This Week (Jan 23 – 25)

Experiment 2: Wheatstone Bridge

Midterm Test

Thursday, March 1, 7 - 9 pm

Location TBA

Chapters 18 – 25

20 multiple choice questions

Formula sheet provided

32Wednesday, January 24, 2007

Resistors in series: Rs = R1 + R2 + . . .

R1 R2

The same current passes through the resistors

Series and Parallel Resistances

Resistors in parallel:1

Rp=

1R1

+1

R2+ . . .

Same potential difference across the resistors

Rs

=

Rp

=

33Wednesday, January 24, 2007

Prob. 20.C12: In one of the circuits, none of the resistors is in series or in

parallel. Which one?

• Resistances must be connected directly end to end with nothing

! between for them to be in series or in parallel

34Wednesday, January 24, 2007

Prob. 20.58: What is the equivalent resistance between points A and B?

2 " 6 " 1 "

4 " 3 " 2 "

3 "

35Wednesday, January 24, 2007

Prob. 20.63: The five resistors are identical. The battery delivers 58 W of

power to the circuit. What is the resistance R of each resistor?

• What is the equivalent resistance in terms of R?

• What does R need to be so that 58 W is dissipated in the equivalent resistance?

R2RR

R

1Rp

=1R

+1

2R+

1R

=5

2R

36Wednesday, January 24, 2007

V = 12 V

I1 = 2 A

Rs V = 12 V

I2 = 9 A

Rp

Prob. 20.-/54: When two resistors are connected in series with a 12 V

battery the current from the battery is 2 A. When they are connected in

parallel the current is 9 A.

Determine the values of the two resistances, R1, R2.

Rs = R1 +R21

Rp=

1R1

+1R2

Rs =VI1

=122

= 6 ! Rp =VI2

=129

=43

!

Series Parallel

37Wednesday, January 24, 2007

Internal Resistance, Terminal Potential Difference

I

I

I

Terminal Potential Difference, TPD = IR and V = I(r + R),

The potential difference that appears across the terminals of the battery

is reduced by the internal resistance of the battery.

TPD

LoadLoad (lights, starter, etc)

I

so, I =V

r+Rand TPD =V

R

r+R emf = V

TPD

Internal resistance of battery, r

38Wednesday, January 24, 2007

Internal resistance, r, increases as a

battery gets older – corrosion, etc.

Suppose r = 0.01 " and V = 12 V and

the starter motor draws 100 A of current.

Then, the terminal potential difference is

reduced by:

I ! r = 100 ! 0.01 = 1 V.

The TPD is 12 – 1 = 11 V.

The internal resistance decreases the

voltage available when the battery is

supplying current.

Terminal Potential Difference

TPD

39Wednesday, January 24, 2007

A 1.2 " resistor is connected across a battery. If the battery had no

internal resistance, the power dissipated in the resistor would be Po.

The battery does have an internal resistance of 0.06 ", so the power

dissipated in the resistor is P.

Find P/Po.

• The power dissipated in the resistor is I2R, so work out I for the two

! cases for an emf of V...

40Wednesday, January 24, 2007

Prob. 20.112: 75 " and 45 " resistors are connected in parallel. When

connected to a battery, the current delivered is 0.294 A.

When the 45 " resistor is removed, the current drops to 0.116 A.

Find the emf and the internal resistance of the battery.

• Draw diagrams!

• For an internal resistance r and an emf V, write expressions involving

! the current for the two cases

! ! simultaneous equations in V and r

• Solve for V and r.

41Wednesday, January 24, 2007

What is the equivalent resistance?

You need Kirchhoff’s rules!

No series or parallel combinations

42Wednesday, January 24, 2007

Kirchhoff’s Rules

1) Junction Rule:

! The sum of the currents entering a junction

! is equal to the sum of currents leaving it.

! 7 = 5 + 2

Conservation of charge and current

2) Loop Rule:

! The sum of potential changes

! around any closed loop is zero.

Potential drops as you go with the

flow of current through a resistor.

12 – 2#5 – 2#1 = 0

Conservation of energy

43Wednesday, January 24, 2007

Thanks to Escher

Loop Rule:

The water apparently always

flows downhill, and arrives

back where it started and

flows forever in a loop with

no energy input.

This cannot be, as there is no

pump to raise the potential

energy of the water back to

its initial value.

Likewise, electrons flowing

around a closed loop cannot

flow forever unless a power

source pumps them back up

to their initial potential.

44Wednesday, January 24, 2007

Kirchhoff’s Rules

1) Junction Rule:

! The sum of currents entering a junction is equal to the sum of

! currents leaving it (conservation of charge and current).

2) Loop Rule:

! Around any closed loop, the sum of potential changes is equal

! to zero (conservation of energy).

The potential decreases when you go with the flow of current

through a resistor.

45Wednesday, January 24, 2007