Transcript Lect19

a b I

e

R C I V R I r

e

R I

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

R

L A

Question 1

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 2

d

.

Compare the resistance of the two cylinders. a)

R

1 >

R

2 b)

R

1 =

R

2 c)

R

1 <

R

2

1. a 2. b 3. c

Question 1

Question 1

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 2

d

.

Compare the resistance of the two cylinders. a)

R

1 >

R

2 b)

R

1 =

R

2 c)

R

1 <

R

2 • Resistance is proportional to Length/Area

Question 2

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 2

d

.

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

v

1 >

v

2 b)

v

1 =

v

2 c)

v

1 <

v

2

1. a 2. b 3. c

Question 2

Question 2

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 2

d

.

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

v

1 >

v

2 b)

v

1 =

v

2 c)

v

1 <

v

2 Current  Area  Current Density Current Density  average velocity of electrons I is the same A 1 v 2

Resistors in Series

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

R

1 I

a b

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

By Ohm’s law, the Voltage difference across resistance

R 1

is

V a

V b

IR

1 •

Across

R 2

is

V b

V c

IR

2 •

Total voltage difference

V a

V c

I

(

R

1 

R

2 ) • 

the effective single resistance is

R effective

 (

R

1 

R

2 )

R

2

c a c R

effective

Another

(intuitive)

way…

R

1

Consider two cylindrical resistors with lengths

L

1 and

L

2

R

1  

L

1

A

V L

1

R

2  

L

2

A

L

2

R

2

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

R effective

 

L

1 

L

2

A

R

1 

R

2

R

R

1 

R

2

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

V

0

R

1

V R

2 By varying R 2 we can controllably adjust the output voltage!

V

 ?

V

IR

2 

R

1

V

0 

R

2  

R

2

R

2

R

2

R

2   

R

1

R

1 V=0 V= V 0 2

R

1 V=V 0

Question 3

Two resistors are connected in series to a battery with emf E.

The resistances are such that

R 1

= 2R

2

. The currents through the resistors are I 1 and I 2 and the potential differences across the resistors Are: V 1 and V 2 .

a) I 1 >I 2 and V 2 =E b) I 1 =I 2 and V 2 = E c) I 1 =I 2 and V 2 =1/3E d) I 1

Resistors in Parallel

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