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Experiment: Wheatstone Bridge

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No labs or tutorials!!

Answers to even-numbered problems are on 16.103 home page

15

Chapter 20 so far...

P = VI = V2/R = I2R

ρ= ρ0[1+α(T !T0)]

Ohm’s law

R= R0[1+α(T !T0)]

Power

R=ρLA

V = IR

Resistance, resistivity

Temperature dependence

16

Prob. 20.C2: The filament of a lamp is made of a tungsten wire. As the filament heats up, does the power increase or decrease? The temperature coefficient of resistance is positive.

• How does the power vary with resistance for constant V?

Prob. 20.C6:

Which bulb filament has the greater resistance, 75 W or 150 W?

P = VI = V2/R

so large R ⇒ small P

17

Prob. 20.27: A tungsten wire is connected to a source of constant voltage via a switch. At the instant the switch is closed, the temperature of the wire is 28º C and the initial power dissipated in the wire is P0.

At what wire temperature has the power dissipated in the wire fallen to P0/2?

α = 0.0045 ºC-1

• How does the power dissipated in the wire vary with the resistance of the wire?

• How does the resistance of the wire vary with temperature?

18

Resistors in Series

R1 R2 R1 + R2

I = 12/9 = 1.33 A

The same current I passes through both resistors (current is conserved and has nowhere else to go).

V1 = IR1 V2 = IR2

I

V = V =

Total potential difference: V = V1 + V2 = I × (R1 + R2)

I = V/Rs, where Rs = R1 + R2

A single resistance Rs = R1 + R2 (equivalent series resistance) would result in the same current.

19

A B

A B

I2

I1

Resistors in Parallel

1Rp

=1R1

+1R2

The potential difference between A and B is V:

VAB = V = I1R1 = I2R2

and I = I1 + I2 (current conserved).

So, I = V/R1 + V/R2 = V(1/R1 + 1/R2)

That is, I = V/Rp or V = IRp

Rp is the equivalent parallel resistance:

The circuits are equivalent

20

Resistors in Parallel

I1 = 6/8= 0.75 A

I2 = 6/4= 1.50 A

A B

1Rp

=14

+18

=38

Rp =83Ω

I

II

I

I2 I1

20.52

I =VRp

=6

8/3= 2.25 A

I = I1 + I2 = 0.75 + 1.50 = 2.25 A

21

Prob. 20.52: A wire of resistance R is cut into three equally long pieces, which are then connected in parallel.

In terms of R, what is the resistance of the parallel combination?

• What is the resistance of each piece?

• What is the equivalent parallel resistance of the pieces?

22

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

23

Combination of Series and Parallel Resistors

1Rp

=1180

+1470

24

I

I = V/R = 24/240 = 0.1 A

Potential drop across 110 Ω resistor is:V = IR = (0.1 A)(110 Ω) = 11 V.

Therefore VAB = 24 – 11 = 13 V.

VAB = I1 × 180, I1 = 13/180 = 0.0722 AVAB = I2 × 470, I2 = 13/470 = 0.0277 AI = I1 + I2 = 0.0999... = 0.1 A

I = 0.1 AI1 I2

Original circuit

25

I = 0.1 AI1 I2 I1 = 0.0722 A

I2 = 0.0277 A

Power dissipated in 180 Ω resistance is I21R= 0.07222!180= 0.94 W

20.5826