are ∝1and ∝2 . The


parallel combinations are nearly

(a) ∝ +∝

, ∝ +∝

(b) ∝ +∝

, ∝ +∝

( c ) ∝ +∝ , ∝ +∝

(d) ∝ +∝ , ∝ ∝ ∝ +∝

19. The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density ρ =

, where A is a constant and r is the distance rom the

centre. At the centre of the spheres is a point charge Q. The

value of A such that the electric field in the region between

the spheres will be constant, is:

(a) π

(b) π − (c)

π − (d) π

20. A combination of capacitors is set up as shown in the figure.

The magnitude of the electric field due to a point charge Q (

having a charge equal to the sum of the charges on the 4μF and 9 μF capacitors), at a point distant 30 m from it, would equal:

(a) 240 N/C (b) 360 N/C

(c) 420 N/C (d) 480 N/C

21. The temperature dependence of resistances of Cu and

undoped Si in the temperature range 300 – 400 K, is best

described by:

(a) Linear increase for Cu, linear increase for Si.

(b) Linear increase for Cu, exponential increase for Si.

(c) Linear increase for Cu, exponential decrease for Si.

(d) Linear decrease for Cu, linear decrease for Si.

22. Two identical wires A and B, each of length ‘l’, carry the

same current I. Wire A is bent into a circle of radius R and

wire B is bent to form a square of side ‘a’, If BA and BB

are the values of magnetic field at the centers of the circle

and square respectively, then the ratio is:

(a) π8 (b)

π√ (c) π

(d) π8√



23. Four capacitors with capacitance C1 = 1 μF, C2 = 1.5 μF, C3 = 2.5 μF, C4 = 0.5 μF, are connected as shown in the figure and are connected to a 30 V source. The potential

difference between points a and b is

(a) 5 V (b) 9 V (c) 10 V (d) 13 V

24. The mass of a proton is 1847 times that of electron. If an

electron and a proton are injected in uniform electric field

at right angle to the direction of the field, with the same

kinetic energy, then

(a) the proton trajectory will be less curved than that of

electron

(b) both the trajectories will be straight

(c ) both the trajectories will be equally curved

(d) the electron trajectory will be less curved than that of

proton.

25. An oil drop of 12 excess electrons is held stationary under a

constant electric field of 2.55 X 104 NC

−1 in Milikan’s oil

drop experiment. The density of the oil is 1.26 g cm-3. .

Estimate the radius of the drop.

(g = 1.26 g cm−2

; e = 1.60 x 10−19

C ).

(a) 4.91 x 10−4

m (b) 4.91 x 10−7

m

(c) 9.82 x 10−4

mm (d) 9.82 x 10−7

m m

26. A charge of 8 mC is located at the origin. Calculate the

work done in taking a small charge of − 2 x 10−9C from a

point P(0,0, 3cm) to a point Q(0,4cm, 0), via a point R (0,6

cm, 9cm).

(a) 5 J (b) 10 J (c) 2.4 J (d) 1.2 J

27. If a resistance R2 is connected in parallel with the

resistance R in the circuit shown, then possible value of

current through R and the possible value of R2 will be

(a) Ι, R (b) Ι,2R ( c) Ι ,2R (d)

Ι ,R

28. A hallow cylinder has a charge q coulomb within it. If Φ is

electric flux in units of volt meter associated with the curved

surface B, the flux linked with the plane surface A in units

of volt meter will be

(a) q/2∈0 b) Φ/3 c) (q/∈0) - Φ d) [(q/∈0) - Φ]/2

29. Two cells A and B are connected in the secondary circuit of

a potentiometer one at a time and the balancing length are

respectively 400 cm and 440 cm. The emf of the cell A is

1.08 V. The emf of the second cell B is

(a) 1.08 V (b) 1.188V (c) 11.88 V (d) 12.8 V

30. Electric field on the axis of a small electric dipole at a

distance r is E 1 and E 2 at a distance of 2r on a line of

perpendicular bisector. Then

(a) E 2 = − E 8 (b) E 2 = − E

(c) E 2 = − E (d) E 2 = E 8


NEET MODEL PAPER 97

NEET – Model Paper

1. A capacitor of 2μF is charged as shown in the diagram.

When the switch S is turned to position 2, the percentage

of its stored energy dissipated is:

(1) 0% (2) 20% (3) 75% (4) 80%

2. Two identical charged spheres suspended from a common

point by two massless strings of lengths l, are initially at a

distance d (d << l ) apart because of their mutual repulsion.

The charges begin to leak from both the spheres at a

constant rate. As a result, the spheres approach each other

with a velocity v. Then v varies as a function of the

distance x between the spheres, as :

(1) v ∝ x1/2 (2) v ∝ x (3) v ∝ x-1/2

(4) v ∝ x–1

3. The charge flowing through a resistance R varies with time

t as Q = at – bt2, where a and b are positive constants. The

total heat produced in R is:

1) a3R /6b 2) a

3R /3b 3) a

3R /2b 4) a

3R /b

4. The electric field in a certain region is acting radially

outward and is given by E = Ar . A charge contained in a

sphere of radius ‗a‘ centred at the origin of the field, will be given by :

(1) εoAa3 (2) 4π εoAa

2 (3) εoAa

2 (4) 4π εoA a

3

5. A, B and C are voltmeters of resistance R, 1.5 R and 3R

respectively as shown in the figure. When some potential

difference is applied between X and Y, the voltmeter

readings are VA, VB and VC respectively. Then :

(1) VA ≠ VB ≠ VC (2) VA = VB = VC

(3) VA ≠ VB = VC (4) VA = VB ≠ VC

6. A potentiometer wire has length 4 m and resistance 8 Ω .

The resistance that must be connected in series with the

wire and an accumulator of e.m.f. 2V, so as to get a

potential gradient 1 mV per cm on the wire is :

(1) 48 Ω (2) 32 Ω (3) 40 Ω (4) 44 Ω

7. Across a metallic conductor of non-uniform cross section a

constant potential difference is applied. The quantity

which remains constant along the conductor is :

(1) electric filed (2) current density

(3) current (4) drift velocity

8. A parallel plate air capacitor of capacitance C is connected

to a cell of emf V and then disconnected from it. A

dielectric slab of dielectric constant K, which can just fill

the air gap of the capacitor, is now inserted in it. Which of

the following is incorrect ?

(1) The charge on the capacitor is not conserved.

(2) The potential difference between the plates decreases

K times.

(3) The energy stored in the capacitor decreases K times.

(4) the change in energy stored is (CV2/2)(K

-1)

9. In an ammeter 0.2% of main current passes through the

galvanometer. If resistance of galvanometer is G, the

resistance of ammeter will be:

(1) 1/500 G (2) 500/499 G (3) 1/499 G (4) 499/500 G

10. The resistance in the two

arms of the meter bridge

respectively. When the

resistance R is shunted

with an equal resistance,

the new balance point is at

1.6l1. The resistance ‘R’ is (2) 25

Ω

\

11. In a region, the potential is represented by V(x, y, z) = 6x -

8xy - 8y + 6yz, where V is in volts and x, y, z are in

meters. The electric force experienced by a charge of 2

coulomb situated at point (1,1,1) is:

(1) 24N (2) 4√35N (3) 6√5N (4) 30N

12. A potentiometer circuit has been set up for finding the

internal resistance of a given cell. The main battery, used

across the potentiometer wire, has an emf of 2.0 V and a

negligible internal resistance. The potentiometer wire itself

is 4 m long. When the resistance, R connected across the

given cell, has value of.

(i) infinity (ii) 9.5Ω, the ‘balancing lengths’ , on the

potentiometer wire are found to be 3m and 2.85m,

respectively. The value of internal resistance of the cell is:

(1) 0.5Ω (2) 0.75 Ω (3) 0.25 Ω (4) 0.95 Ω

13. Two thin dielectric slabs of

dielectric constants K1 and

K2 (K1 < K2) are inserted

between plates of a parallel

plate capacitor, as shown in

the figure. The variation of

electric field ‘E’ between the plates with distance ‘d’ as measured from plate P is

correctly shown by

(1) (2)

(3) (4)

14. A conducting sphere of radius R is given a charge Q. The

electric potential and the electric field at the centre of the

sphere respectively are :

(1) Q/4πεoR2 and Q/4πεoR

2 (2) Both are zero


NEET MODEL PAPER 98

(3) Zero and Q/4πεoR2 (4) Q/4πεoR

2 and zero

15. A, B and C are three

points in a uniform

electric field. The

electric potential is –

(1) maximum at A

(2) maximum at B

(3) maximum at C

(4) same at all the three points A, B and C

16. A wire of resistance 4Ω is stretched to twice its original

length. The resistance of stretched wire would be -

(1) 2Ω (2) 4Ω (3) 8Ω (4) 16Ω

17. The internal resistance of a 2.1 V cell which gives a

current of 0.2A through a resistance of 10Ω is -

(1) 0.2Ω (2) 0.5Ω (3) 0.8Ω (4) 1.0Ω

18. The resistances of the four arms P, Q, R and S in a

Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the

cell are 7 volt and 5 ohm respectively. If the galvanometer

resistance is 50 ohm, the current drawn from the cell will

be -

(1) 1.0 A (2) 0.2 A (3) 0.1 A (4) 2.0 A

19. Two metallic spheres of radii 1 cm and 3 cm are given

charges of –1 × 10–2

C and 5 × 10–2

C, respectively. If

these are connected by a conducting wire, the final charge

on the bigger sphere is

(1)1 × 10–2

C (2)2 × 10–2

C (3)3 × 10–2

C (4)4 × 10–2

C

20. The power dissipated in the

circuit shown in the figure is 30

Watts. The value of R is

(1) 30Ω (2) 20 Ω (3) 15 Ω (4) 10 Ω

21. A cell having an emf ε and internal resistance r is

connected across a variable external resistance R. As the

resistance R is increased, the plot of potential difference V

across R is given by

1) 2)

3) 4)

22. A charge Q is enclosed by a Gaussian spherical surface of

radius R. If the radius is doubled, then the outward electric

flux will

(1) Be doubled (2) Increase four times

(3) Be reduced to half (4) Remain the same

23. Four electric charges +q, +q, –q and –q are placed at the

corners of a square of side 2L. The electric potential at

point A, midway between the two charges +q and +q, is:

1) Zero

24. A parallel plate condenser has a uniform electric field E

(V/m) in the space between the plates. If the distance

between the plates is d (m) and area of each plate is A(m)2

the energy (joules) stored in the condenser is

25. If power dissipated in the 9 W

resistor in the circuit shown is

36 Watt, the potential

difference across the 2 Ω resistor is

1) 2V 2) 4V

3) 8V 4) 10V

26. A current of 2 A flows through a 2 Ω resistor when

connected across a battery. The same battery supplies a

current of 0.5 A when connected across a 9 Ω resistor. The

internal resistance of the battery is

(1) 1 Ω (2) 0.5 Ω (3) 1/3 Ω (4) 1/4 Ω

27. Consider the following two statements:

A. Kirchhoff's junction law follows from the conservation

of charge.

B. Kirchhoff's loop law follows from the conservation of

energy

Which of the following is correct?

1. Both (A) and (B) are wrong

2. (A) is correct and (B) is wrong

3. (A) is wrong and (B) is correct

4. Both (A) and (B) are correct

28. A series combination of n1 capacitors, each of value C1, is

charged by a source of potential difference 4 V. When

another parallel combination of n2 capacitors, each of

value C2, is charged by a source of potential difference V,

it has the same (total) energy stored in it, as the first

combination has. The value of C2, in terms of C1, is then:

1) 2C1/n1n2 2) 16 n1 C1/n2 3) 2 n1 C1/n2 4) 16C1/n1n2

29. See the electric circuit shown in this Figure. Which of the

following equations is a correct equation for it?


NEET MODEL PAPER 99

30. A wire of resistance 12 ohms per meter is bent to form a

complete circle of radius 10 cm. The resistance between its

two diametrically opposite points, A and B as shown in the

Figure, is:

(1) 3 Ω (2) 6π Ω (3) 6 Ω (4) 0.6 π Ω


CET MODEL PAPER 100

CET Model Paper

I Answer the following:

1. The angle between the dipole movement and electric field at

any point on the equatorial plane is

(a) 900 (b) 45

0 (c) 0

0 (d) 180

0

2. Pick out the statement which is incorrect.

(a) The electric field lines forms closed loop.

(b) Field lines never intersect.

( c)The tangent drawn to a line of force represents the

direction of electric field.

(d) A negative test charge experiences a force opposite to the

direction of the field.

3. Two spheres carrying charges + 6μC and + 9μC separated

by a distance d, experiences a force of repulsion F. When a

charge of − 3μC is given to both the spheres and kept at the

same distance as before, the new force of repulsions is

(a) 3F (b) 9 (c) F (d)

4. A parallel plate capacitor is charged and then isolated. The

effect of increasing the plate separation on charge, potential

and capacitance respectively are

(a) increases, decreases, decreases

(b) constant, increases, decreases

(c) constant, decreases, decreases

(d) constant, decreases, increases

5. A spherical shell radius 10 cm is carrying a charge q. If the

electric potential at distances 5 cm, 10 cm and 15 cm from

the centre of the spherical shell is V1V2 and V3 respectively,

then

(a) V1 < V2 < V3 (b) V1 = V2 < V3

(c) V1 > V2 > V3 (d) V1 = V2 > V3

6. Three point charges 3 nC, 6nC and 9nC are placed at the

corners of an equilateral triangle of side 0.1 m The potential

energy of the system is

(a) 89100 J (b) 99100 J (c) 8910 J (d) 9910 J

7. The resistance of the bulb filament is 100 Ω at temperature

of 1000C. If its temperature coefficient of resistance be 0.005

per 0C, its resistance will become 200Ω at a temperature

(a) 4000C (b) 200

0C (c) 300

0C (d) 500

0C

8. In Wheatstone’s network P = 2 Ω, Q= 2 Ω, R = 2 Ω and S = 3 Ω. the resistance with which S is to be shunted in order

that the bridge may be balanced is

(a) 2 Ω (b) 6 Ω (c) 1 Ω (d) 4 Ω

Three resistances 2 Ω, 3 Ω and 4 Ω are connected in parallel.

The ratio of currents passing through them when a potential

difference is applied across its ends will be

(a) 6 : 4 : 3 (b 4 : 3 : 2 (c) 6 :3 : 2 (d) 5: 4: 3

9. Four identical cells of emf E and internal resistance r are to

be connected in series. Suppose if one of the cell is connected

wrongly, the equivalent emf and effective internal resistance

of the combination is

(a) 4E and 2r (b) 2E and 2r

(c) 4E and 4r (d) 2E and 4r

10. In the circuit shown below, the ammeter and the voltmeter

readings are 3 A and 6 V respectively . Then the value of the

resistance R is

(a) > 2 Ω (b) ≥ 2 Ω (c) 2 Ω (d) < 2 Ω

11. Two cells of emf E1 and E2 are joined in opposition (such

that E1 > E2 ). If r1 and r2 be the internal resistance and R be

the external resistance, then the terminal potential difference

is

(a) + r + r + R x R (b) − r + r + R x R

(c ) + r + r x R (d)

− r + r x R

12. What is the value of shunt resistance required to covert

a galvanometer of resistance 100 Ω into an ammeter of range 1 A ? Given full scall deflection is 5 m A .

(a) 0.5 Ω (b) 9.9 Ω (c) 0.05 Ω (d) 9.9

Ω

13. When an additional charge of 2C is given to a capacitor,

energy stored in it is increased by 21%. The original charge

of the capacitor is

(a) 30 C (b) 40 C (c) 10 C (d) 20 C

14. When a potential difference of 103 V is applied between A

and B, a charge of 0.75 mC is stored in the system of

capacitors as shown. The value of C is (in μF )


CET MODEL PAPER 101

(a) (b) 2 (c) 2.5 (d) 3

15. See the diagram. Area of each plate is 2.0 m2 and d = 2 ×

10−3

m. A charge of 8.85 × 10−8

C is given to Q. Then the

potential of Q becomes

(a) 13 V (b) 10 V (c) 6.67 V (d) 8.825 V

16. Three conductors draw currents of 1 A, 2 A and 3 A

respectively, when connected in turn across a battery. If they

are connected in series and the combination is connected

across the same battery, the current drawn will be

(a) 7 A (b) 7 A (c) 7 A (d) 7 A

17. In the circuit, R1 = R2. The value of E and R1 are _______ (E

– EMF , R1 – resistance)

(a) 180

V, 60 Ω (b) 120 V, 60 Ω

(c ) 180 V, 10 Ω (d) 120 V, 10 Ω

18. Masses of three wires of copper are in the ratio of 1 : 3 : 5

and their lengths are in the ratio of 5: 3 : 1. the ratio of their

electrical resistances is

(a) 1 : 3 : 5 (b) 5: 3 : 1

(c) 1 :15 : 125 (d) 125 : 15 : 1

In the circuit diagram, heat produces in R, 2 R and 1.5R are

in the ratio of

(a) 4:2 :3 (b) 8: 4 : 27 (c) 2: 4 : 3 (d) 27 : 8 : 4

19. Two metal spheres of radii 0.01 m and 0.02 m are given a

charge of 15mC and 45mC respectively. They are then

connected by a wire. The final charge on the first sphere is

_______× 10−3

C

(a) 40 (b) 30 (c) 20 (d) 10

20. Two concentric spheres of radii R and r have positive charges

q1 and q2 with equal surface charge densities. What is the

electric potential at their common centre?

(a) σϵ (R + r ) (b)

σϵ (R −r )

(c) σϵ R + r (d)

σϵ Rr

21. Two capacitors of 10 PF and 20 PF are connected to 200 V

and 100 V sources respectively. If they are connected by

the wire, what is the common potential of the capacitors?

(a) 300 volt (b) 133.3 volt (c) 400 volt (d) 150 volt

22. If a charge on the body is 1 nC, then how many electrons are

present on the body ?

(a) 6.25 x 10 27

(b) 1.6 X 1019

(c) 6.25 X 1028

(d) 6.25 X 109

23. Two equal and opposite charges of masses m1 and m2 are

accelerated in an uniform electric field through the same

distance. What is the ratio of their accelerations if their ratio

of masses is mm = 0.5 ?

(a) aa = 2 (b)

aa = 0.5 (c) aa = 3 ( d)

aa = 1

24. What is the nature of Gaussian surface involved in Gausss

law of electrostatic?

(a) Magnetic (b) Scalar (c) Vector (d) Electrical

25. What is the electric potential at a distance of 9 cm from 3

nC?

(a) 300 V (b) 270 V (c) 30 V (d) 3 V

26. A voltmeter reads 4 V when connected to a parallel plate

capacitor with air as a dielectric. When a dielectric slab is

introduced between plates for the same configuration,

voltmeter reads 2 V. What is the dielectric constant of the

material ?

(a) 8 (b) 0.5 (c) 10 (d) 2

27. A spherical conductor of radius 2 cm is uniformly charged

with 3 nC. What is the electric field at a distance of 3 cm

from the centre of the sphere?

(a) 3 x 104 V m

−1 (b) 3 x 10

6 V m

−1

(c ) 3 x 10−4 V m

−1 (d) 3 V m

−1

28. A carbon film resistor has color code Green Black Violet

Gold. The value of the resistor is

(a) 500 ± 5% MΩ (b) 50 MΩ ( c) 500 ± 10% MΩ (d) 500 MΩ

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