Homework Set 3: Solutions
Due: Wednesday, September 15, 2010
Chapter 18: Questions
(6) 8 points
Can a copper wire and an aluminum wire of the same length have the same resistance?
Explain.
The resistance of a uniform wire is given by R = ρ`/A, where ρ is the resistivity of the material, ` is the
length of the wire, and A is its cross-sectional area. Therefore, even though the two wires, which are made
of different materials, have the same length, they can have the same resistance if their cross-sectional
areas are different. In this case, the cross-sectional area of the aluminum wire should be larger than the
cross-sectional area of the copper wire. (Another way of accomplishing this would be by changing the
resistivity of the wires, which can be accomplished by changing the temperature.)
(9) 12 points
The equation P = V 2 /R indicates that the power dissipated in a resistor decreases if the
resistance is increased, whereas the equation P = I 2 R implies the opposite. Is there a
contradiction here? Explain.
There is no contradiction. More precisely, P = V 2 /R states that the power dissipated in a resistor
decreases if the resistance is increased and the voltage drop across the resistor is held constant. The
equation P = I 2 R states that the power dissipated in a resistor increases if the resistance is increased
and the current through the resistor is held constant. However, using V = IR, if we increase the resistance
while keeping the current constant, the voltage drop across the resistor will also increase. Therefore, both
equations are in direct agreement.
(11) 10 points
Explain why light bulbs almost always burn out just as they are turned on and not after
they have been on for some time.
When a light bulb is first turned on, it will be cool and the filament will have a lower resistance than
when it is hot. This lower resistance means that there will be more current through the bulb while it is
cool. This momentary high current will make the filament quite hot. If the temperature is too high, the
filament will vaporize, and the current will no longer be able to flow in the bulb.
(13) 10 points
Electric power is transferred over large distances at very high voltages. Explain how the
high voltage reduces power losses in transmission lines.
The rate at which electrical energy is transmitted is given by Ptrans = IV , while the the rate at which
electrical energy is “lost” to thermal energy is given by Plost = I 2 R. For a given value of Ptrans , the
higher the voltage is, the lower the current has to be. As the current decreases, so too does Plost .
(18) 10 points
Is current used up in a resistor? Explain.
Current is not used up in a resistor. The same current flows into the resistor as flows out of the resistor
(conservation of charge). What does get “used up” is potential energy. The charges that come out of
a resistor have lower potential energy than the charge that go into the resistor (this “lost” energy is
converted into thermal energy).
Chapter 18: Problems
(6) 8 points
(II) A hair dryer draws 7.5 A when plugged into a 120-V line. (a) What is its resistance?
(b) How much charge passes through it in 15 min? (Assume direct current.)
(a) Using ∆V = IR, and solving for R
R=
∆V
120 V
=
= 16 Ω.
I
7.5 A
(b) Using I = ∆Q/∆t, and solving for ∆Q
∆Q = I∆t = (7.5 A)(900 s) = 6.8 × 103 C.
(10) 10 points
(II) An electric device draws 6.50 A at 240 V. (a) If the voltage drops by 15%, what will
be the current, assuming nothing else changes? (b) If the resistance of the device were
reduced by 15%, what current would be drawn at 230 V?
(a) If the voltage drops by n %, then the current through the device will be In = ∆Vn /R. In other
words, the current will drop by the same percentage. Therefore, if the voltage drops by n = 15%, so too
will the current (in this case, the new current would be (0.85)(6.50 A) = 5.53 A).
(b) If the resistance of the device decreases by m %, then then current through the device will be
Im = ∆V /Rm ; in other words, decreasing the resistance by, say, 15% (which is the same as multiplying
it by 0.85), then the current is increased by a factor of 1/0.85 = 1.18 (in this case, the new current would
be (1.18)(6.50 A) = 7.65 A).
(14) 10 points
(II) Calculate the ratio of the resistance of 10.0 m of aluminum wire 2.0 mm in diameter,
to 20.0 m of copper wire 2.5 mm in diameter.
The resistance of a uniform wire is given by R = ρ`/A, where ρ is the resistivity of the material of the
wire, ` is the length of the wire, and A is its cross-sectional area. Referencing the aluminum wire as wire
1 and the copper wire as wire 2, we find that
R1
R2
=
=
(32) 10 points
ρAl `1
ρAl r22 `1
πr12
=
ρCu `1
ρCu r12 `2
πr22
(2.65 × 10−8 Ωm)(2.5 × 10−3 m)2 (10.0 m)
= 1.23.
(1.68 × 10−8 Ωm)(2.0 × 10−3 m)2 (20.0 m)
(II) 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.]
Using P = ∆V 2 /R, we find that
PU.S
PE.U.
2
∆VU.S
=
∆VE.U.
2
1
120 V
= .
=
240 V
4
In other words, the lightbulb would only appear 1/4 as bright as a standard 75-W lightbulb purchased
here in the United States.
(82) 12 points
If a wire of resistance R is stretched uniformly so that its length doubles, by what factor
does the power dissipated in the wire change, assuming it remains hooked up to the same
voltage source?
The volume of the wire remains constant as the wire is stretched from the original length of L0 to the
final length of L = 2L0 . Thus the cross-sectional area change from A0 to A = 21 A0 (because the volume
of the wire is equal to the product of its length and its cross-sectional area). The resistance R0 of the wire
before it was stretched was R0 = ρL0 /A0 ; after it is stretched, its resistance becomes R = ρL/A = 4R0 .
Therefore, the power dissipated in the wire is
P =
P0
1 ∆V 2
∆V 2
=
=
.
R
4 R0
4
Therefore, the power is reduced by a factor of 4.