a

b

R

C

II

R

RI I

rV

Lecture 10, ACT 1

• Consider the circuit shown:

– What is the relation between Va -Vd and Va -Vc ?

(a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc)

12VI1 I2

a

b

d c

50

20 80

(a) I1 < I2 (b) I1 = I2 (c) I1 > I2

1B – What is the relation between I1 and I2? 1B

1A

Lecture 10, ACT 1• Consider the circuit shown:

– What is the relation between Va -Vd and Va -Vc ?

(a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc)

12VI1 I2

a

b

d c

50

20 80

1A

• Do you remember that thing about potential being independent of path?

Well, that’s what’s going on here !!!

(Va -Vd) = (Va -Vc)

Point d and c are the same, electrically

Lecture 10, ACT 1

(a) I1 < I2 (b) I1 = I2 (c) I1 > I2

1B – What is the relation between I1 and I2? 1B

• Consider the circuit shown:

– What is the relation between Va -Vd and Va -Vc ?

(a) (Va -Vd) < (Va -Vc) (b) (Va -Vd) = (Va -Vc) (c) (Va -Vd) > (Va -Vc)

12VI1 I2

a

b

d c

50

20 80

1A

• Note that: Vb -Vd = Vb -Vc • Therefore,

)80()20( 21 II 21 4II

Summary of Simple Circuits

...321 RRRReffective• Resistors in series:

• Resistors in parallel: ...1111

321

RRRReffective

Current thru is same; Voltage drop across is IRi

Voltage drop across is same; Current thru is V/Ri

Kirchhoff’s laws: (Tipler problems on Kirchhoff’’s rules)

loop

nV 0

outin II

Batteries(“Nonideal” = cannot output arbitrary current)

RI I

rV

• Parameterized with "internal resistance"

IrV

0Ir IR

rRI

rR

RV

Internal Resistance DemoAs # bulbs increases, what happens to “R”??

R

I

rV

How big is “r”?

PowerBatteries & Resistors Energy expended

chemical to electrical

to heat

What’s happening?

Assert:

sJpower

time

energyRate is:

Charges per time

VIP Potential difference per charge

Units okay? WattsJ

secondCoulomb

CoulombJoule

For Resistors: RIIIRP 2 RVRVVP 2

or you can write it as

More complex now…

R

C

IILet’s try to add a Capacitor to our simple circuit

Recall voltage “drop” on C?

C

QV

Write KVL:

0dQ Q

Rdt C

KVL gives Differential Equation !

0Q

IRC

What’s wrong here?

dt

dQI Consider that and substitute. Now eqn has only “Q”.

The Big Idea• Previously:

– Analysis of multi-loop circuits with batteries and resistors.

– Main Feature: Currents are attained instantaneously and do not vary with time!!

• Now:– Just added a capacitor to the circuit.

– What changes??

• KVL yields a differential equation with a term proportional to Q and a term proportional to

I = dQ/dt.

0C

Q

dt

dQR

The Big Idea

•Physically, what’s happening is that the final charge cannot be placed on a capacitor instantly.

•Initially, the voltage drop across an uncharged capacitor = 0 because the charge on it is zero ! (V=Q/C)

•As current starts to flow, charge builds up on the capacitor, the voltage drop is proportional

to this charge and increases; it then becomes more difficult to add more charge so the current slows

2

0C

Q

dt

dQR

Lecture 10, ACT 2

• At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged.– What is the value of the current I0+

just after the switch is thrown?

(a) I0+ = 0 (b) I0+ = /2R (c) I0+ = 2/R

a

b

R

C

II

R

2A

(a) I = 0 (b) I = /2R (c) I > 2/R

– What is the value of the current I after a very long time?

2B

Lecture 10, ACT 2

• At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged.– What is the value of the current I0+

just after the switch is thrown?

(a) I0+ = 0 (b) I0+ = /2R (c) I0+ = 2/R

a

b

R

C

II

R

2A

•Just after the switch is thrown, the capacitor still has no charge, therefore the voltage drop across the capacitor = 0!

•Applying KVL to the loop at t=0+, IR 0IR = 0 I = /2R

Lecture 10, ACT 2• At t=0 the switch is thrown from

position b to position a in the circuit shown: The capacitor is initially uncharged.– What is the value of the current I0+

just after the switch is thrown?

2A

(a) I0+ = 0 (b) I0+ = /2R (c) I0+ = 2/R

(a) I = 0 (b) I = /2R (c) I > 2/R

– What is the value of the current I after a very long time?

2B

• The key here is to realize that as the current continues to flow, the charge on the capacitor continues to grow.

• As the charge on the capacitor continues to grow, the voltage across the capacitor will increase.

• The voltage across the capacitor is limited to ; the current goes to 0.

a

b

R

C

II

R

Behavior of Capacitors

• Charging

– Initially, the capacitor behaves like a wire.

– After a long time, the capacitor behaves like an open switch.

• Discharging

– Initially, the capacitor behaves like a battery.

– After a long time, the capacitor behaves like a wire.

Discharging Capacitor

C

a

b+

- -

R+

I I•The capacitor is initially fully charged, Q = Q0. At t = 0 the switch is thrown from position a to position b in the circuit shown.

•From KVL: 0dQ Q

Rdt C

QRCdt

dQ

1

atetQ )(

Q(t) must be an exponential function of the form:

Therefore, aQaedt

dQ at whereRC

a1

From initial condition, Q(0) = Q0, we get:

RCt

eQtQ

0)(

Summary• Kirchhoff’s Laws

– KCL: Junction Rule (Charge is conserved)

– Review KVL (V is independent of path)

• Non-ideal Batteries & Power

• Discharging of capacitor through a Resistor:

Reading Assignment: Chapter 26.6

RC

t

eQtQ

0)(

Examples: 26.17,18 and 19

Two identical light bulbs are represented by the resistors R2 and R3 (R2 = R3 ). The switch S is initially open.

2) If switch S is closed, what happens to the brightness of the bulb R2?

a) It increases b) It decreases c) It doesn’t change

3) What happens to the current I, after the switch is closed ?

a) Iafter = 1/2 Ibefore

b) Iafter = Ibefore

c) Iafter = 2 Ibefore

I

E

R1

R4

R2

R3

Four identical resistors are connected to a battery as shown in the figure.

5) How does the current through the battery change after the switch is closed ?

a) Iafter > Ibefore

b) Iafter = Ibefore

c) Iafter < Ibefore

Before: Rtot = 3R Ibefore = 1/3 E/R

After: R23 = 2R R423 = 2/3 R Rtot = 5/3 R

Iafter = 3/5 E/R

– determine which KCL equations are algebraically independent (not all are in this circuit!)

– I1=I2+I3

– I4=I2+I3

– I4=I5+I6

– I1=I5+I6

Appendix: A three-loop KVL example

• Identify all circuit nodes - these are where KCL eqn’s are found

• Analyze circuit and identify all independent loops where Vi = 0 KVL

I1=I4

I1=I2+I3

I4+I5+I6

I1

I2

I3

I4

I5

I6

A three-loop KVL example

• Now, for Kirchoff’s voltage law: (first, name the resistors)

I1=I4

I1=I2+I3

I4+I5+I6

I1

I2

I3

I4

I5

I6

• Here are the node equations from applying Kirchoff’s current law:

• There are simpler ways of analyzing this circuit, but this illustrates Kirchoff’s laws

R1

R2a

R2b

R3

R4R6

R5

I6R6+I5R5=0

I2R2b+I2R2a- I3R3 =0

VB-I1R1-I2R2a- I2R2b-I4R4-I5R5 = 0Six equations, six unknowns….