Lecture 7 Circuits Ch. 27

• Cartoon -Kirchhoff's Laws• 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

• Elmo

Transmission line demo

Kirchhoff's Laws

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)

How do we apply Kirchhoff’s Rule 1?

• Must assume the direction of the current – assume clockwise.• Choose a starting point and sum all voltages drops around the circuit using

Ohm’s Law

Example (Single Loop Circuit)

21321

21

1132221 0

rrRRR

EEi

irEiRirEiRiR

++++−

=

=−+−−−−−

a. Potential drop across resistors is negative when traversing in the same direction as current.

b. Sign of E for a battery is positive when traversing in the same direction as current and

assumed current direction is negative to positive in the battery.

c. Sign of E for a battery is negative when traversing in the same direction as current and

assumed current direction is positive to negative in the battery.

d. Reverse the sign for b and c when you are traversing the circuit in the opposite direction as the assumed current.

21321

21

rrRRR

EEi

++++−

=

Now let us put in numbers.

VE

VE

rr

RRR

5

10

1

10

2

1

21

321

==

Ω==Ω===Suppose:

325

11101010510 =

Ω++++−= Vi amp

VE

VE

10

5

2

1

==Suppose:

325

32)105( −=

Ω−= 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

.=

Ω==

VEi

q

eqamp

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

==

Ω=

=+−=

∑

∑

=

=

Question: What is the current in the circuit?

AampsVi

iV

625.0625.01610

0)311551()2412(

==Ω

=

=Ω+++++−+−+

Write down Kirchhoff’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 =Ω⋅−=−=εV375.11A625.0V2irV =Ω⋅−=−=εV625.41A625.0V4irV =Ω⋅+=−=ε

2V:

4V:

12V:

Multiloop Circuits

Kirchoff’s Rules1. 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)

1.

2.

Rule 2

3.

12 −4i1 −3i =0−2i2 −5 + 4i1 =0

21iii +=

We now have 3 equations with 3 unknowns.

024503712

0)(3412

21

211

21

=−+−=−−

=+−−

iiii

iii

061215061424

21

21

=−+−=−−ii

ii

AiAi

Ai

i

0.25.0

5.12639

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

524012043

0

21

1

21

=−+−=+−−

=−−

iiiiiii

Cramer’s Rule says if :

a1i +b1i1 + c1i2 =d1

a2i +b2i1 + c2i2 =d2

a3i +b3i1 + c3i2 =d3

Then,

i =

d1 b1 c1d2 b2 c2

d3 b3 c3

a1 b1 c1a2 b2 c2

a3 b3 c3

i1 =

a1 d1 c1a2 d2 c2

a3 d3 c3

a1 b1 c1a2 b2 c2

a3 b3 c3

i2 =

a1 b1 d1

a2 b2 d2

a3 b3 d3

a1 b1 c1a2 b2 c2

a3 b3 c3

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 cofactorsAlso use this to remember how to evaluate cross products of two vectors.

Another example

Find all the currents including directions.

+8V + 4V − 4V − 3i − 2i1

= 0

8 − 3i1− 3i

2− 2i

1= 0

012120010166

0246

1

12

12

=+−=−+−

=++−

iii

ii0)1(246 2 =++− Ai

Loop 1

Loop 2

Ai 11 =

Ai

Ai

2

12

==

Loop 1

Loop 2

i i

i

i

i1i2

i2

21iii +=

−6i2

+ 4 + 2i1

= 0

Multiply red eqn of loop 1 by 2 and subtract from the red eqn of loop 2

Start here

8 −5i1 −3i2 =0

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:

R23

2RRR eq =+=

R3V2

RVI

231 ==

RIP,Power 2=RVI =

For Bulb #1

For Bulb #2

For Bulb #3

RV44.

R9V4RIP

22211 ===

R3V

2I

I 12 ==

RV11.

R9VRIP

22222 ===

R3V

2I

I 13 == R

V11.R9

VRIP22

233 ===

1I

2I

I 12 = 2

II 1

3 =

R2RRR eq =+=

R2VI1 =

RIP,Power 2=RVI =

For Bulb #1

For Bulb #2

For Bulb #3

RV25.

R4VRIP

22211 ===

0I2 = 0RIP 222 ==

R2VII 13 == R

V25.R4

VRIP22

233 ===

1I

13 II =

So, Bulb #1 gets dimmer and bulb #3 gets brighter. And the total power decreases.

f) is the answer.

Before total power wasRV66.

RV

RVP

2

23

2

eq

2

b ===

RV50.

R2V

RVP

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..

Remove battery and 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

Just after you throw the switch

RC

QiRi

C

Q ooo

o=⇒=

C

Qo

What is the current I at time t?RC

)t(Q)t(i =

dtdQi but ,

RCQi ,So −==

RCdt

QdQ

RCQ

dtdQ

=−

=−

Integrating both the sides

∫ ∫=−RCdt

QdQ

Q =Q0e−

tRC

Time constant =RC

Q

0Q

RCt =

7.2Q

eQ =

t

What is the charge Q at time t?

RCt

0eQQ−

=

dtdQi = Ignore - sign

i

tRC

RV0

What is the current?

RCt

0RCt

0 eRV

eRCQ −−

−=−=

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

What about charging the capacitor?

t

Q

i

t

00 QCV =

Same as before

)e1(CVQ RCt

0

−−=

γ−

=t

0 eRV

i

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 Ω=20Rg

milliAmp5.00Ig →=

Ohmmeter

gs RRRVi

++= Adjust Rs so when R=0 the

galvanometer read full scale.

Ammeter

V10A5 Ω2

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 s3

gg ≈−=×=Ω= −

Ω=Ω×==−

002.0)20(A5

A105.0RI

IR ,So

3

gs

gs

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 Ω2

sR

gRI

Ω=×= −

20R

A105.0I

g

3g

What is the value of RS now?

)RR(IV10 gsg +=

A105V10

IV10RR

4g

gs −×==+

Ω=+ 000,20RR gs

Ω= 980,19Rs

So, the shunt resistor needs to be about 20K

Note: the voltmeter is in parallel with the battery.

We need

Chapter 27 Problem 19 In Figure 27-34, R1 = 100 , R2 = 30 , and the ideal batteries have emfs script e1 = 6.0 V, script e2 = 5.0 V, script e3 = 3.0 V.

Fig. 27-34(a) Find the current in resistor 1.(b) Find the current in resistor 2.(c) Find the potential difference between points a and b.

Fig. 27-34

Chapter 27 Problem 27 In Figure 27-40, the resistances are R1 = 0.5 , R2 = 1.7 , and the ideal batteries have emfs script e1 = 2.0 V, and script e2 = script e3 = 3 V.

Fig. 27-40(a) What is the current through each battery? (Take upward to be positive.)battery 1battery 2battery 3(b) What is the potential difference Va - Vb?

Chapter 27 Problem 38 A simple ohmmeter is made by connecting a 4.0 V battery in series with a resistance R and an ammeter that reads from 0 to 1.00 mA, as shown in Figure 27-47. Resistance R is adjusted so that when the clip leads are shorted together, the meter deflects to its full-scale value of 1.00 mA.

Fig. 27-47(a) What external resistance across the leads results in a deflection of 10% of full scale?(b) What resistance results in a deflection of 50% of full scale?(c) What resistance results in a deflection of 90% of full scale?(d) If the ammeter has a resistance of 40.0 and the internal resistance of the battery is negligible, what is the value of R?