1. What is the diameter of a 1.00-m length of tungsten wire whose resistance is
ID: 1445085 • Letter: 1
Question
1. What is the diameter of a 1.00-m length of tungsten wire whose resistance is 0.29 ?
d = _______ m
2. For some applications, it is important that the value of a resistance not change with temperature. For example, suppose you made a 18 k resistor from a carbon resistor and a Nichrome wire-wound resistor connected together so the total resistance is the sum of their separate resistances. The temperature coefficients of resistivity of carbon and Nichrome at 20C are -0.0005 (C)1 and0.0004 (C)1, respectively.
a) What value should the carbon resistor have (at 20C) so that the combination is temperature independent?
R0C = ________________
b) What value should the Nichrome wire-wound resistor have (at 20 C) so that the combination is temperature independent?
R0N = ________________
3. Determine at what temperature aluminum will have the same resistivity as tungsten does at 24 C
T = ______ C
4. How much would you have to raise the temperature of a copper wire (originally at 20 C) to increase its resistance by 19 % ? The temperature coefficient of resistivity of copper is 0.0068 (C)1.
T = ___________ C
5. An extension cord made of two wires of diameter 0.129 cm (no. 16 copper wire) and of length 2.6 m is connected to an electric heater which draws 14.5 A on a 120-V line. The resistivity of copper is 1.68×108m.
How much power is dissipated in the cord?
P = ___________
6. How many coulombs are there in a 70ampere-hour car battery?
Q = __________
7. What is the resistance of a toaster if 120 V produces a current of 4.5 A ?
R = _____________
8. Each channel of a stereo receiver is capable of an average power output of 100W into an 8- loudspeaker
a) What is the rms voltage fed to the speaker at the maximum power of 100W?
Vrms = _____________
b) What is the rms current fed to the speaker at the maximum power of 100W?
Irms = ______________
c) What is the rms voltage fed to the speaker at 1.0W when the volume is turned down
d) What is the rms current fed to the speaker at 1.0W when the volume is turned down
Irms = _____________
9. A small immersion heater can be used in a car to heat a cup of water for coffee or tea. If the heater can heat 120 mL of water from 25C to 95C in8.5 min . Assume the manufacturer's claim of87 % efficiency. The specific heat of water is4186 J/(kgC).
a) Approximately how much current does it draw from the car's 12-V battery?
I = ____________
b) What is its resistance?
R = ___________
10. You buy a 75-W lightbulb in Europe, where electricity is delivered to homes at 240 V.
If you use the lightbulb in the United States at 120 V (assume its resistance does not change), how bright will it be relative to 75-W 120-V bulbs? [Hint: Assume roughly that brightness is proportional to power consumed.]
brightness in the United States/brightness in Europe=
11. An ordinary flashlight uses two D-cell 1.5-V batteries connected in series as in (Figure 1) and in (Figure 2) . The bulb draws 420 mA when turned on.
a) Calculate the resistance of the bulb.
R = _______
b) Calculate the power dissipated.
P = _______
c) By what factor would the power increase if four D-cells in series were used with the same bulb? (Neglect heating effects of the filament.)
12. The peak value of an alternating current in a1800-W device is 7.0 A .
What is the rms voltage across it?
_____________
13. What voltage will produce 0.25 A of current through a 4800- resistor?
V = ____________
14. An electric device draws 4.60 A at 240 V.
a) If the voltage drops by 15 % , what will be the current, assuming nothing else changes?
Ifinal = __________
b) If the resistance of the device were reduced by 15 % , what current would be drawn at 240V?
Ifinal = __________
15. A 120-V fish-tank heater is rated at 130 W .
a) Calculate the current through the heater when it is operating.
I =_____________
b) Calculate its resistance.
R = ____________
Vrms = _____________Explanation / Answer
1.
Resistivity of tungsten
p=5.6*10-8 ohm-m
Since Resistance
R=pL/A
=>A=pL/R=(5.6*10-8)*1/0.29
A=1.93*10-7 m2
SInce A=pi*D2/4
1.93*10-7=pi*D2/4
D=4.95*10-4 m or 0.495 mm
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5.
Area
A=pi*D2/4 =pi*(0.129*10-2)2/4 =1.306*10-6 m2
Resistance
R=pL/A =(1.68*10-8)*2.6/(1.306*10-6 )
R=0.0334 ohms
Power dissipated
P=I2R=14.52*0.0334
P=7.03 Watts
6.
Charge
Q=70 A-H
1 hour =3600 s
Q=70*3600 = 2.52*105 C
7.
R=V/I =120/4.5
R=26.67 ohms
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