3-Current-Electricity.pdf

8/10/2019 3-Current-Electricity.pdf

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Question 3.1:

The storage battery of a car

0.4, what is the maximum

Answer

Emf of the battery,E= 12 V

Internal resistance of the batt

Maximum current drawn fro

According to Ohms law,

The maximum current draw

Question 3.2:

A battery of emf 10 V and inin the circuit is 0.5 A, what i

of the battery when the circu

Answer



as an emf of 12 V. If the internal resistance of t

urrent that can be drawn from the battery?

ery, r= 0.4

the battery =I

from the given battery is 30 A.

ternal resistance 3 is connected to a resistor. Ithe resistance of the resistor? What is the term

t is closed?

ery, r= 3

e battery is

f the currentnal voltage


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Current in the circuit,I= 0.5

Resistance of the resistor =

The relation for current usin

Terminal voltage of the resis

According to Ohms law,

V=IR

= 0.5 17

= 8.5 V

Therefore, the resistance of t

8.5 V.

Question 3.3:

Three resistors 1 , 2 , and

the combination?

If the combination is connec

resistance, obtain the potenti

Answer

Three resistors of resistances

the combination is given by t

A

Ohms law is,

or = V

e resistor is 17 and the terminal voltage is

3 are combined in series. What is the total re

ed to a battery of emf 12 V and negligible inter

l drop across each resistor.

1 , 2 , and 3 are combined in series. Tota

he algebraic sum of individual resistances.

sistance of

al

resistance of


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Total resistance = 1 + 2 + 3

Current flowing through the


Total resistance of the circui

The relation for current usin

Potential drop across 1 res

From Ohms law, the value

V1 = 2 1= 2 V (i)


Again, from Ohms law, the

V2 = 2 2= 4 V (ii)


Again, from Ohms law, the

V3 = 2 3= 6 V (iii)

Therefore, the potential drop

respectively.

Question 3.4:

Three resistors 2 , 4 andthe combination?

If the combination is connec

resistance, determine the cur

the battery.

Answer

6

circuit =I

,R = 6

Ohms law is,

istor = V1

f V1 can be obtained as

istor = V2

alue of V2 can be obtained as

istor = V3

alue of V3 can be obtained as

across 1 , 2 , and 3 resistors are 2 V, 4 V,

5 are combined in parallel. What is the total r

ed to a battery of emf 20 V and negligible inter

ent through each resistor, and the total current

and 6 V

esistance of

al

rawn from


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There are three resistors of resistances,

R1 = 2 ,R2 = 4 , andR3 = 5

They are connected in parallel. Hence, total resistance (R) of the combination is given by,

Therefore, total resistance of the combination is .

Emf of the battery, V= 20 V

Current (I1) flowing through resistorR1 is given by,



Total current,I=I1 +I2 +I3 = 10 + 5 + 4 = 19 A


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Therefore, the current throug

total current is 19 A.

Question 3.5:

At room temperature (27.0 temperature of the element i

temperature coefficient of th

Answer

Room temperature, T= 27C

Resistance of the heating ele

Let T1 is the increased tempe

Resistance of the heating ele

Temperature co-efficient of t

Therefore, at 1027C, the res

Question 3.6:

h each resister is 10 A, 5 A, and 4 A respectivel

) the resistance of a heating element is 100 .the resistance is found to be 117 , given that

material of the resistor is

ent at T,R = 100

rature of the filament.

ent at T1,R1 = 117

he material of the filament,

istance of the element is 117.

y and the

hat is thehe


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A negligibly small current is

section 6.0 107

m2, and its

the material at the temperatu

Answer

Length of the wire, l=15 m

Area of cross-section of the

Resistance of the material of

Resistivity of the material of

Resistance is related with the

Therefore, the resistivity of t

Question 3.7:

A silver wire has a resistance

Determine the temperature c

Answer

Temperature, T1 = 27.5C

Resistance of the silver wire

passed through a wire of length 15 m and unifo

resistance is measured to be 5.0 . What is the

e of the experiment?

ire, a = 6.0 107

m2

the wire,R = 5.0

the wire =

resistivity as

e material is 2 107

m.

of 2.1 at 27.5 C, and a resistance of 2.7 a

efficient of resistivity of silver.

at T1,R1 = 2.1

m cross-

resistivity of

t 100 C.


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Temperature, T2 = 100C

Resistance of the silver wire

Temperature coefficient of si

It is related with temperature

Therefore, the temperature c

Question 3.8:

Aheating element using nich

3.2 A which settles after a fe

temperature of the heating el

coefficient of resistance of ni

1.70 104

C1

.

Answer

Supply voltage, V= 230 V

Initial current drawn,I1 = 3.

Initial resistance =R1, which

Steady state value of the curr

Resistance at the steady state

at T2,R2 = 2.7

lver =

and resistance as

efficient of silver is 0.0039C1

.

ome connected to a 230 V supply draws an init

seconds toa steady value of 2.8 A. What is th

ement if the room temperature is 27.0 C? Tem

chrome averaged over the temperature range in

A

is given by the relation,

ent,I2 = 2.8 A

=R2, which is given as

ial current of

steady

erature

olved is


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Temperature co-efficient of

Initial temperature of nichro

Study state temperature reac

T2 can be obtained by the rel

Therefore, the steady temper

Question 3.9:

Determine the current in eac

Answer

Current flowing through vari

ichrome, = 1.70 104

C1

e, T1= 27.0C

ed by nichrome = T2

tion for ,

ature of the heating element is 867.5C

branch of the network shown in fig 3.30:

ous branches of the circuit is represented in the given figure.


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I1 = Current flowing through the outer circuit

I2 = Current flowing through branch AB

I3

= Current flowing through branch AD

I2 I4 = Current flowing through branch BC

I3 +I4 = Current flowing through branch CD

I4 = Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,

10I2 + 5I4 5I3 = 0

2I2 +I4 I3 = 0

I3 = 2I2 +I4 (1)

For the closed circuit BCDB, potential is zero i.e.,

5(I2 I4) 10(I3 +I4) 5I4 = 0

5I2 + 5I4 10I3 10I4 5I4 = 0

5I2 10I3 20I4 = 0

I2 = 2I3 + 4I4 (2)

For the closed circuit ABCFEA, potential is zero i.e.,

10 + 10 (I1) + 10(I2) + 5(I2 I4) = 0

10 = 15I2 + 10I1 5I4


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3I2 + 2I1 I4 = 2 (3)

From equations (1) and (2), we obtain

I3 = 2(2I3 + 4I4) +I4

I3 = 4I3 + 8I4 +I4

3I3 = 9I4

3I4 = +I3 (4)

Putting equation (4) in equation (1), we obtain

I3 = 2I2 +I4

4I4 = 2I2

I2 = 2I4 (5)

It is evident from the given figure that,

I1 =I3 +I2 (6)

Putting equation (6) in equation (1), we obtain

3I2 +2(I3 +I2) I4 = 2

5I2 + 2I3 I4 = 2 (7)

Putting equations (4) and (5) in equation (7), we obtain

5(2I4) + 2( 3I4) I4 = 2

10I4 6I4 I4 = 2

17I4 = 2

Equation (4) reduces to

I3 = 3(I4)


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Therefore, current in branch

In branch BC =

In branch CD =

In branch AD

In branch BD =

Total current =

Question 3.10:

In a metre bridge [Fig. 3.27],when the resistor Y is of 12.5

between resistors in a Wheat

Determine the balance point

What happens if the galvano

bridge? Would the galvanom

the balance point is found to be at 39.5 cm fro. Determine the resistance ofX. Why are the

stone or meter bridge made of thick copper strip

of the bridge above ifX and Y are interchanged.

eter and cell are interchanged at the balance p

eter show any current?

the endA,connections

s?

int of the


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Answer

A metre bridge with resistorsXand Yis represented in the given figure.

Balance point from end A, l1 = 39.5 cm

Resistance of the resistor Y= 12.5

Condition for the balance is given as,

Therefore, the resistance of resistorXis 8.2 .

The connection between resistors in a Wheatstone or metre bridge is made of thick copper

strips to minimize the resistance, which is not taken into consideration in the bridge

formula.

IfX and Yare interchanged, then l1 and 100l1 get interchanged.

The balance point of the bridge will be 100l1 from A.

100l1 = 100 39.5 = 60.5 cm

Therefore, the balance point is 60.5 cm from A.


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When the galvanometer and

galvanometer will show no d

galvanometer.

Question 3.11:

A storage battery of emf8.0

dc supply using a series resis

during charging? What is the

Answer

Emf of the storage battery,E


DC supply voltage, V= 120

Resistance of the resistor,R

Effective voltage in the circu

R is connected to the storage

V1

= VE

V1

= 120 8 = 112 V

Current flowing in the circui

Voltage across resistorR giv

DC supply voltage = Termin

Terminal voltage of battery

cell are interchanged at the balance point of the

eflection. Hence, no current would flow throug

and internal resistance 0.5 is being charged

tor of 15.5 . What is the terminal voltage of th

purpose of having a series resistor in the chargi

= 8.0 V

ery, r= 0.5

15.5

it = V1

battery in series. Hence, it can be written as

=I, which is given by the relation,

en by the product,IR = 7 15.5 = 108.5 V

al voltage of battery + Voltage drop acrossR

120 108.5 = 11.5 V

bridge, the

the

by a 120 V

e battery

ng circuit?


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15/30

Number density of free electcopper wire, l= 3.0 m

Area of cross-section of the

Current carried by the wire,

I = nAeVd

Where,

e = Electric charge = 1.6 1

Vd = Drift velocity

Therefore, the time taken by

2.7 104

s.

Question 3.14:

The earths surface has a neg

difference of 400 kV betwee

low conductivity of the loweglobe. If there were no mech

time (roughly) would be req

practice because there is a m

thunderstorms and lightning

m.)

Answer

ons in a copper conductor, n = 8.5 1028

m3

L

ire,A = 2.0 106

m2

= 3.0 A, which is given by the relation,

19

C

an electron to drift from one end of the wire to

ative surface charge density of 109

C m2

. The

the top of the atmosphere and the surface resu

atmosphere) in a current of only 1800 A over tanism of sustaining atmospheric electric field, h

ired to neutralise the earths surface? (This nev

chanism to replenish electric charges, namely t

in different parts of the globe). (Radius of earth

ength of the

he other is

potential

lts (due to the

he entireow much

er happens in

he continual

= 6.37 106


16/30

Surface charge density of the

Current over the entire globe

Radius of the earth, r= 6.37

Surface area of the earth,

A = 4r2

= 4 (6.37 106)2

= 5.09 1014 m2

Charge on the earth surface,

q = A

= 109

5.09 1014

= 5.09 105

C

Time taken to neutralize the

Current,

Therefore, the time taken to

Question 3.15:

Six lead-acid type of second

are joined in series to provid

drawn from the supply and it

earth, = 109

C m2

,I= 1800 A

106

m

arths surface = t

eutralize the earths surface is 282.77 s.

ry cells each of emf 2.0 V and internal resistan

a supply to a resistance of 8.5 . What are the

s terminal voltage?

e 0.015

current


17/30

A secondary cell after long u

. What maximum current c

motor of a car?

Answer

Number of secondary cells,

Emf of each secondary cell,

Internal resistance of each ce

series resistor is connected t

Resistance of the resistor,R

Current drawn from the supp

Terminal voltage, V=IR = 1.

Therefore, the current drawn

11.87 A.

After a long use, emf of the s

Internal resistance of the cell

Hence, maximum current

Therefore, the maximum cur

required to start the motor of

se has an emf of 1.9 V and a large internal resis

n be drawn from the cell? Could the cell drive

= 6

= 2.0 V

ll, r= 0.015

the combination of cells.

8.5

ly =I, which is given by the relation,

.39 8.5 = 11.87 A

from the supply is 1.39 A and terminal voltage

econdary cell,E= 1.9 V

, r= 380

ent drawn from the cell is 0.005 A. Since a larg

a car, the cell cannot be used to start a motor.

ance of 380

he starting

is

e current is


18/30

Question 3.16:

Two wires of equal length, one of aluminium and the other of copper have the same

resistance. Which of the two wires is lighter? Hence explain why aluminium wires are

preferred for overhead power cables. (Al = 2.63 108

m, Cu = 1.72 108

m,

Relative density of Al = 2.7, of Cu = 8.9.)

Answer

Resistivity of aluminium, Al = 2.63 108

m

Relative density of aluminium, d1 = 2.7

Let l1be the length of aluminium wire and m1be its mass.

Resistance of the aluminium wire =R1

Area of cross-section of the aluminium wire =A1

Resistivity of copper, Cu = 1.72 108

m

Relative density of copper, d2 = 8.9

Let l2be the length of copper wire and m2be its mass.

Resistance of the copper wire =R2

Area of cross-section of the copper wire =A2

The two relations can be written as

It is given that,


19/30


20/30

Current

A

Volta

V

0.2 3.94

0.4 7.87

0.6 11.8

0.8 15.7

1.0 19.7

2.0 39.4

Answer

It can be inferred from the gi

which is equal to 19.7. Henc

Ohms law. According to Oh

the conductor. Hence, the res

Question 3.18:

Answer the following questi

A steady current flows in a

these quantities is constant al

drift speed?

Is Ohms law universally ap

If not, give examples of ele

A low voltage supply from

resistance. Why?

A high tension (HT) supply

Why?

Answer

e Current

A

Voltage

V

3.0 59.2

4.0 78.8

5.0 98.6

6.0 118.5

7.0 138.2

8.0 158.0

ven table that the ratio of voltage with current i

, manganin is an ohmic conductor i.e., the allo

ms law, the ratio of voltage with current is the

istance of manganin is 19.7 .

ns:

etallic conductor of non-uniform cross- section

ong the conductor: current, current density, ele

licable for all conducting elements?

ents which do not obey Ohms law.

hich one needs high currents must have very lo

f, say, 6 kV must have a very large internal resi

a constant,

obeys

resistance of

. Which of

tric field,

internal

stance.


21/30


22/30


23/30

Hence, maximum resistance of the combination,R1 = nR

When n resistors are connected in parallel, the effective resistance (R2) is the minimum,

given by the ratio .

Hence, minimum resistance of the combination,R2 =

The ratio of the maximum to the minimum resistance is,

The resistance of the given resistors is,

R1 = 1 ,R2 = 2 ,R3 = 3 2

Equivalent resistance,

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,


Consider the following combination of the resistors.


Equivalent resistance,R= 6


24/30

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by the sum,

R = 1 + 2 + 3 = 6


Consider the series combination of the resistors, as shown in the given circuit.


(a) It can be observed from the given circuit that in the first small loop, two resistors of

resistance 1 each are connected in series.

Hence, their equivalent resistance = (1+1) = 2

It can also be observed that two resistors of resistance 2 each are connected in series.

Hence, their equivalent resistance = (2 + 2) = 4 .

Therefore, the circuit can be redrawn as

It can be observed that 2 and 4 resistors are connected in parallel in all the four

loops. Hence, equivalent resistance (R) of each loop is given by,

The circuit reduces to,


25/30

All the four resistors are con

Hence, equivalent resistance

It can be observed from the

connected in series.

Hence, equivalent resistance

= 5R

Question 3.21:

Determine the current drawn

infinite network shown in Fi

Answer

The resistance of each resist

Equivalent resistance of the

The network is infinite. Hen

ected in series.

of the given circuit is

iven circuit that five resistors of resistanceR ea

of the circuit =R +R +R +R +R

from a 12 V supply with internal resistance 0.5

. 3.32. Each resistor has 1 resistance.

r connected in the given circuit,R = 1

iven circuit =R

e, equivalent resistance is given by the relation,

ch are

by the


26/30

Negative value ofRcannot

Internal resistance of the circ

Hence, total resistance of the

Supply voltage, V= 12 V

According to Ohms Law, c

3.72 A

Question 3.22:

Figure 3.33 shows a potentio

maintaining a potential drop

a constant emf of 1.02 V (for

point at 67.3 cm length of th

cell, a very high resistance o

balance point. The standard

balance point found similarl

e accepted. Hence, equivalent resistance,

uit, r= 0.5

given circuit = 2.73 + 0.5 = 3.23

rrent drawn from the source is given by the rati

meter with a cell of 2.0 V and internal resistanc

across the resistor wire AB. A standard cell whi

very moderate currents up to a few mA) gives

wire. To ensure very low currents drawn from

600 k is put in series with it, which is shorte

ell is then replaced by a cell of unknown emf

, turns out to be at 82.3 cm length of the wire.

, =

e 0.40

ch maintains

balance

the standard

close to the

and the


27/30

What is the value ?

What purpose does the high resistance of 600 k have?

Is the balance point affected by this high resistance?

Is the balance point affected by the internal resistance of the driver cell?

Would the method work in the above situation if the driver cell of the

potentiometer had an emf of 1.0 V instead of 2.0 V?

(f ) Would the circuit work well for determining an extremely small emf, say of the order

of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify

the circuit?

Answer

Constant emf of the given standard cell,E1 = 1.02 V

Balance point on the wire, l1 = 67.3 cm

A cell of unknown emf, ,replaced the standard cell. Therefore, new balance point on the

wire, l= 82.3 cm

The relation connecting emf and balance point is,

The value of unknown emfis 1.247 V.

The purpose of using the high resistance of 600 k is to reduce the current through thegalvanometer when the movable contact is far from the balance point.

The balance point is not affected by the presence of high resistance.

The point is not affected by the internal resistance of the driver cell.


28/30

The method would not work

instead of 2.0 V. This is beca

than the emf of the other cell

The circuit would not work

would be unstable, the balan

large percentage of error.

The given circuit can be mo

The potential drop across A

error would be small.

Question 3.23:

Figure 3.34 shows a potentio

point with a standard resistor

unknown resistanceX is 68.5failed to find a balance point

Answer

Resistance of the standard re

Balance point for this resista

Current in the potentiometer

Hence, potential drop across

Resistance of the unknown r

if the driver cell of the potentiometer had an e

use if the emf of the driver cell of the potentio

, then there would be no balance point on the w

ell for determining an extremely small emf. As

e point would be close to end A. Hence, there

ified if a series resistance is connected with the

is slightly greater than the emf measured. The

meter circuit for comparison of two resistances.

R = 10.0 is found to be 58.3 cm, while that

cm. Determine the value ofX. What might youwith the given cell of emf ?

sistor,R = 10.0

ce, l1 = 58.3 cm

wire = i

R,E1 = iR

sistor =X

f of 1.0 V

eter is less

re.

the circuit

ould be a

wire AB.

ercentage

The balance

ith the

do if you


29/30

Balance point for this resisto

Hence, potential drop across

The relation connecting emf

Therefore, the value of the u

If we fail to find a balance p

acrossR andXmust be redu

potential drop acrossR orXi

wire AB, a balance point is o

Question 3.24:

Figure 3.35 shows a 2.0 V p

of a 1.5 V cell. The balance

9.5 is used in the external

of the potentiometer wire. D

Answer

, l2 = 68.5 cm

X,E2 = iX

and balance point is,

known resistance,X, is 11.75 .

int with the given cell of emf, , then the poten

ed by putting a resistance in series with it. Onl

s smaller than the potential drop across the pote

btained.

tentiometer used for the determination of inter

oint of the cell in open circuit is 76.3 cm. Whe

ircuit of the cell, the balance point shifts to 64.

termine the internal resistance of the cell.

ial drop

if the

ntiometer

al resistance

a resistor of

cm length


30/30

Internal resistance of the cell

Balance point of the cell in o

An external resistance (R) is

New balance point of the cir

Current flowing through the

The relation connecting resis

Therefore, the internal resist

= r

pen circuit, l1 = 76.3 cm

connected to the circuit withR = 9.5

uit, l2 = 64.8 cm

circuit =I

tance and emf is,

nce of the cell is 1.68.

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3-Current-Electricity.pdf