Lecture 7 Circuits Chp. 28

• Cartoon -Kirchoff’s Laws• Opening Demo- transmission lines• Physlet • Topics

– Direct Current Circuits– Kirchoff’s Two Rules– Analysis of Circuits Examples– Ammeter and voltmeter– RC circuits

• Demos– Ohms Law– Power loss in transmission lines– Resistivity of a pencil– Blowing a fuse

• Warm-up problems

Transmission line demo

Direct Current Circuits

1. The sum of the potential charges 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

++++−

=

Put in numbers.

VE

VE

rr

RRR

5

10

1

10

2

1

21

321

==

Ω==Ω===Suppose:

32

5

11101010

510=

++++−

=i amp

VE

VE

10

5

2

1

==Suppose:

32

5

32

105 −=

−=i

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?

Write down Kirchoff’s loop equation.

Loop equation

Assume current flow is clockwise.

Do the batteries first – Then the current. AampsV

i

i

625.0625.016

10

0)311551()2412(

==Ω

=

=+++++−+−+

Example with numbers (continued)

Question: What are the terminal voltages of each battery?

12V:

2V:

4V:

€

V =ε − ir =12V − 0.625A • 1Ω =11.375V

V =ε − ir = 2V − 0.625A • 1Ω =1.375V

V =ε − ir = 4V + 0.625A • 1Ω = 4.625V

Multiloop Circuits

Kirchoff’s Rules1. in any loop

2. at any junction

0=∑i

iV

∑∑ = outin ii

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

21 iii +=

We now have 3 equations with 3 unknowns.

0245

03712

0)(3412

21

21

211

=−+−=−−

=+−−

iiii

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

=−+−=+−−

=−−

iiiiiii

For example solve for i

Ai 226

52

6128

242048

240

043

111

245

0412

110

==+++−

=

−+−−

−−−+

−−−−

=

You try it for i1 and i2.

See Appendix in your book on how to use Cramer’s Rule.

Another example

Find all the currents including directions.

21

121

1

3580

23380

234480

ii

iii

iiVVV

−−=−−−=

−−−++=

012120

010166

0246

1

12

12

=+−=−+−

=++−

iii

ii 0)1(2426 =++− AiLoop 1 Loop 2

Ai 11 =Ai

Ai

2

12

==

multiply by 2

i = i1+ i2

Loop 1

Loop 2

i i

i

i

i1i2

i2

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.

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?

€

i(t) =Q(t)

RC

Time constant = RC

What is the current?

€

Q =Q0e−t

RC

€

i =dQ

dt= −

Q0

RCe

−t

RC = −V0

Re

−t

RCIgnore - sign

RC

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

Ohmmeter

Ammeter

Voltmeter

Warm up set 7Warm up set 7 Due 8:00 am Tuesday 1. HRW6 28.TB.05. [119859] In the context of the loop and junctions rules for electrical circuits a junction is:

where a wire is connected to a battery where three or more wires are joined where a wire is bent where a wire is connected to a resistor where only two wires are joined

2. HRW6 28.TB.18. [119872] Two wires made of the same material have the same length but different diameter. They are connected in parallel to a battery. The quantity that is NOT the same for the wires is:

the electric field the electron drift velocity the current the current density the end-to-end potential difference

3. HRW6 28.TB.26. [119880] The emf of a battery is equal to its terminal potential difference:

only when there is no current in the battery only when a large current is in the battery under all conditions under no conditions only when the battery is being charged