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Solutions to Homework #3
Chapter
21
6.
Picture the Problem
: Electric current is delivered to a television set at a specified voltage.
Strategy:
The power delivered to the television is the energy per charge (voltage) multiplied by the charge per time
(current) as given by equation 214.
For part (b) we can use the definition of current (equation 211) and the charge
on an electron to find the required time for 10 million electrons to pass through the circuit.
Solution:
1. (a)
Solve
equation 214 for
I
:
78 W
0.65 A
120 V
P
I
ε
=
=
=
2.
(b)
Solve equation 211 for
t
∆
:
(
29
(
29
7
19
12
e
1 10 e
1.6 10
C/e )
2.5 10
s
2.5 ps
0.65 A
N e
Q
t
I
I




∆
∆ =
=
=
=
=
Insight:
Ten million electrons in 2.5 picoseconds!
It’s a good thing electric companies don’t charge by the
electron!
16.
Picture the Problem
: The four conducting cylinders shown in
the figure are all made of the same material, though they differ
in length and/or diameter. They are connected to four different
batteries, which supply the necessary voltages to give the
circuits the same current,
I
.
Strategy:
Use Ohm’s law,
V
I R
=
(equation 212) together
with the expression for resistance,
R
L A
ρ
=
(equation 213)
to determine the ranking of the voltages.
Solution:
Combining equations 212 and 213 we find that
2
2
4
I
L
I L
L
V
B
A
D
D
π
=
=
=
, where
B
is a constant.
We may
then calculate
the voltages:
(
29
1
2
2
3
3
,
4
2
L
L
V
B
B
D
D
=
=
2
2
2
2
2
,
L
L
V
B
B
D
D
=
=
3
2
,
L
V
B
D
=
and
(
29
4
2
2
1
.
4
2
L
L
V
B
B
D
D
=
=
From
these calculations we can determine the ranking,
V
4
<
V
1
<
V
3
<
V
2
.
Insight:
The smallest resistance (case 4) needs the smallest voltage in order to flow current
I
.
20.
Picture the Problem
: A wire conducts current when a potential difference is applied between the two ends.
Strategy:
Ohm’s Law (equation 212) gives the relationship between potential difference, current, and resistance
for any circuit element. To solve this problem we can combine equation 213 with Ohm’s Law to find the
resistivity of the material from which the wire was made.
Solution:
Combine
equations 212
and
213 to find
:
(
29
(
29
(
29
(
29
2
3
2
8
2
1
4
0.33 10 m
12 V
7.1 10
m
4
4 2.1 A
6.9 m
L
L
D V
V
IR
I
I
A
I L
D


=
=
=
=
=
=
Ω
Insight:
The resistivity of this material is a little higher than tungsten but lower than iron.
If it is a pure metal we
might suspect it is made out of one of the transition metals.
32.
Picture the Problem
: A light bulb consumes electric power by drawing current at a certain voltage.
Strategy:
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