a

b

e

R

C

II

R

e

RI I

rV

Resistors in Parallel

a

d

I

I

R1 R2

I1 I2V

Ia

dI

RV

• But current through R1 is not I ! Call it I1. Similarly, R2 «I2.

KVL Þ1 1 0V I R

2 2 0V I R

• What to do? IRV • Very generally, devices in parallel

have the same voltage drop

• How is I related to I 1 & I 2 ?

Current is conserved!

21 III

Þ21 R

V

R

V

R

V Þ

21

111

RRR

Another (intuitive) way…

Consider two cylindrical resistors with cross-sectional areas A1 and A2

11 A

LR

22 A

LR

V R1

R2

A1A2

21 AA

LReffective

Put them together, side by side … to make one “fatter”one,

21

21 111

RRL

A

L

A

Reffective

Þ21

111

RRR

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

50W

20W 80W

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

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

1A

• 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

50W

20W 80W

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

(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

50W

20W 80W

1A

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

)80()20( 21 II 21 4II

Kirchhoff’s Second Rule“Junction Rule” or “Kirchhoff’s Current Law (KCL)”

• In deriving the formula for the equivalent resistance of 2 resistors in parallel, we applied Kirchhoff's Second Rule (the junction rule).

"At any junction point in a circuit where the current can divide (also called a node), the sum of the currents into the node must equal the sum of the currents out of the node."

• This is just a statement of the conservation of charge at any given node.

outin II

• The currents entering and leaving circuit nodes are known as “branch currents”.

• Each distinct branch must have a current, Ii assigned to it

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

Old Preflight

How to use Kirchhoff’s LawsA two loop example:

•Analyze the circuit and identify all circuit nodes and use KCL.

(2) e1 - I1R1 - I2R2 = 0(3) e1 - I1R1 - e2 - I3R3 = 0(4) I2R2 - e2 - I3R3 = 0

e 1

e 2

R1

R3R2

I1 I2

I3

(1) I1 = I2 + I3

• Identify all independent loops and use KVL.

How to use Kirchoff’s Laws

e 1

e 2

R1

R3R2

I1 I2

I3

•Solve the equations for I1, I2, and I3:

First find I2 and I3 in terms of I1 :

1 1 2 1 11 1

2 3 2 3

( )R R

I IR R R R

1 1 2

2 31

1 1

2 3

1

R RI

R RR R

Þ

2 1 1 1 2( ) /I I R R

3 1 2 1 1 3( ) /I I R R

From eqn. (2)

From eqn. (3)

Now solve for I1 using eqn. (1):

Let’s plug in some numbers

e 1

e 2

R1

R3R2

I1 I2

I3

e1 = 24 V e 2 = 12 V R1= 5 W R2=3 W R3=4W

Then I1=2.809 A and I2= 3.319 A

and I3= -0.511 A

See Appendix for a more complicated example, with three loops.

The sign means that the direction of I3 is opposite to what’s shown in the circuit

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: (for further discussion see online “tutorial essay”)

loop

nV 0

outin II

Batteries(“Nonideal” = cannot output arbitrary current)

e

RI I

rV

• Parameterized with "internal resistance"

IrV

0Ir IR

rRI

Þ

rR

RV

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

eR

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 2P V V R V R or you can write it as

Power Transmission• Why do we use “high tension” lines to transport power?

– Note: for fixed input power and line resistance, the inefficiency µ 1/V2

Example: Quebec to Montreal 1000 km Þ R= 220W suppose Pin = 500 MW

With Vin=735kV, e = 80%.The efficiency goes to zero quickly if Vin were lowered!

2

21 1out in in in

inin in in in

P IV I R V P RIR

P IV V V V

Keep R small

Make Vin big

–Transmission of power is typically at very high voltages (e.g., ~500 kV)

• But why? –Calculate ohmic losses in the transmission lines–Define efficiency of transmission:

Why do we use AC (60 Hz)? Easy to generate high voltage (water/steam → turbine in magnetic field → induced EMF) [Lecture 16]

We can use transformers [Lecture 18] to raise the voltage for transmission and lower the voltage for use

Resistor- capacitor circuits

e

R

C

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

Recall voltage “drop” on C?

C

QV

Write KVL: 0Q

IRC

What’s wrong here?

dt

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

KVL gives Differential Equation ! 0C

Q

dt

dQR

We will solve this next time. For now, look at qualitative behavior…

Capacitors Circuits, Qualitative

• Charging (it takes time to put the final charge on)– Initially, the capacitor behaves like a wire (DV = 0, since Q = 0).– As current starts to flow, charge builds up on the capacitor

it then becomes more difficult to add more charge

the current slows down– 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.

Basic principle: Capacitor resists rapid change in Q

resists rapid changes in V

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+ = e /2R (c) I0+ = 2 e /R

a

b

e

R

C

II

R

2A

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

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

2B

• 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+ = e /2R (c) I0+ = 2 e /R

a

b

e

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+, e -IR - 0 - IR = 0 Þ I = e /2R

• 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+ = e /2R (c) I0+ = 2 e /R

(a) I¥ = 0 (b) I¥ = e /2R (c) I¥ > 2 e /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 e ; the current goes to 0.

a

b

e

R

C

II

R

Summary• Kirchhoff’s Laws

– KCL: Junction Rule (Charge is conserved)

– Review KVL (V is independent of path)

• Non-ideal Batteries & Power

– Effective “internal resistance” limits current

– Power generated ( ) = Power dissipated ( )

– Power transmission most efficient at low current high voltage

• Resistor-Capacitor Circuits

– Capacitors resist rapid changes in Q resist changes in V

2 2I R V RI V

– 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 S DVi = 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….