•

Very generally, devices in parallel have the same voltage drop

• •

Current through

R

1 is

I

1 . Current through

R

2 is

I

2 .

 1 1

V



I R

2 2

V a d I I I

1

R

1

I a

•

But current is conserved

I



I

1 

I

2 

V R



V R

1 

V R

2  1

R

 1

R

1  1

R

2

V d I R I

2

R

2

Another

(intuitive)

way…

Consider two cylindrical resistors with cross-sectional areas

A

1 and

A

2

R

1  

L A

1

R

2  

L A

2

V R

1

A

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

R effective

 

A

1 

L



A

2   1

R effective

 

A

1

L



A

2 

L

 1

R

1  1

R

2

A

2

R

2

 1

R

 1

R

1  1

R

2

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." KVL:



loop

V

n

 0 •

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

e 1

I R

1

R

2

e 2  e 1 

IR

1



IR

2

 e 2 

0

Rules of the Road

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.

e 1

I R

1

R

2

e 2  e 1 

IR

1



IR

2

 e 2 0

Current in a Loop

b R

1

a

e 1 Start at point a (could be anywhere) and assume current is in direction shown (could be either)

I R

2

c

e 2

f d R

4

R

3

e

KVL:



loop V n

 0    1

IR

2 e 2

IR

3 

IR

4 e 1 

I



R

1 e 1 

R

2  e 2 

R

3 

R

4 0

I

Question 3

•

Consider the circuit shown.

– –

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

I

0 .

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

I

1 .

12V

–

What is the relation between

I

0 and

I

1 ?

(a)

I

1 <

I

0 (b)

I

1 =

I

0

R R a b

(c)

I

1 >

I

0

I

12V 12V

1. a 2. b 3. c

Question 3

Question 3

•

Consider the circuit shown.

– –

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

I

0 .

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

I

1 .

12V

–

What is the relation between

I

0 and

I

1 ?

(a)

I

1 <

I

0 (b)

I

1 =

I

0

R R a b

(c)

I

1 >

I

0

I

12V 12V

•

From symmetry the potential (

V

a -

V

b ) before the switch is closed is

V

a -

V

b = +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.

Question 3

•

Consider the circuit shown.

–

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

I

0 .

–

After the switch is closed, the current flowing through the bottom resistor is

I

1 .

12V

–

What is the relation between

I

0 and

I 1

?

(a)

I

1 <

I 0

(b)

I

1 =

I

0

•

Write a loop law for original loop: 12V +12V



I

0

R



I

0 R = 0

•

I

0 = 12V/R Write a loop law for the new loop: 12V



I

1 R = 0

I

1 = 12V/R

R a R b I

(c)

I

1 >

I

0 12V 12V

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."

I in

 

I out

• • •

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

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

Each distinct branch must have a current,

I i

assigned to it

How to use Kirchhoff’s Laws

A two loop example:

R

1

I

3

e

1

I

1

R

2

I

2

R

3

e

2

•

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

(1)

I

1 =

I

2 +

I

3

•

Identify all independent loops and use KVL.

2 e

1



(3) I

2

R

2

4 e 1 

I

1

R

1

 e

2

I

1

R

1

 

I

2

 e 2

R I

3

R

3



2

= 0 = 0

I

3

R

3

 0

How to use Kirchoff’s Laws

R

1

I

3

e

1

I

1

I

2

e

2

R

2

R

3

•

Solve the equations for

I

1 ,

I

2 , and

I

3 : First find

I

2 and

I

3 in terms of

I 1

:

I

2  ( e 1 

I R

1 1 ) /

R

2

From eqn. (2)

I

3  ( e e 1   2

I R

1 1 ) /

R

3

From eqn. (3) Now solve for

I

1 using eqn. (1):

I

1  e

R

2 1  e e 1  2

R

3 

R I

1 (

R

2 1 

R R

3 1 ) 

I

1  e

R

2 1 1   e e 1  2

R

3

R

1 

R

1

R

2

R

3

Let’s plug in some numbers

R

1

I

3

e

1

I

1

I

2

e

2

R

2

R

3

e

1 = 24 V

e

2 = 12 V

R

1 = 5

W

R

2 =3

W

R

3 =4

W

Then,

I

1 =2.809 A and

I

2 = 3.319 A,

I

3 = -0.511 A

Junction:

I

1 

I

2 

I

3

Junction Demo

e 1

R

Outside loop:

e 1  1  3 e 3 0

I

1

I

2

e 2

Top loop:

e 1 

I R

1 e 2

I R

2  0

I

1  2 e 1  e 2  e 3 3

R I

2  e 1  e 3  2 e 2 3

R

I

3

R

e 3

I

3  e 1  e 2  2 e 3 3

R

R

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:

Q

(

t

)  

t

Q

0

e

RC

Reading Assignment: Chapter 26.6

Examples: 26.17,18 and 19

Two identical light bulbs are represented by the resistors

R

2

and

R

3

(

R

2

switch

S

=

R

3

).

The is initially open.

2) If switch

S

is closed, what happens to the brightness of the bulb

R 2

?

a) It increases b) It decreases c) It doesn’t change 3) What happens to the current

I

, after the switch is closed ?

a)

I

after = 1/2

I

before b)

I

after =

I

before c)

I

after = 2

I

before

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

E

I R

1

R 4 R

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

a)

I

after >

I

before Before:

R

tot

=

3

R I

before = 1/3

E

/

R

b)

I

after =

I

before After: c)

I

after <

I

before

R

23 = 2

R R

423

R

tot = 2/3 = 5/3

R R I

after = 3/5

E

/

R R 3