Experiment: Wheatstone Bridge - Department of Physics and ...Next Week (Jan 24 – 26) Experiment:...
Transcript of Experiment: Wheatstone Bridge - Department of Physics and ...Next Week (Jan 24 – 26) Experiment:...
Next Week (Jan 24 – 26)
Experiment: Wheatstone Bridge
The Week After (Jan 31 – Feb 2)
No labs or tutorials!!
Answers to even-numbered problems are on 16.103 home page
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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
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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
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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?
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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.
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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
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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
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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?
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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
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Combination of Series and Parallel Resistors
1Rp
=1180
+1470
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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
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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