a

b

R

C

II

R

RI I

rV

Yesterday• Ohm’s Law V=IR• Ohm’s law isn’t a true law but a good approximation for

typical electrical circuit materials

• Resistivity =1/ (Conductivity): Property of the material

• Resistance proportional to resistivity and length, inversely proportional to area L

RA

Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d.

Compare the resistance of the two cylinders.

a) R1 > R2 b) R1 = R2 c) R1 < R2

Question 1

Question 1

1. a

2. b

3. c

Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d.

Compare the resistance of the two cylinders.

a) R1 > R2 b) R1 = R2 c) R1 < R2

Question 1

• Resistance is proportional to Length/Area

Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d.

If the same current flows through both resistors, compare the average velocities of the electrons in the two resistors:

a) v1 > v2 b) v1 = v2 c) v1 < v2

Question 2

Question 2

1. a

2. b

3. c

Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d.

If the same current flows through both resistors, compare the average velocities of the electrons in the two resistors:

a) v1 > v2 b) v1 = v2 c) v1 < v2

Question 2

Current Area Current Density

Current Density average velocity of electrons

I is the same A1<A2 v1>v2

•By Ohm’s law, the Voltage difference across resistance R1 is

•Across R2 is

•Total voltage difference

• the effective single resistance is

Resistors in Series

a

c

Reffective

a

b

c

R1

R2

I

1IRVV ba

2IRVV cb

)( 21 RRIVV ca

)( 21 RRReffective

•What is the same effective single resistance to two resistances in series?

•Whenever devices are in SERIES, the current is the same through both.

Another (intuitive) way…

•Consider two cylindrical resistors with lengths L1 and L2

V

R1

R2

L2

L1

A

LR 1

1

A

LR 2

2

21 RRR

2121 RR

A

LLReffective

•Put them together, end to end to make a longer one...

The World’s Simplest (and most useful) circuit:Voltage Divider

?V0

2 21 2

VV IR R

R R

02 1

V V=

2R R

2 1 0 V=VR R

2 1 V=0R R

By varying R2 we can controllably adjust the output voltage!

V0

R1

R2

V

Two resistors are connected in series to a battery with emf E. The resistances are such that R1 = 2R2. The currents through the resistors are I1 and I2 and the potential differences across the resistors V1 and V2.

Are:

Question 3

a) I1>I2 and V2=E

b) I1=I2 and V2= E

c) I1=I2 and V2=1/3E

d) I1<I2 and V2=1/2E

e) I1<I2 and V2=1/3E

Resistors in Parallel

a

d

I

I

R1 R2

I1 I2V

Ia

dI

RV

• Current through R1 is I1.

• Current through R2 is I2.

1 1V I R 2 2V I R

• Very generally, devices in parallel have the same voltage drop

• But current is conserved

21 III

21 R

V

R

V

R

V

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,

1 2

1 2

1 1 1

effective

A A

R L L R R

21

111

RRR

Kirchhoff’s First Rule“Loop Rule” or “Kirchhoff’s Voltage Law (KVL)”

"When any closed circuit loop is traversed, the algebraic sum of the changes in potential must equal zero."

loop

nV 0KVL:

• This is just a restatement of what you already know: that the potential difference is independent of path!

R1

R2I

IR1 IR2 0

Rules of the Road

R1

R2I

IR1 IR2

Our convention:•Voltage gains enter with a + sign, and voltage drops

enter with a sign.•We choose a direction for the current and move around

the circuit in that direction.•When a battery is traversed from the negative terminal to

the positive terminal, the voltage increases, and hence the battery voltage enters KVL with a + sign.

•When moving across a resistor, the voltage drops, and hence enters KVL with a sign.

Current in a Loop

b a

d ec

f

R1

I

R2 R3

R4

I

4321

21

RRRRI

1 2 2 3 4 1 0IR IR IR IR

loopnV 0KVL:

Start at point a (could be anywhere) and assume current is in direction shown (could be either)

Question 3

(a) I1 < I0 (b) I1 = I0 (c) I1 > I0

• Consider the circuit shown. – The switch is initially open and the current

flowing through the bottom resistor is I0.

– Just after the switch is closed, the current flowing through the bottom resistor is I1.

– What is the relation between I0 and I1?

R

12V

12V

R

12V

Ia

b

Question 3

1. a

2. b

3. c

Question 3

(a) I1 < I0 (b) I1 = I0 (c) I1 > I0

•From symmetry the potential (Va-Vb) before the switch is closed is Va-Vb = +12V.

• Therefore, when the switch is closed, potential stays

the same and NO additional current will flow!

• Therefore, the current before the switch is closed is equal to the current after the switch is closed.

• Consider the circuit shown. – The switch is initially open and the current

flowing through the bottom resistor is I0.

– Just after the switch is closed, the current flowing through the bottom resistor is I1.

– What is the relation between I0 and I1?

R

12V

12V

R

12V

Ia

b

Question 3• Consider the circuit shown.

– The switch is initially open and the current flowing through the bottom resistor is I0.

– After the switch is closed, the current flowing through the bottom resistor is I1.

– What is the relation between I0 and I1?

(a) I1 < I0 (b) I1 = I0 (c) I1 > I0

• Write a loop law for original loop:

12V I1R = 0

I1 = 12V/R

• Write a loop law for the new loop:

12V +12V I0R I0R = 0

I0 = 12V/R

R

12V

12V

R

12V

Ia

b

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, Iiassigned to it

How to use Kirchhoff’s LawsA two loop example:

•Assume currents in each section of the circuit, identify all circuit nodes and use KCL.

1 I1R1 I2R2 = 0(3) I2R2 2 I3R3 = 04 I1R1I3R3

1

2

R1

R3R2

I1 I2

I3

(1) I1 = I2 + I3

• Identify all independent loops and use KVL.

How to use Kirchoff’s Laws

1

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

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

1

2

R1

R3R2

I1 I2

I3

1 = 24 V 2 = 12 V R1= 5R2=3R3=4

Then, and

I1=2.809 A I2= 3.319 A, I3= -0.511 A

Junction Demo

I1

R

R

RI2

I3

Outside loop:

1 1 3 3 0I R I R

Top loop:

1 1 2 2 0I R I R

Junction:

321 III

RI

3

2 3211

RI

3

2 2312

R

I3

2 3213

1

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