Lecture 7 Circuits Ch. 27

• Cartoon -Kirchhoff's Laws

• Warm-up problems

• Topics

– Direct Current Circuits

– Kirchhoff's Two Rules

– Analysis of Circuits Examples

– Ammeter and voltmeter

– RC circuits

• Demos

– Three bulbs in a circuit

– Power loss in transmission lines

– Resistivity of a pencil

– Blowing a fuse

Transmission line demo

Direct Current Circuits

1. The sum of the potential drops

around a closed loop is zero.

This follows from energy

conservation and the fact that

the electric field is a

conservative force.

2. The sum of currents into any

junction of a closed circuit must

equal the sum of currents out of

the junction. This follows from

charge conservation.

Example (Single Loop Circuit)

No junction so we don’t need that rule.

How do we apply Kirchoff’s rule?

Must assume the direction of the current –

assume clockwise.

Choose a starting point and apply Ohm’s Law as

you go around the circuit.

a. Potential across resistors is negative

b. Sign of E for a battery depends on assumed

current flow

c. If you guessed wrong on the sign, your

answer will be negative

Start in the upper left hand corner.

21321

21

1132221 0

rrRRR

EEi

irEiRirEiRiR

++++

!=

=!+!!!!!

21321

21

rrRRR

EEi

++++

!=

Now let us put in numbers.

VE

VE

rr

RRR

5

10

1

10

2

1

21

321

=

=

!==

!===Suppose:

32

5

11101010

510=

!++++

"=

Vi amp

VE

VE

10

5

2

1

=

=Suppose:

32

5

32

)105( !=

"

!=

Vi amp

We get a minus sign. It means our

assumed direction of current must be

reversed.

Note that we could have simply added all

resistors and get the Req. and added the EMFs

to get the Eeq. And simply divided.

32

5

)(32

)(5

R .e

.=

!==

VEi

q

eq

amp

Sign of EMF

Battery 1 current flows from - to + in battery +E1

Battery 2 current flows from + to - in battery -E2

In 1 the electrical potential energy increases

In 2 the electrical potential energy decreases

Example with numbers

Quick solution:

AE

I

R

VVVVE

q

eq

i

i

i

i

16

10

R

16

102412

.e

.

6

1

3

1

==

!=

=+"=

#

#

=

=

Question: What is the current in the circuit?

AampsV

i

iV

625.0625.016

10

0)311551()2412(

==!

=

=!+++++"+"+

Write down Kirchoff’s loop equation.

Loop equation

Assume current flow is clockwise.

Do the batteries first – Then the current.

Example with numbers (continued)

Question: What are the terminal voltages of each battery?

V375.111A625.0V12irV =!"#=#= $

V375.11A625.0V2irV =!"#=#= $

V625.41A625.0V4irV =!"+=#= $

2V:

4V:

12V:

Multiloop Circuits

Kirchoff’s Rules

1. in any loop

2. at any junction

0=!i

iV

!! =outinii

Find i, i1, and i2

Rule 1 – Apply to 2 loops (2 inner loops)

a.

b.

Rule 2

a.

0452

03412

12

1

=+!!

=!!

ii

ii

21iii +=

We now have 3 equations with 3

unknowns.

0245

03712

0)(3412

21

211

21

=!+!

=!!

=+!!

ii

ii

iii

061215

061424

21

21

=!+!

=!!

ii

ii

Ai

Ai

Ai

i

0.2

5.0

5.126

39

02639

2

1

1

=

=

==

=!

multiply by 2

multiply by 3

subtract them

Find the Joule

heating in each

resistor P=i2R.

Is the 5V battery

being charged?

Method of determinants for solving simultaneous equations

5240

12043

0

21

1

21

=!+

!=+!!

=!!

ii

ii

iii

Cramer’s Rule says if :

3332313

2322212

1312111

dicibia

dicibia

dicibia

=++

=++

=++

Then,

333

222

111

333

222

111

1

cba

cba

cba

cbd

cbd

cbd

i =

333

222

111

333

222

111

2

cba

cba

cba

cda

cda

cda

i =

333

222

111

333

222

111

3

cba

cba

cba

dba

dba

dba

i =

i =

0 !1 !1

!12 !4 0

5 +4 !2

1 !1 !1

!3 !4 0

0 +4 !2

=

0!4 0

4 !2

"#$

%&' !1

0 !12

!2 5

"#$

%&' !1

!12 !4

5 4

"#$

%&'

1

!4 0

4 !2

"#$

%&' !1

0 !3

!2 0

"#$

%&' !1

!3 !4

0 4

"#$

%&'

=

24 + 48 ! 20

8 + 6 + 12

=52

26

= 2A

You try it for i1 and i2.

See inside of front cover in your book on how to use Cramer’s Rule.

For example solve for i

Method of determinants using Cramers Rule and cofactors

Also use this to remember how to evaluate cross products

of two vectors.

Another example

Find all the currents including directions.

21

121

1

3580

23380

234480

ii

iii

iiVVV

!!=

!!!=

!!!++=

012120

010166

0246

1

12

12

=+!

=!+!

=++!

i

ii

ii

0)1(246 2 =++! Ai

Loop 1

Loop 2

Ai 11 =

Ai

Ai

2

12

=

=

Loop 1

Loop 2

i i

i

i

i1

i2

i2

21iii +=

024612=++! ii

Multiply eqn of loop 1 by

2 and subtract from the

eqn of loop 2

Rules for solving multiloop circuits

1. Replace series resistors or batteries with their equivalent values.

2. Choose a direction for i in each loop and label diagram.

3. Write the junction rule equation for each junction.

4. Apply the loop rule n times for n interior loops.

5. Solve the equations for the unknowns. Use Cramer’s Rule if

necessary.

6. Check your results by evaluating potential differences.

3 bulb question

The circuit above shows three identical light bulbs attached to an ideal

battery. If the bulb#2 burns out, which of the following will occur?

a) Bulbs 1 and 3 are unaffected. The total light emitted by the circuit decreases.

b) Bulbs 1 and 3 get brighter. The total light emitted by the circuit is unchanged.

c) Bulbs 1 and 3 get dimmer. The total light emitted by the circuit decreases.

d) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit is

unchanged.

e) Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit is

unchanged.

f) Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit

decreases.

g) Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit

decreases.

h) Bulb 1 is unaffected, but bulb 3 gets brighter. The total light emitted by the circuit

increases.

i) None of the above.

When the bulb #2 is not burnt out:

R2

3

2

RRReq =+=

R3

V2

R

VI

2

31==

RIP,Power2

=R

VI =

For Bulb #1

For Bulb #2

For Bulb #3

R

V44.

R9

V4RIP

22

2

11===

R3

V

2

II

1

2==

R

V11.

R9

VRIP

22

2

22===

R3

V

2

II

1

3==

R

V11.

R9

VRIP

22

2

33===

1I

2

II

1

2= 2

II

1

3=

R2RRReq =+=

R2

VI1=

RIP,Power2

=R

VI =

For Bulb #1

For Bulb #2

For Bulb #3

R

V25.

R4

VRIP

22

2

11===

0I2= 0RIP

2

22==

R2

VII13==

R

V25.

R4

VRIP

22

2

33===

1I

13II =

So, Bulb #1 gets dimmer and bulb #3 gets

brighter. And the total power decreases.

f) is the answer.

Before total power wasR

V66.

R

V

R

VP

2

23

2

eq

2

b ===

R

V50.

R2

V

R

VP

22

eq

2

a ===After total power is

When the bulb #2 is burnt out:

How does a capacitor behave in a circuit with a resistor?

Charge capacitor with 9V battery with switch open, then remove battery.

Now close the switch. What happens?

Discharging a capacitor through a resistor

V(t)

Potential across capacitor = V =

just before you throw switch at time t = 0.

Potential across Resistor = iR

at t > 0.

RC

QiRi

C

Q ooo

o=!=

C

Qo

What is the current I at time t?

RC

)t(Q)t(i =

RC

Qi =or

dt

dQi but ,

RC

Qi ,So !==

RC

dt

Q

dQ

RC

Q

dt

dQ

=!

=!

Integrating both the sides

! !="RC

dt

Q

dQ

ARC

tQln +=!

ARC

tA

RC

t

eeeQ ,So!

!!!

==

ARC

tQln !!=

At t=0, Q=Q0

RC

t

0eQQ

!

="

Time constant =RC

Q

0Q

RCt =

7.2

Q

e

Q=

t

What is the current I at time t?

AA

RC

0

0 eeQ ,So!

!!

==

RC

t

0eQQ

!

=

dt

dQi = Ignore - sign

i

tRC

R

V0

What is the current?

RC

t

0RC

t

0 eR

Ve

RC

Q !!

!=!=

How the charge on a capacitor varies with time as it is being charged

What about charging the capacitor?

t

Q

i

t

00QCV =

Same as before

)e1(CVQ RC

t

0

!

!=

!"

=

t

0e

R

Vi

Note that the current is zero when either the

capacitor is fully charged or uncharged. But the

second you start to charge it or discharge it, the

current is maximum.

Instruments

Galvanometers: a coil in a magnetic field that senses current.

Ammeters: measures current.

Voltmeter: measures voltage.

Ohmmeters: measures resistance.

Multimeters: one device that does all the above.

Galvanometer is a needle mounted to a coil that rotates in a magnetic field.

The amount of rotation is proportional to the current that flows

through the coil.

Symbolically we write

gRUsually when != 20Rg

milliAmp5.00Ig !=

Ohmmeter

gs RRR

Vi

++=

Adjust Rs so when R=0 the

galvanometer read full scale.

Ammeter

V10A5

!2

gI

gR

sR

sI

A5I = A5I =!

A5III sg =+=

ssgg RIRI =

The idea is to find the value of RS that will give a full

scale reading in the galvanometer for 5A

A5A0005.A5I So, ,A105.0I and 20R s

3

gg !"=#=$="

!=!"

==

#

002.0)20(A5

A105.0R

I

IR ,So

3

gs

g

s

Ammeters have very low resistance when put in series in a circuit.

You need a very stable shunt resistor.

Very small

Voltmeter

Use the same galvanometer to construct a voltmeter for which full scale reading

in 10 Volts.

!V10 !2

sR

gRI

!=

"=#

20R

A105.0I

g

3

g

What is the value of RS now?

)RR(IV10 gsg +=

A105

V10

I

V10RR

4g

gs !"

==+

!=+ 000,20RR gs

!= 980,19Rs

So, the shunt resistor needs to

be about 20K!.

Note: the voltmeter is in

parallel with the battery.

We need