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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 1
Chapter 21: Electric Current and DirectCurrent Circuits
Answers to EvenNumbered Conceptual Questions
2. No. An electric current is produced when a net charge moves. If your body is electrically neutral, no current
is produced when you walk.
4. Yes. There is a net charge on the comb, and by moving it from one place to another you have created an
electric current.
6. Car headlights are wired in parallel, as we can tell by the fact that some cars have only one working
headlight.
8. Yes. Just connect two of these resistors in parallel and you will have an equivalent resistance of R/2.
10. Resistors connected in parallel have the same potential difference across their terminals.
12. Resistivity is an intrinsic property of a particular substance. In this sense it is similar to density, which has a
particular value for each particular substance. Resistance, however, is a property associated with a given
resistor. For example, the resistance of a given wire can be large because its resistivity is large, or because it
is long. Similarly, the weight of a ball can be large because its density is large, or because it has a large
radius.
14. Magnetic resonance imaging (MRI) machines would definitely benefit from roomtemperature
superconductivity. As it is, they must cool their magnets to low temperature. Similarly, electrical power
transmission would benefit if the resistance of the wires could be eliminated. On the other hand, a toaster or
an electric oven requires resistance to do its job; superconductivity would not be a help.
16. Capacitors connected in series all have the same magnitude of charge on their plates. This is illustrated in
Figure 2118 (a).
18. The light shines brightest immediately after the switch is closed. With time, the intensity of the light
diminishes. Eventually, the light stops glowing altogether.
Solutions to Problems and Conceptual Exercises
1. Picture the Problem: This is a units conversion problem.
Strategy: Use equation 211 to find the amount of charge that flows by a point in a circuit when one ampere of current
is sustained for one hour.
Solution: Solve equation 211 for
QΔ
:
1 C 3600 s
1 h 3600 C
s h
Q I t
⎛ ⎞⎛ ⎞
Δ = Δ = × =
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
Insight: Dry cell batteries are sometimes rated by the number of milliamphours they can deliver, not as a measure of
the total charge they can supply but the total energy they deliver, which is given by the charge times the voltage.
2. Picture the Problem: A current flows through the filament of a flashlight bulb.
Strategy: Use equation 211 to find the amount of charge that flows through the filament, then divide by the charge on
an electron to find the number of electrons.
Solution: 1. (a) Solve equation 211 for
QΔ
:
(
)
(
)
0.18 A 78 s 14 CQ I tΔ = Δ = =
2. (b) Divide by e to find the number of electrons:
19
e
19
14 C
8.8 10 electrons
1.60 10 C/electron
N
−
= = ×
×
Insight: If the flashlight contains two 1.5 V batteries connected in series, then the total energy delivered to the bulb is
given by equation 202:
( )
(
)
14 C 3.0 V 42 J.U q V= Δ = =
Chapter 21: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 2
3. Picture the Problem: An electric current is created by a stream of electrons that hit the viewing screen of a TV set.
Strategy: The current of 15.0 amperes equals 15.0 coulombs of charge per second. Divide by the charge on an electron
to find the number of electrons that flow through the picture tube per second.
Solution: Divide the current
by the charge on an electron:
19
e
19
15 C/s
9.4 10 electrons/s
1.6 10 C/electron
N
Q I
t e t e
−
= = = = ×
Δ Δ ×
Insight: These electrons are accelerated to a high voltage, more than 10 kV for a typical color television, so a beam
with 15.0 A of current at this voltage would deliver 150 kW of power (equation 214) to the viewing screen! Typical
color television sets actually have beam currents of a few nanoamperes and deliver a few tens of microwatts to the
viewing screen.
4. Picture the Problem: A car battery does work on the charge that passes through it.
Strategy: The work done by a battery is
W Q
ε
=
Δ
, as illustrated in active example 211. Use this relationship to
determine the amount of charge that passes through the automotive battery.
Solution: 1. (a) Solve
W Q
ε
= Δ
for
QΔ
:
260 J
22 C
12 V
W
Q
ε
Δ = = =
2.
(b)
If the battery emf is doubled to 24 V, yet it still does 260 J of work on the charges that pass through it, the amount
of charge required will decrease by a factor of 2.
Insight:
The amount of energy given to the charges and hence delivered to the circuit is the energy per charge (voltage)
times the amount of charge.
5.
Picture the Problem
: An ammeter measures a tiny amount of electric current.
Strategy:
The current of 10.0 fA equals 10.0×10
−15
coulombs of charge per second. Divide by the charge on an
electron to find the number of electrons per second that the ammeter detects.
Solution:
Divide the current
by the charge on an electron:
15
4
e
19
10.0 10 C/s
6.25 10 electrons/s
1.60 10 C/electron
N
Q I
t e t e
−
−
×
= = = = ×
Δ Δ ×
Insight:
If a single electron were to flow through a circuit each second the current would be 1.60×10
−19
A or 160 zA!
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
tΔ
:
(
)
(
)
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
− − −
−
× ×
Δ
Δ = = = = × =
Insight:
Ten million electrons in 2.5 picoseconds! It’s a good thing electric companies don’t charge by the electron!
7.
Picture the Problem
: A battery for a pacemaker is rated for a certain number of amperehours.
Strategy:
Convert the number of amperehours to coulombs of charge, then use equation 211 to determine the time
required for the specified average current to deliver that much charge.
Solution:
1. (a)
Find
QΔ
from equation 211:
(
)
0.42 A h 3600 s/h 1500 C 1.5 kCQ I tΔ = Δ = ⋅ × = =
2.
(b)
Solve equation 211 for
tΔ
:
8
6 7
1500 C 1.00 y
2.7 10 s 8.5 y
5.6 10 A 3.16 10 s
Q
t
I
−
Δ
Δ = = = × × =
× ×
Insight:
The long life span of the battery is helpful in order to minimize the frequency of invasive surgery required to
replace the batteries. The major risk, however, is the toxicity of the chemicals used to make the battery.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 3
8.
Picture the Problem
: A conducting wire is quadrupled in length and tripled in diameter.
Strategy:
Use the expression for resistance,
R L A
ρ
=
(equation 213) to answer the conceptual questions.
Solution:
1. (a)
The length L of the wire increases by a factor of 4, and its crosssectional area A increases by a factor
of 3
2
= 9. The resistance will therefore decrease because it is proportional to L/A.
2.
(b)
Use a ratio with equation 213:
( )
2
2
old
new new new new old new old old
4
2 2
old old old old new old old
new
4
old
4
4
9
3
D
R L A L A L L D
R L A L A L L
D
D
π
π
ρ
ρ
= = = = =
Insight:
The new resistance is a little less than half of the original resistance.
9.
Picture the Problem
: The figure at right shows a plot of current versus voltage for
two different materials, A and B.
Strategy:
Use Ohm’s law,
V I R=
(equation 212) to answer the question.
Solution:
Rearranging Ohm’s law to
( )
1,
I
R V=
we can see that an I vs. V plot
will be linear with a slope of 1/R for any material that obey Ohm’s law. Material A
satisfies Ohm’s law because the relationship between current and voltage for this
material is linear; that is, current is proportional to voltage.
Insight:
The resistance of material B decreases as the voltage and current increases. This behavior resembles that of a
semiconductor, where the heat generated by increased current flow generates new charge carriers and decreases the
effective resistance of the material. Semiconductors do not obey Ohm’s law; they are sometimes called “nonOhmic.”
10.
Picture the Problem
: The figure at right shows a plot of current versus voltage for
two different materials, A and B.
Strategy:
Use Ohm’s law,
V I R=
(equation 212) to answer the question.
Solution:
1. (a)
Rearranging Ohm’s law to
(
)
1,
I
R V=
we can see that an I vs. V
plot will be linear with a slope of 1/R for any material that obey Ohm’s law. This
means that the resistance of a material is the inverse of the slope of the I vs. V plot.
At voltage V
1
the slope of curve B is greater than the slope of curve A, implying
that the resistance of material A is greater than the resistance of material B at the
voltage V
1
.
2.
(b)
The best explanation is
II
. A larger slope means a larger value of I/V, and hence a smaller value of R. Statement I
is true, but irrelevant, and statement III is false because the resistance is proportional to the inverse of the slope.
Insight:
You can draw the same conclusion from the value of the graph instead of its slope. Material A in the figure
allows less current to flow at the voltage V
1
than does material B. It follows that material A has the greater resistance at
this voltage.
11.
Picture the Problem
: Two cylindrical wires are made of the same material and have the same length but different radii.
Strategy:
Use a ratio together with the expression for resistance,
R L A
ρ
=
(equation 213) to answer the question.
Solution:
1.
If wire B is to have the greater resistance with the same length, its crosssectional area must be smaller than
that of wire A. This is because resistance is linearly proportional to L/A.
2.
Use a ratio with equation 213:
B B
B A A
A B A
A A
1
9 3
A L R
r R R
r R R
A L R
π ρ
π ρ
= = = = =
Insight:
A narrow wire has a larger resistance because it constricts the flow of current in a manner analogous to the
constriction of water flow by a narrow pipe, although the physical phenomena involved are fundamentally different.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 4
12.
Picture the Problem
: A silver wire of known dimensions has an intrinsic resistance.
Strategy:
Use Table 211 together with equation 213 to determine the resistance of the wire.
Solution:
Apply equation 213, using
ρ
=
晲潭⁔慢=攠e11nd=
㈲
1
4
A
r D
π π
= =
:
( )
( )
8
2
3
4
5.9 m
1.59 10 m 0.50
0.49 10 m
L
R
A
π
ρ
−
−
=
= × Ω⋅ = Ω
×
Insight:
Metals typically have a very low resistance. In this case, a wire only half a millimeter thick and 19 ft long has
a resistance of less than half an ohm.
13.
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. In this problem we can solve Ohm’s Law for the resistance R of the wire.
Solution:
Solve Ohm’s Law for R:
18 V
51
0.35 A
V
R
I
=
= = Ω
Insight:
This is a large resistance for a wire. Even if it were made out of steel, which has a high resistivity (see Table
211) and was 0.50 mm thick, it would have to be over 100 m long to offer 51 Ω of resistance.
14.
Picture the Problem
: A tungsten wire of unknown diameter has an intrinsic resistance.
Strategy:
Use Table 211 together with equation 213 to determine the resistance of the wire.
Solution:
Solve equation 213
for D, using
ρ
晲潭⁔慢=攠e11=
湤=
㈲
1
4
A
r Dπ π= =
:
( )
( )
( )
2
1
4
8
4 5.6 10 m 0.27 m
4
0.5 mm
0.07
L L
R
A
D
L
D
R
ρ ρ
π
ρ
π π
−
= =
× Ω⋅
= = =
Ω
Insight:
A tungsten wire of this diameter would have to be 3.5 m long in order to offer 1.0 Ω of resistance.
15.
Picture the Problem
: A copper wire of known dimensions has an intrinsic resistance.
Strategy:
Use Table 211 together with equation 213 to determine the resistance of the wire.
Solution:
Apply equation 213 , using
ρ
晲潭⁔慢=攠e11nd=
㈲
1
4
A
r D
π π
= =
:
( )
(
)
( )
8
2
3
4
6.0 mi 1609 m/mi
1.68 10 m 0.68 k
0.55 10 m
L
R
A
π
ρ
−
−
×
=
= × Ω⋅ = Ω
×
Insight:
This is a large resistance, and copper wires are often used for electric power and communication lines.
However, the choice of a larger diameter can make a huge difference. For instance, if the diameter of the wire were
3.0 mm, the resistance would only be 23 Ω, a reduction by a factor of 30!
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
ρ ρ
= = =
, where B is a constant. We may then calculate
the voltages:
( )
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
( )
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.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 5
17.
Picture the Problem
: The section of copper wire between a bird’s feet has an intrinsic resistance. If any current is
flowing in the wire, there is a small potential difference across the two ends of the wire section.
Strategy:
Ohm’s Law (equation 212) gives the relationship between potential difference, current, and resistance for
any circuit element. In this problem we first apply equation 213 and the data from Table 211 to find the resistance of
the wire section, and then apply Ohm’s Law to find the potential difference.
Solution:
1. (a)
Combine
eqs. 212 and 213 to find V:
( )
( )
8
4 2
0.060 m
32 A 1.68 10 m 2.5 mV
0.13 10 m
L
V IR I
A
ρ
−
−
= = = × Ω⋅ =
×
2.
(b)
Since V is directly proportional to the separation L of the bird’s feet, it will increase if L increases.
Insight:
Landing on a high voltage wire will not kill a bird because the potential difference between its feet is small.
However, if another part of the bird’s body gets too close to a grounded conductor, a spark could leap from the bird’s
body to the grounded conductor and kill the bird.
18.
Picture the Problem
: A length of copper wire has an intrinsic resistance. If any current is flowing in the wire, there is
a potential difference across the two ends of the wire.
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, using resistivity data from Table 211, with
Ohm’s Law to find the length of the wire.
Solution:
Combine eqs. 212
and 213 to find L:
( )
( )
( )
( )
2
1
4
2
3
2
8
0.44 10 m 15 V
0.14 km
4
4 0.96 A 1.68 10 m
L L
V I R I I
A
D
D V
L
I
ρ ρ
π
π
π
ρ
−
−
⎛ ⎞
= = =
⎜ ⎟
⎝ ⎠
×
= = =
× Ω⋅
Insight:
This relatively short wire has a large resistance, almost 16 Ω. Copper wires are often used for electric power
and communication lines, and such a large resistance would reduce the efficiency of power transmission. However, if
the diameter of the wire were 3.0 mm, the resistance would only be 0.34 Ω, a reduction by a factor of 46!
19.
Picture the Problem
: A potential difference drives electric current across the membrane of a cell wall.
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, using the given resistivity of the cell wall,
with Ohm’s Law to find the electric current that passes through and area
(
)
2
6 12 2
1.0 10 m 1.0 10 m.A
− −
= × = ×
Solution:
1. (a)
Combine
eqs. 212 and 213 to find I:
(
)
(
)
( )( )
3 12 2
13
7 9
75 10 V 1.0 10 m
7.2 10 A
1.3 10 m 8.0 10 m
V V VA
I
R L A Lρ ρ
− −
−
−
× ×
= = = = = ×
× Ω⋅ ×
2.
(b)
If the thickness L of the membrane is doubled, but
ρ
慮搠 V remain constant, the current through the membrane
will decrease by a factor of 2.
Insight:
A current of 0.72 pA doesn’t seem like much, but it is equivalent to 4.5 million electrons passing through a
2
1.0 m
μ
牥愠=∝敲礠獥s→湤℠
=
㈰.†
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
:
ρ
=
(
)
( )
( )( )
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 LD
π
π
ρ ρ ρ
π
−
−
×
= = = ⇒ = = = × 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.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 6
21.
Picture the Problem
: An aluminum wire of known crosssectional area has an intrinsic resistance per unit length.
Strategy:
Use Table 211 together with equation 213 to determine the resistance per unit length of the wire.
Solution:
1. (a)
Apply equation 213,
using
ρ
晲潭⁔慢=攠e1ㄺ=
8
㜲
㈮㘵 m
〮ㄱ =
㈮4 10 m
R
L A
ρ
−
−
× Ω⋅
= = = Ω
×
2. (b)
The resistance per unit length is inversely proportional to the crosssectional area, and the crosssectional area is
directly proportional to the square of the diameter. If the diameter were increased the answer to part (a) would decrease.
3. (c)
Repeat part (a) for the new A:
8
7 2
2.65 10 m
0.074 /m
3.6 10 m
R
L A
ρ
−
−
× Ω⋅
= = = Ω
×
Insight:
A crosssectional area of 3.6×10
−7
m
2
is equivalent to a diameter of 0.34 mm. You would need a length of this
wire almost 14 m long in order to have a resistance of 1.0 Ω.
22.
Picture the Problem
: A human finger has an intrinsic electrical resistance. The application of an electric potential will
drive a current through the finger.
Strategy:
Use equation 213 and an estimate that the finger is a cylinder of radius 1.0 cm and length 10 cm to estimate
the electrical resistance of the finger. Then use Ohm’s Law (equation 212) to find the potential difference required to
drive a 15 mA current through the finger.
Solution:
1. (a)
Apply
equation 213 to estimate R:
( )
( )
2
2
0.10m
0.15 m 50 10
0.010
L
R
A
ρ
π
=
= Ω⋅ = Ω≅ Ω
2.
(b)
Use Ohm’s Law to estimate V:
(
)
( )
3
15 10 A 50 0.75 1 VV IR
−
= = × Ω = Ω≅
Insight:
An important factor in determining how much current will actually flow through your finger if you attach it to
a battery using electrical wires is the resistance of your skin, which strongly depends on whether the skin is wet or dry.
23.
Picture the Problem
: A battery is connected across the faces of a
rectangular block of metal in two different ways as shown.
Strategy:
Use Ohm’s Law (equation 212) together with equation 21
3 to form a ratio of the currents between the A×B and B×C faces, and
use the ratio to find
BC
I
in terms of
AB
I
.
Solution:
Combine equations 212 and
213 to form a ratio of the currents:
( )( )
( )( )
2
2
BC BC AB BC
AB AB AB
AB AB BC BC BC BC AB
BC AB
C BC
I V R L A
R L A
C
I
V R R L A L A A AB A
C
I I
A
ρ
ρ
⎛ ⎞
= = = = = =
⎜ ⎟
⎝ ⎠
⎛ ⎞
=
⎜ ⎟
⎝ ⎠
Insight:
The ability to form ratios is a useful skill to develop when faced with problems such as this one. However,
there are other ways to solve the problem, such as finding an expression for
(
)
AB
V I C AB
ρ
=
and then using Ohm’s
Law to find
( )
AB
I
V A BC
ρ
=
.
24.
Picture the Problem
: Light A has four times the power rating of light B when operated at the same voltage.
Strategy:
Use the expression
2
P V R=
(equation 216) to answer the conceptual question.
Solution:
1. (a)
The resistance of light A is less than the resistance of light B because light A has the greater power
rating and
2
.R V P=
2.
(b)
Note that the power is inversely proportional to the resistance. It follows that the ratio of the resistance of light A
to the resistance of light B is
1 4.
Insight:
The solution to part (b) can also be presented as a ratio:
2
A A A B B
2
B A BB B
1
4 4
R V P P P
R P PV P
= = = =
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 7
25.
Picture the Problem
: Light A has four times the power output of light B when operated at the same voltage.
Strategy:
Use the expression
P I V=
(equation 214) to answer the conceptual question.
Solution:
1. (a)
The current
I
P V=
passing through bulb A is greater than that passing through bulb B because they
operate at the same voltage V but the power output P of bulb A is greater than the power output of bulb B.
2.
(b)
Because power is proportional to current, the current in bulb A is four times greater than the current in bulb B.
Insight:
The solution to part (b) can also be presented as a ratio:
A A A B
B B B B
4
4
I P V P P
I P V P P
=
= = =
26.
Picture the Problem
: Light A has four times the power output of light B when the same current flows through each.
Strategy:
Use the expression
P I V=
(equation 214) to answer the conceptual question.
Solution:
1. (a)
The potential difference
V P I
=
across bulb A is greater than that across bulb B because they each
have the same current I passing through them but the power output P of bulb A is greater than that of bulb B.
2.
(b)
The ratio of the potential difference across bulb A to that across bulb B is 4 because voltage is directly
proportional to the power output.
Insight:
The solution to part (b) can also be presented as a ratio:
A A A B
B B B B
4
4
V P I P P
V P I P P
=
= = =
27.
Picture the Problem
: A generator produces power by delivering current at a certain voltage.
Strategy:
The power produced by the generator is the current it delivers times the voltage at which it delivers it
(equation 214). Solve this equation for the current produced.
Solution:
Solve equation 214 for I:
3
3.8 10 W
51 A
75 V
P
I
V
×
= = =
Insight:
If the output voltage were to be increased to 240 V, the generator could produce the same power while
supplying only 16 A.
28.
Picture the Problem
: A portable CD player consumes electric power by drawing current at a certain voltage.
Strategy:
The power consumed by the CD player is the current it draws times the voltage at which it operates.
Solution:
Apply equation 214 directly:
(
)( )
0.022A 4.1 V 0.090 WP IV= = =
Insight:
Power consumed at this rate will drain two 2850mA∙h AA alkaline batteries in about 260 hours.
29.
Picture the Problem
: An electric heater converts electric power to heat energy.
Strategy:
Equation 216 gives the rate at which power is consumed by a circuit element of known resistance when the
potential difference across it is V.
Solution:
Apply equation 216 directly:
( )
2
2
120 V
580 W 0.58 kW
25
V
P
R
= = = =
Ω
Insight:
In order to create a heating element that consumes more power, you would have to make it with a lower
resistance.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 8
30.
Picture the Problem
: A reading lamp consumes electric power by drawing current at a certain voltage.
Strategy:
The power consumed by the lamp is given by equation 214. First find the power P consumed and then
determine the energy
UΔ
consumed by multiplying the power by the time over which the lamp is operated. The cost is
then the energy consumed multiplied by the cost per kilowatthour.
Solution:
1.
Calculate the power delivered to the lamp:
(
)( )
2.6 A 120 V 310 WP IV= = =
2.
Multiply P by
tΔ
to find
:
UΔ
(
)( )
0.31 kW 1.0 h 0.31 kWhU P tΔ = Δ = =
3.
Multiply by the cost per kilowatthour:
(
)
( )
cost 0.31 kWh $0.075 kWh $0.023= =
Insight:
If you fall asleep reading and leave the lamp on for the entire 10hour night, it has cost you only $0.23.
31.
Picture the Problem
: A battery charger consumes electric power by drawing current at a certain voltage.
Strategy:
The power consumed by the battery charger is given by equation 214. First find the power P consumed and
then determine the energy
U
Δ
consumed by multiplying the power by the time over which the charger is operated (120
minutes or 2.0 hours). The cost per kilowatthour is then the total cost divided by the energy consumed.
Solution:
1.
Calculate the power delivered to the battery:
(
)
(
)
15 A 12 V 180 WP IV= = =
2.
Multiply P by
tΔ
to find
:
UΔ
(
)( )
0.18 kW 2.0 h 0.36 kWh
U P tΔ = Δ = =
3.
Divide the total cost by
U
Δ
in kilowatthours:
(
) ( )
cost/kWh $0.026/0.36 kWh $0.072 kWh= =
Insight:
At this cost you could operate the charger for 77 hours and charge almost 39 batteries for $1.00.
32.
Picture the Problem
: A light bulb consumes electric power by drawing current at a certain voltage.
Strategy:
Use equation 214 to find the current drawn by the light bulb given its power rating and operating voltage.
We can then use either equation 215 or 216 to find the resistance of the filament.
Solution:
1. (a)
Solve equation 214 for
I
:
75 W
0.79 A
95 V
P
I
V
= = =
2.
(b)
Solve equation 216 for
R
:
( )
2
2
95 V
120 0.12 k
75 W
V
R
P
=
= = Ω= Ω
3. (c)
Form a ratio of the power ratings:
2
new new old old
2
1
old new oldold
2
2
P V R R R
P R R
V R
=
= = =
4.
The power rating of the lamp with half the resistance would be greater by a factor of 2, or 150 W.
Insight:
Note that we used equation 216 to form the ratio in part (c) instead of equation 215 because the voltage
across each lamp will be the same but the current drawn by each will be different.
33.
Picture the Problem
: Two different rating schemes measure the energy stored in a car battery.
Strategy:
The two rating schemes each measure the amount of current the battery can supply at a specified voltage for a
period of time. Find the power produced by the battery by multiplying the current by the voltage (equation 214) and
then the energy delivered by multiplying the power by the time elapsed.
Solution:
1.
Find the energy delivered
under the first rating scheme:
(
)
(
)
(
)
5
1
905 A 7.2 V 30.0 s 2.0 10 J
E P t I V t= Δ = Δ = = ×
2.
Repeat for the second rating scheme:
(
)
(
)
(
)
6
1
25 A 10.5 V 155 min 60 s/min 2.4 10 J
E P t I V t= Δ = Δ = × = ×
3.
The 155minute reserve capacity rating represents the greater amount of energy delivered by the battery.
Insight:
During the cold cranking test the battery pours out a huge amount of current, but for a short period of time. It
requires 12 times more energy to sustain the much lower current for over 2.5 hours!
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 9
34.
Picture the Problem
: A dozen identical light bulbs are connected to a given emf.
Strategy:
Use the characteristics of series and parallel circuits to answer the question.
Solution:
1. (a)
In a parallel circuit the full potential difference is placed across each light bulb, driving a large current
through each. In a series circuit only a fraction of the full emf is placed across each bulb. With more current flowing
and with a larger potential difference across each bulb, the lights will be brighter if they are connected in parallel.
2.
(b)
The best explanation is
I
. When connected in parallel each bulb experiences the maximum emf, and dissipates the
maximum power. Statements II and III are each false because power is not dissipated by resistance but by a
combination of voltage and current.
Insight:
Let’s try a numerical example. Suppose the dozen lights each have a resistance of 100 Ω and are connected to
a 120V source of emf. In a parallel circuit there is a 120V potential difference across each bulb and they produce
( ) ( )
2
2
120 V 100 144 WP V R= = Ω =
of heat and light. In a series circuit there is only
120 V 12 bulbs 10 V
=
=
慣牯a猠敡s栠扵b戠bnd⁴桥=p→睥爠潵→灵琠潦慣栠楳h
( ) ( )
2
10 V 100 1.0 W.
P = Ω =
35.
Picture the Problem
: A fuse is a small strip of metal that burns through when the current in it exceeds a certain value,
thus producing an open circuit.
Strategy:
Use the characteristics of series and parallel circuits to answer the question.
Solution:
1. (a)
The same current must flow through each element in a series circuit. Therefore, the fuse should be
connected in series with the circuit it protects. Then, if the fuse burns out, no current will flow in the circuit.
2.
(b)
The best explanation is
III
. With the fuse connected in series the current in the circuit drops to zero as soon as the
fuse burns through. Statements I and II are false because an open circuit (blown fuse) in one branch of a parallel circuit
will not prevent current from flowing in other branches.
Insight:
Although household circuits are wired in parallel so that the same voltage appears across each outlet, a fuse or
circuit breaker is connected in series with the power source to shut down the entire circuit if two much current flows.
36.
Picture the Problem
: A circuit consists of three resistors,
1 2 3
R R R
<
<
, connected in series to a battery.
Strategy:
Use Ohm’s Law and the characteristics of series circuits to answer the question.
Solution:
1. (a)
The same current must flow through each element in a series circuit. Therefore, the ranking of currents
is I
1
= I
2
= I
3
.
2.
(b)
The potential difference across each circuit element is
V I R
=
(equation 212). With the same current I flowing
through each element, the potential difference V is largest for the largest resistor. We conclude that V
1
< V
2
< V
3
.
Insight:
Because
2
P I R=
(equation 215) we can also conclude that the circuit element with the largest resistance will
also dissipate the greatest amount of power.
37.
Picture the Problem
: Two resistors are connected in parallel, and then a third resistor is connected in parallel with the
original two.
Strategy:
Use the characteristics of parallel circuits to answer the question.
Solution:
1. (a)
The equivalent resistance of a circuit will decrease each time an additional resistor is connected in
parallel.
2.
(b)
The best explanation is
III
. The third resistor gives yet another path for current to flow in the circuit, which
means that the equivalent resistance is less. Statement I is false, and statement II is true only for series circuits.
Insight:
It seems counterintuitive that adding a resistor to a circuit will lower its equivalent resistance, but the effect is
to create a new path for the current to flow, allowing more current for the same voltage.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 10
38.
Picture the Problem
: Three resistors are connected as shown in the circuit at the right:
Strategy:
The top two resistors are connected in series with each other, and their
resistances add (equation 217). That pair of resistors is connected in parallel with the
bottom resistor, and the resistances combine in the manner described by equation 2110.
Solution:
Apply equations 217
and 2110 to find
eq
:
R
eq
eq top 3
1 1 1 1 1
33
35 82 45
R
R R R
=
+ = + ⇒ = Ω
Ω+ Ω Ω
Insight:
The equivalent resistance is smaller than the smallest (35 Ω) resistor in the group!
39.
Picture the Problem
: A number of identical resistors are connected in parallel.
Strategy:
The resistances of resistors connected in parallel combine in the manner described by equation 2110. Write
this equation such that N identical resistors of 65 Ω each combine to create an equivalent resistance of 11 Ω.
Solution:
Set
eq
11 R = Ω and solve for N:
1 1 1 1
...
11 65 65 65 65
65 11 5.9 or 6 resistors
N
N
= + + + =
Ω
Ω Ω Ω Ω
= Ω Ω=
Insight:
If only five resistors were connected in parallel, the equivalent resistance would be 13 Ω. The six resistors
produce an equivalent resistance of 10.8 Ω, just under the maximum allowed equivalent resistance of 11 Ω.
40.
Picture the Problem
: Four light bulbs are connected together with a battery in a circuit of unknown arrangement.
Stategy:
Study the chart provided with the problem statement to find clues about how the bulbs must be connected.
Solution:
First note that bulb C is always on unless it is removed, and when it is
removed, all other bulbs go out. We conclude it must be connected in series with
the battery. Secondly, we note that bulbs B and D mirror each other; when one is
off, the other is off, and when one is on, the other is on as well. We conclude that
B and D must be connected in series. Finally, we see that removing A has no
effect on the other lamps, so it must be connected in parallel with B and D but in
series with C. The subsequent circuit diagram is indicated at right.
Insight:
Brightness information was not included in this problem, but removing bulb A would in fact have the effect of
dimming bulb C slightly and brightening bulbs B and D slightly. This is because A draws extra current and lowers the
effective resistance of the portion of the circuit that is past bulb C. Removing it will increase the effective resistance of
the circuit, reducing the current through C and decreasing the voltage drop across C. That in turn would increase the
voltage drop across B and D, slightly brightening them.
41.
Picture the Problem
: The power cord of a toaster represents a small additional
resistance in the circuit, as indicated in the circuit diagram at right.
Strategy:
The current in the entire circuit is
eq
I V R
=
(equation 212). Since the
power cord is connected in series with the heating element, the equivalent resistance
is simply
eq 1 2
R R R= +
(equation 217). Once we know
eq
R
we can find the current
and then the power dissipated in each circuit element by using equation 215,
2
P I R=
.
Solution:
1. (a)
Combine eqs. 212,
215, and 217 to find
cord
P
:
( )
2
2
2
cord cord cord
eq
120 V
0.020 3.1 W
0.020 9.6
V
P I R R
R
⎛ ⎞
⎛ ⎞
= = = Ω =
⎜ ⎟
⎜ ⎟
⎜ ⎟
+ Ω
⎝ ⎠
⎝ ⎠
2.
(b)
Repeat part (a) to find
he
P
:
( )
2
2
2
he he he
eq
120 V
9.6 1.5 kW
0.020 9.6
V
P I R R
R
⎛ ⎞
⎛ ⎞
= = = Ω =
⎜ ⎟
⎜ ⎟
⎜ ⎟
+ Ω
⎝ ⎠
⎝ ⎠
Insight:
In this simple circuit the heating element has 480 times more resistance and dissipates 480 times more power
than does the resistance in the power cord.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 11
42.
Picture the Problem
: A hobbyist connects three resistors of known values in order
to produce a certain effective resistance.
Strategy:
The configuration of resistors can be determined with logical reasoning.
Connecting all three of the resistors in series would make
eq
391
R
=
Ω
, higher than
the 150Ω desired effective resistance. Connecting the three in parallel would make
eq
36
R = Ω
, smaller than the desired resistance. We could connect two resistors in
series, then connect the pair in parallel with the third resistor. The three possible
arrangements produce
eq
R
values of 63 Ω, 70 Ω, and 96 Ω, all too small. Therefore,
we must connect two resistors in parallel and then place that pair in series with the
third resistor. The three possible arrangements yield
eq
R
values of 144 Ω, 150 Ω,
and 263 Ω.
Solution: 1.
Connect the 220Ω and 79Ω resistors in parallel. Then connect this pair in series with the 92Ω resistor.
2.
Check that the solution produces the correct
eq
R
:
1
eq
1 1
92 150
220 79
R
−
⎛ ⎞
=
+ + Ω= Ω
⎜ ⎟
Ω Ω
⎝ ⎠
Insight:
One of the three possible configurations, the one with the two smaller resistances in parallel and connected in
series with the 220Ω resistor, can be ruled out because the effective resistance would always be greater than 220 Ω.
43.
Picture the Problem
: Three resistors are connected in series with a battery, as
indicated in the diagram at the right.
Strategy:
First find the equivalent resistance of the circuit using equation 217 and
then apply Ohm’s Law to find the current flowing through the circuit. Since the
same current flows through each resistor, Ohm’s Law can be applied again to find
the potential difference across each resistor.
Solution:
1. (a)
Apply Ohm’s Law
with equation 217 to find I:
eq
12.0 V
0.071 A 71 mA
42 17 110
V
I
R
= = = =
Ω+ Ω+ Ω
2. (b)
Combine the expression from
step 1 with Ohm’s Law to find
1
V
:
(
)
(
)
1 1
0.071 A 42 3.0 V
V I R= = Ω =
3.
Repeat to find
2
V
:
(
)
(
)
2 2
0.071 A 17 1.2 V
V I R= = Ω =
4.
Repeat to find
3
V
:
(
)
(
)
3 3
0.071 A 110 7.8 V
V I R= = Ω =
Insight:
The sum of the potential drops is
3.0 1.2 7.8 V 12.0 V
+
+ =
so that the net
V
Δ
around the whole circuit is zero.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 12
44.
Picture the Problem
: Three resistors are connected in series with a battery as
indicated in the diagram at right.
Strategy:
First write an expression for the equivalent resistance of the circuit using
equation 217 and then apply Ohm’s Law to find the unknown resistance R. Since
the same current flows through each resistor, Ohm’s Law can be applied again to
find the potential difference across each resistor.
Solution:
1. (a)
Combine Ohm’s
Law with equation 217:
eq 1 2
V
R R R R
I
=
= + +
2.
Solve the expression for R:
1 2
24.0 V
11 53 86
0.16 A
V
R R R
I
=
− − = − Ω− Ω= Ω
3.
(b)
Now apply Ohm’s Law to
1
R
to find
1
V
:
(
)
(
)
1 1
0.16 A 11 1.8 V
V I R= = Ω =
4.
Repeat to find
2
V
:
(
)
(
)
2 2
0.16 A 53 8.5 V
V I R= = Ω =
5.
Repeat to find
3
V
:
(
)
(
)
3
0.16 A 86 ) 14 V
V I R= = Ω =
6. (c)
Since
,
R V∝
the answer to part (a) would have been larger if
24.0 V.
V >
Insight:
To within the allowed number of significant digits, the total potential drop across the three resistors is
1.8 8.5 14 V 24 V,+ + =
so that the net
VΔ
around the whole circuit is zero.
45.
Picture the Problem
: Three resistors are connected in parallel with a battery as
indicated in the diagram at the right.
Strategy:
First determine the equivalent resistance of the circuit by combining the
resistances of the three resistors according to equation 2110. The known current
and equivalent resistance will then yield the emf of the battery, and once the emf is
known, Ohm’s Law can be applied to find the current through each resistor.
Solution:
1. (a)
Determine
eq
:
R
1
eq
1 1 1
16
65 25 170
R
−
⎛ ⎞
=
+ + = Ω
⎜ ⎟
Ω Ω Ω
⎝ ⎠
2.
Apply Ohm’s Law to find
:
ε
=
(
)( )
eq
1.8 A 16 29 V
I R
ε
= = Ω =
3.
(b)
Now apply Ohm’s Law to
1
R
to find
1
:
I
1
29 V
0.45 A
65
I
= =
Ω
4.
Repeat step 3 to find
2
:
I
2
29 V
1.2 A
25
I = =
Ω
5.
Repeat step 3 to find
3
:
I
3
29 V
0.17 A
170
I = =
Ω
Insight:
Note that for this parallel circuit the voltages are all the same, but the currents are different and sum to 1.8 A.
46.
Picture the Problem
: Three resistors are connected in parallel with a battery as
indicated in the diagram at right.
Strategy:
First write an expression for the equivalent resistance of the circuit using
equation 2110 and then apply Ohm’s Law to find the unknown resistance
R
. Since
the same potential difference exists across each resistor, Ohm’s Law can be applied
again to find the current through each resistor.
Solution:
1. (a)
Combine Ohm’s
Law with Eq. 2110:
eq 1 2
1 1 1 1
I
R R R R
ε
=
= + +
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 13
2.
Solve the expression for
R
:
1
1
1 2
1 1 0.88 A 1 1
77
12.0 V 22 67
I
R
R R
ε
−
−
⎛ ⎞
⎛ ⎞
=
− − = − − = Ω
⎜ ⎟
⎜ ⎟
Ω Ω
⎝ ⎠
⎝ ⎠
3. (b)
Now apply Ohm’s Law to
1
R
to find
1
I
:
1
12.0 V
0.55 A
22
I = =
Ω
4.
Repeat step 3 to find
2
I
:
2
12.0 V
0.18 A
67
I = =
Ω
5.
Repeat step 3 to find
3
I
:
3
12.0 V
0.16 A
77
I
= =
Ω
6. (c)
Since
1,
R I∝
the answer to part (a) would have been smaller if
0.88 A.
I >
Insight:
An alternative method to solve this problem involves first finding
1 2
and
I
I
as in steps 3 and 4 and then
subtracting their sum from the total current to get 0.15 A of current through the unknown resistor
R
, then applying
Ohm’s Law to find
80
R = Ω
. Besides the rounding error, using that method makes it more difficult to answer part (c)
because you would not have found an algebraic expression for
R
in terms of
I
.
47.
Picture the Problem
: Two resistors are connected in series with a battery as shown
in the diagram at right.
Strategy:
The current will be the same through the two resistors and their
resistances will add according to equation 217 because they are connected in series.
Use the equivalent resistance of the circuit together with Ohm’s Law to determine
the emf of the battery.
Solution:
Apply Ohm’s Law to find
ε
㨠
(
)
(
)
eq
0.72 A 89 130
160 V 0.16 kV
I R
ε
=
= Ω+ Ω
= =
Insight:
Verify for yourself that if the two resistors were instead connected in parallel, and that 0.72 A flows through the
89Ω resistor, the emf of the battery that is connected across the pair would be only 64 V.
48.
Picture the Problem
: Three resistors are connected in the manner indicated by the
diagram at the right.
Strategy:
Use the rules concerning resistors in series and in parallel to write an
expression for the equivalent resistance of the circuit in terms of
R
, and then solve
the expression for
R
. Begin by finding the equivalent resistance of the 55Ω resistor
connected in parallel with
R
, then add the equivalent resistance of the pair to 12 Ω
to find
eq
R
for the entire circuit.
Solution:
1.
Use equations 217 and
2110 to find an expression for
eq
R
:
1
eq
1
1 1
12 26
55
1 1 1 1 1
14
55 55 14
R
R
R R
−
−
⎛ ⎞
= Ω+ + = Ω
⎜ ⎟
Ω
⎝ ⎠
⎛ ⎞
+ = Ω ⇒ + =
⎜ ⎟
Ω
Ω Ω
⎝ ⎠
2.
Solve the expression for
R
:
1
1 1 1
14 55
1 1
19
14 55
R
R
−
= −
Ω Ω
⎛ ⎞
=
− = Ω
⎜ ⎟
Ω Ω
⎝ ⎠
Insight:
The 19Ω and 55Ω resistor pair combine to make an effective resistance of 14 Ω, which adds to 12 Ω to make
an effective resistance of 26 Ω for the entire circuit.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 14
49.
Picture the Problem
: Six resistors are connected in the manner indicated by the
diagram at the right.
Strategy:
Use the rules concerning resistors in series and in parallel to write an
expression for the equivalent resistance of the entire circuit Begin by finding the
equivalent resistance of the three uppermost resistors (4.8 Ω, 3.3 Ω, and 8.1 Ω),
then add their equivalent resistance to the 6.3Ω resistor. The equivalent resistance
of those four can then be added to the 1.5Ω and 2.5Ω resistors according to
equation 2110 because the three are connected in parallel.
Solution:
Add the uppermost three resistors in
parallel, then add them in series with the 6.3Ω
resistor, then add the result in parallel with the
1.5Ω and 2.5Ω resistors to find
eq
:
R
( )
1
1 1 1
eq
4.8 3.3 8.1
eq
1 1 1 1
1.5 2.5
6.3
0.84
R
R
−
Ω Ω Ω
= + +
Ω Ω
Ω+ + +
= Ω
Insight:
The equivalent resistance of the entire circuit is less than the smallest resistor (1.5 Ω) in the network.
50.
Picture the Problem
: A number of light bulbs are connected in parallel across a given potential difference.
Strategy:
Because the lamps are connected in parallel, the total current drawn in the circuit is the sum of the currents
drawn by each lamp. Find the current drawn by each lamp by using its power rating together with equation 214 and
then divide the total allowed current by the current per lamp to find the number of lamps
N
.
Solution:
1.
Use equation 214 to find
lamp
I
:
lamp
lamp
65 W
0.765 A
85 V
P
I
V
= = =
2.
Find
N
using the total allowed current:
total
lamp
2.1 A
2.7 lamps
0.765 A
I
N
I
= = =
3.
Adding a third lamp will cause the circuit current to exceed 2.1 A, so only two lamps can be connected in parallel.
Insight:
If the voltage were increased to 125 V, each lamp would only draw 0.52 A and four lamps could be connected
before the circuit current would exceed 2.1 A.
51.
Picture the Problem
: Five resistors are connected to a battery that has an internal
resistance, as indicated in the circuit diagram at the right.
Strategy:
We must first find the equivalent resistance of the entire circuit and then
apply Ohm’s Law to determine the current
I
that is drawn from the battery. To
accomplish this we note that the 7.1Ω and 5.8Ω resistors are connected in series,
and their 12.9Ω equivalent resistance is connected in parallel with the 3.2Ω
resistor. The combination of those three resistors is connected in series with
r
and
the 4.5Ω and 1.0Ω resistors. Once the current
I
is known, the voltage drop across
r
and the 4.5Ω and 1.0Ω resistors can be found, and the remaining potential
difference across the 3.2Ω resistor and 12.9Ω combination can be used together
with Ohm’s Law to find the two branch currents
1 2
and .
I
I
The current
I
is the
current through the battery (see the diagram) and can be used to find the potential
drop across the internal resistance
r
and hence the potential across the terminals of
the battery by using Ohm’s Law.
Solution:
1. (a)
Find
eq
R
by applying
equations 127 and 1210 as appropriate:
1
eq
1 1
0.50 4.5 1.0 8.6
3.2 7.1 5.8
R
−
⎛ ⎞
=
Ω+ Ω+ Ω+ + = Ω
⎜ ⎟
Ω Ω+ Ω
⎝ ⎠
2.
Apply Ohm’s Law to find
I
:
eq
12.0 V
1.4 A
8.6
V
I
R
= = =
Ω
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 15
3.
Determine
VΔ
across
r
and the 4.5Ω
and 1.0Ω resistors, and therefore
VΔ
across the 3.2Ω resistor:
(
)
( )( )
3.2
4.5 1.0
12.0 V 1.4 A 0.50 4.5 1.0 3.6 V
V I r
ε
Ω
Δ = − + Ω+ Ω
= − Ω+ Ω+ Ω =
4.
The 3.6 V potential difference
drives current through the 3.2Ω
and 7.1Ω resistors:
3.2
1 3.2
3.2
2 7.1
3.6 V
1.1 A
3.2
3.6 V
0.29 A
7.1 5.8
V
I I
R
V
I
I
R
Ω
Ω
Ω
Ω
Δ
= = = =
Ω
Δ
= = = =
+ Ω
5. (b)
The current
I
was found in step 2:
1.4 AI =
6.
(c)
Use
I
to find
b
attery
VΔ
:
(
)
(
)
battery
12.0 V 1.4 A 0.50 11.3 VV Ir
ε
= − = − Ω =
Insight:
Kirchoff’s Rules could also be applied to two circuit loops in this problem to find
1 2
and ,
I
I
but the solution is
no simpler than the one presented here.
52.
Picture the Problem
: Three resistors are connected in the manner indicated by the
diagram at right, and a 12V battery is connected to terminals A and B.
Strategy:
Use the rules concerning resistors in series and in parallel to determine
the equivalent resistance of the circuit. Begin by finding the equivalent resistance
of the 55Ω resistor connected in parallel with
R
, and then add the equivalent
resistance of the pair to 12 Ω to find
eq
R
for the entire circuit. Apply Ohm’s Law to
the entire circuit in order to find the current
3
I
through the 12Ω resistor. Ohm’s
Law applied to the 12Ω resistor by itself gives the voltage drop across the resistor, which can be subtracted from the
total 12V potential across A and B to give the voltage across the two resistors connected in parallel. Ohm’s Law can
then be applied to each of those resistors in order to find the individual currents.
Solution:
1. (a)
Find
eq
R
for the circuit if
R
= 85 Ω:
1
eq
1 1
12 45.4
55 85
R
−
⎛ ⎞
=
Ω+ + = Ω
⎜ ⎟
Ω Ω
⎝ ⎠
2.
Apply Ohm’s Law to find
3
:
I
3
12 V
0.26 A
45.4
I = =
Ω
3.
Find
VΔ
across the parallel resistors:
( )( )
2 3 3
12.0 V 0.26 A 12 8.9 V
V I R
ε
Δ = − = − Ω =
4.
Apply Ohm’s Law to the 55Ω resistor to find
1
I
:
2
1
1
8.9 V
0.16 A
55
V
I
R
Δ
= = =
Ω
5.
Repeat for the 85Ω resistor:
2
2
8.9V
0.10 A
85
V
I
R
Δ
= = =
Ω
6.
(b)
Increasing
R
causes the resistance of the parallel path and of the entire circuit to increase, which in turn reduces
the total current
3
.
I
Because
R
is now larger,
1
I
will get a proportionately larger share of the total current and
2
I
a
smaller share. As an extreme example, suppose
R
increases to infinity. Then
3
12 V 67 0.18 AI
=
Ω=
and
1 3
I
I=
because no current can pass through
R
. Thus, the current through the 12Ω resistor and
R
decreases, and the
current through the 55Ω resistor increases.
Insight:
Another way to accomplish step 5 is to realize that
2
I
must be the difference between
3
I
and
2
,
I
or
0.260.16 A 0.10 A.
− =
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 16
53.
Picture the Problem
: Six resistors are connected in the manner indicated
in the diagram, and a 9.0V battery is connected to terminals A and B.
Strategy:
Find the equivalent resistance
eq
R
of the righthand portion of
this circuit by combining resistors 4, 5, and 6 in parallel, and then
combining that group in series with resistor 3. The voltage drops across
resistors 1, 2, and
eq
R
are each 9.0 V. Ohm’s Law can then be used to find
1 2 3
, , and .
I
I I
Use Ohm’s Law together with
3
I
to find the voltage drop
across resistor 3, then subtract that result from 9.0 V to find the voltage
drop across resistors 4, 5, and 6. Ohm’s Law can then be applied to each
of those resistors to find
4 5 6
, , and .
I
I I
Solution:
1. (a)
Use Ohm’s Law to find
1
:
I
1
9.0 V
6.0 A
1.5
I
= =
Ω
2.
Repeat step 1 to find
2
:
I
2
9.0 V
3.6 A
2.5
I
= =
Ω
3.
Find
eq
R
for the righthand portion of the circuit:
1
eq
1 1 1
6.3 7.9
4.8 3.3 8.1
R
−
⎡ ⎤
⎛ ⎞
=
+ + + Ω = Ω
⎢ ⎥
⎜ ⎟
Ω Ω Ω
⎝ ⎠
⎢ ⎥
⎣ ⎦
4.
Apply Ohm’s Law to the righthand branch:
3
9.0 V
1.14 A 1.1 A
7.9
I
= = =
Ω
5.
Find
VΔ
across resistors 4, 5, and 6:
( )( )
3 3
9.0 V 1.14 A 6.3 1.8 VV I R
ε
Δ = − = − Ω =
6.
Apply Ohm’s Law to find
4
:
I
4
1.8 V
0.38 A
4.8
I = =
Ω
7.
Repeat step 6 to find
5
:
I
5
1.8 V
0.55 A
3.3
I
= =
Ω
8.
Repeat step 6 to find
6
:
I
6
1.8 V
0.22 A
8.1
I
= =
Ω
9. (b)
There is a potential drop across resistors 4, 5, and 6 (1.8 V in this case), resulting in a potential difference across
the 6.3Ω resistor that is less than the 9.0 V that is across the 1.5Ω resistor.
Insight:
The total current in the circuit is
1 2 3
10.7 AI I I
+
+ =
, which is consistent with the equivalent resistance of
0.84 Ω found in problem 49 if you apply Ohm’s Law to the entire circuit:
9.0 V 0.84 10.7 A.I V R
=
= Ω=
54.
Picture the Problem
: Five resistors are connected to a battery that has an internal
resistance, as indicated in the circuit diagram at the right.
Strategy:
We must first find the equivalent resistance of the entire circuit and then
apply Ohm’s Law to determine the current
I
that is drawn from the battery. To
accomplish this we note that the 7.1Ω and 5.8Ω resistors are connected in series,
and their 12.9Ω equivalent resistance is connected in parallel with the 3.2Ω
resistor. The combination of those three resistors is connected in series with
r
and
the 4.5Ω and 1.0Ω resistors. Once the current
I
through the battery is known, it
can be used to find the potential drop across the internal resistance
r
and hence the
potential across the terminals of the battery by using Ohm’s Law.
Solution:
1. (a)
Find
eq
R
by applying
equations 127 and 1210 as appropriate:
1
eq
1 1
0.25 4.5 1.0 8.3
3.2 7.1 5.8
R
−
⎡ ⎤
=
Ω+ Ω+ Ω+ + = Ω
⎢ ⎥
Ω Ω+ Ω
⎣ ⎦
2.
Apply Ohm’s Law to find
I
:
12 V
1.4 A
8.3
I = =
Ω
3.
(b)
Use
I
to find
b
attery
V
Δ
:
(
)
(
)
battery
12.0 V 1.4 A 0.25 11.6 VV Ir
ε
= − = − Ω =
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 17
4.
(c)
Increasing the resistance of the 3.2 Ω resistor will increase
eq
R
and decrease
I
, hence the current
1
I
will decrease.
Insight:
The current in the circuit is essentially the same as that found in problem 51 because decreasing the internal
resistance from 0.50 Ω to 0.25 Ω represents a small change compared to the overall resistance of more than 8 Ω. If the
3.2Ω resistance were increased to infinity,
eq
R
would be 18.7 Ω and
I
would be reduced to 0.64 A.
55.
Picture the Problem
: Six resistors are connected to a battery in the
manner indicated in the diagram.
Strategy:
The voltage drop from A to B can be determined from the given
current
2
I
by applying Ohm’s Law to resistors 5 and 6, both of which
have current
2
I
flowing through them. Once
AB
V
Δ
is known, Ohm’s Law
will allow us to calculate
3 4
and
I
I
, and since
1 2 3 4
I
I I I
=
+ +
, we can find
the total current
1
I
and the voltage drop across resistors 1 and 2. The
voltage of the battery is the sum of the voltage drops across resistors 1 and
2 plus the voltage drop
AB
VΔ
from A to B.
Solution:
1. (a)
Apply Ohm’s Law to
resistors 5 and 6 to find
AB
VΔ
:
(
)
(
)
AB
1.52 A 8.45 4.11 19.1 VVΔ = Ω+ Ω =
2.
Use
AB
VΔ
to find
3
:
I
AB
3
3
19.1 V
1.38 A
13.8
V
I
R
Δ
= = =
Ω
3.
Use
AB
VΔ
to find
4
I
:
4
19.1 V
1.11 A
17.2
I = =
Ω
4.
Add the currents to find
1
I
:
1 2 3 4
1.521.381.11 A 4.01 AI I I I
=
+ + = + + =
5.
Find
VΔ
across
1 2
and R R
:
(
)
(
)
(
)
12 1 1 2
4.01 A 15.0 12.5 110 VV I R RΔ = + = + Ω =
6.
Find the battery voltage:
12 AB
110 19.1 V 129 VV V
ε
= Δ + Δ = + =
7.
(b)
Increasing the resistance of
4
R
increases the total resistance of the circuit, causing the current
1
I
to decrease.
Insight:
Verify for yourself that if
4
R
were increased to infinity, the current
1
I
provided by the battery would decrease
from 4.01 A to 3.80 A.
56.
Picture the Problem
: Six resistors are connected to a battery in the
manner indicated in the diagram.
Strategy:
The current
3
I
through the 13.8 Ω resistor determines the
voltage
AB
VΔ
because of Ohm’s Law. That potential difference can
then be used to find
2 4
and
I
I
, and the current
1
I
is the sum of the
other three currents.
Solution:
1. (a)
Find
AB
VΔ
using Ohm’s Law:
(
)
(
)
AB 3 3
0.795 A 13.8 11.0 VV I RΔ = = Ω =
2.
Apply Ohm’s Law to
4
R
to find
4
I
:
AB
4
4
11.0 V
0.640 A
17.2
V
I
R
Δ
= = =
Ω
3.
Apply Ohm’s Law to
5 6
and R R
to find
2
I
:
AB
2
5 6
11.0 V
0.876 A
8.454.11
V
I
R R
Δ
= = =
+ + Ω
4.
Add the currents to find
1
I
:
1 2 3 4
0.8760.7950.640 A 2.31 A
I I I I= + + = + + =
Insight:
Ohm’s Law is a powerful tool for analyzing circuits because it applies to the entire circuit as a whole as well as
to each individual element in the circuit.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 18
57.
Picture the Problem
: Four resistors and a switch are connected to a battery as
shown in the diagram at the right.
Strategy:
The symmetric nature of the circuit gives clues to its behavior. Because
the total resistances for the two paths on either side of the switch are the same, equal
currents will flow through each branch. Because the top two resistors in each
branch are the same, the voltage drops across each will be the same as well. That
means the electric potential at either side of the switch will be the same, and
therefore the potential difference across the switch is zero and no current will flow
through the switch when it is closed. Use Ohm’s Law and the rules for calculating
equivalent resistance (equations 217 and 2110) to verify these observations.
Solution:
1. (a)
Because the resistors are identical, the potential is the same at both sides of the switch regardless of
whether it is open or closed. Therefore the current through the battery will stay the same when the switch is closed.
2.
(b)
Find the current
0
I
through the battery when the switch is open.
The circuit is two parallel branches of resistors connected in series:
0
eq
1
2
I
R R R R
ε
ε
ε
1
⎛ ⎞
=
= + =
⎜ ⎟
2
⎝ ⎠
3.
Find the current
I
through the battery when the
switch is closed. The circuit is a series connection
of two sets of resistors that are connected in parallel:
( )
( )
0
1 1
1 1 1 1
eq
R R R R
I
I
R R
ε
ε ε
− −
= = = =
+ + +
Insight:
If the bottom two resistors were each doubled to 2
R
, the current
0
I
would decrease but the current
I
when the
switch is closed would still equal
0
I
.
58.
Picture the Problem
: A single loop circuit is constructed from two batteries and
three resistors as indicated in the diagram at the right.
Strategy:
Use Kirchoff’s Loop Rule to determine the current in the circuit. Both
batteries are pushing the current clockwise, so that is the expected direction of
current flow. Sum the changes in potential across each circuit element as you go
clockwise around the circuit and set the sum equal to zero in order to find the
current. If the current turns out to be negative, it must be flowing counterclockwise
after all.
Solution:
1.
Set
0VΔ =
as you go clockwise
around the entire loop, starting at point B:
(
) ( )
(
)
15.0 V 8.50 11.5 V 6.22 15.1 0
15.0 11.5 V
0.889 A
8.50 6.22 15.1
I I I
I
+
− Ω + − Ω − Ω =
+
= =
+ + Ω
2.
Since the calculated current is positive, we conclude our prediction was correct that the current flows clockwise.
Insight:
A shortcut solution for a singleloop circuit like this one is to construct an equivalent circuit with a 26.5V
battery connected to a 29.8Ω resistor and then apply Ohm’s Law.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 19
59.
Picture the Problem
: A single loop circuit is constructed from two batteries and
three resistors as indicated in the diagram at the right.
Strategy:
Use Kirchoff’s Loop Rule to determine the current in the circuit. The
15.0V battery is pushing the current clockwise, and the 11.5V battery is pushing
counterclockwise. The net effect is a 3.5V battery pushing clockwise, so that is the
expected direction of current flow. Sum the changes in potential across each circuit
element as you go clockwise around the circuit, and then set the sum equal to zero
in order to find the current. If the current turns out to be negative, it must be
flowing counterclockwise after all.
Solution:
1. (a)
Since the two batteries now oppose each other, the net emf is less than before, and so the current will
decrease.
2. (b)
Set
0VΔ =
as you go clockwise
around the entire loop, starting at point B:
(
) ( )
(
)
15.0 V 8.50 11.5 V 6.22 15.1 0
15.0 11.5 V
0.12 A
8.50 6.22 15.1
I I I
I
+
− Ω − − Ω − Ω =
−
= =
+ + Ω
3.
Since the calculated current is positive, we conclude our prediction was correct that the current flows clockwise.
Insight:
A shortcut solution to a singleloop circuit like this one is to construct an equivalent circuit with a 3.5V battery
connected to a 29.8Ω resistor and then apply Ohm’s Law. Note that there are only two significant digits in the answer
due to the rules of subtraction:
15.0 11.5 V 3.5 V− =
(only two significant figures).
60.
Picture the Problem
: A single loop circuit is constructed from two batteries and
three resistors as indicated in the diagram at the right.
Strategy:
Both batteries are pushing the current clockwise, so that is the expected
direction of current flow. Sum the changes in potential across each circuit element
as you go clockwise around the circuit, and then set the sum equal to zero in order
to find the current. Once the current is known, Ohm’s Law can be applied to each
circuit element to determine the potential difference across that element.
Solution:
1. (a)
The direction of current flow produces a potential drop from
A
to
B
, so the potential at
B
is less than
zero.
2.
(b)
The direction of current flow produces a potential drop from
C
to
A
, so the potential at
C
is greater than zero.
3. (c)
Apply Kirchoff’s loop rule to find
I
:
(
) ( )
(
)
15.0 V 8.50 11.5 V 6.22 15.1 0
15.0 11.5 V
0.889 A
8.50 6.22 15.1
I I I
I
+
− Ω + − Ω − Ω =
+
= =
+ + Ω
4.
Starting with
A
V
, subtract the drop across
the 15.1Ω resistor and add the 15.0V emf of
the battery to find
D
V
:
( )( )
D A 15.1 15.0
D
0 0.889 A 15.1 15.0 V
1.6 V
V V IR
V
ε
=
− +
= − Ω +
=
Insight:
An analysis similar to that in part (a) reveals that
B C
13.4 V and 5.53 V.V V
=
− = +
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 20
61.
Picture the Problem
: Four resistors and a battery are connected as shown
in the circuit diagram at the right.
Strategy:
The circuit can be analyzed by combining resistors according to
the rules of equations 217 and 2110 or by applying Kirchoff’s rules. For
the former case, let
2 4
and R R
add in series, and add that combination in
parallel with
3
.R
The effective resistance of those three are then added in
series with
1
R
to get the equivalent resistance of the entire circuit, from
which the current
1
I
can be determined. Then Ohm’s Law can be applied
to find
2
I
and
3
I
. When applying Kirchoff’s rules, the equations of two loops and one junction can be combined to
algebraically find
12 3
, , and .
I
I I
Solution:
1. (a)
Find
eq
R
of the circuit:
1
1
eq 1
3 2 4
1 1 1 1
11
7.5 6.2 12
16.3
R R
R R R
−
−
⎡ ⎤
⎡ ⎤
=
+ + = + + Ω
⎢ ⎥
⎢ ⎥
+ Ω + Ω
⎣ ⎦
⎣ ⎦
= Ω
2.
Find
1
I
by applying
Ohm’s Law to the circuit:
1
eq
15 V
0.92 A
16.3
V
I
R
= = =
Ω
3.
Find
AB
VΔ
by applying Ohm’s Law to
1
R
:
(
)( )
AB 1 1
15.0 V 0.92 A 11 4.9 VV I R
ε
Δ = − = − Ω =
4.
Determine
2
I
by applying
Ohm’s Law to
2
R
and
4
R
:
AB
2
2 4
4.9 V
0.27 A
6.2 12
V
I
R R
Δ
= = =
+ + Ω
5.
Determine
3
I
by applying Ohm’s Law to
3
R
:
AB
3
3
4.9 V
0.65 A
7.5
V
I
R
Δ
= = =
Ω
6. (b)
Apply the junction rule to point B:
1 2 3
0I I I
−
− =
7.
Apply the loop rule to loop 1 starting at C:
3 3 1 1 3 3 1 1
0
I
R I R I R I R
ε
ε
+
− − = ⇒ = −
8.
Apply the loop rule to loop 2 starting at A:
3 3 2 2 2 4
0I R I R I R
+
− − =
9.
Substitute the loop 1 expression
for
3 3
I
R
into the loop 2 expression:
(
)
(
) ( )
(
)
1 1 2 2 4 2 1 1 2 4
I
R I R R I I R R R
ε
ε
− = + ⇒ = − +
10.
Substitute the expressions for
2 3
and
I
I
from steps 7 and 9 into the
j
unction rule of step 6 and rearrange:
(
)
( )
1 1
1 1
1
2 4 3
1 1
1
2 4 3 2 4 3
0
1 1
1
I R
I R
I
R R R
R R
I
R R R R R R
ε
ε
ε
−
−
− − =
+
⎡
⎤ ⎡ ⎤
+ + = +
⎢
⎥ ⎢ ⎥
+ +
⎣
⎦ ⎣ ⎦
11.
Isolate
1
I
on the left side and then divide
top and bottom by the term in brackets:
2 4 3
1
1
1
1
2 4 3
2 4 3
1 1
1 1
1 1
1
R R R
I
R
R
R R R
R R R
ε
ε
−
⎡ ⎤
+
⎢ ⎥
+
⎣ ⎦
= =
⎡ ⎤
⎡ ⎤
+ +
+ +
⎢ ⎥
⎢ ⎥
+
+
⎣ ⎦
⎣ ⎦
12.
Insert numerical values:
1
1
15 V
0.92 A
1 1
11
6.2 12 7.5
I
−
= =
⎡ ⎤
+ + Ω
⎢ ⎥
+ Ω Ω
⎣ ⎦
13.
Substitute
1
I
into the expression from step 9:
(
)
( )
(
)( )
1 1
2
2 4
15.0 V 0.92 A 11
0.27 A
6.2 12
I R
I
R R
ε
− − Ω
= = =
+ + Ω
14.
Substitute
1 2
and
I
I
into the loop rule from step 6:
3 1 2
0.92 0.27 A 0.65 AI I I= − = − =
Insight:
For this circuit the application of Kirchoff’s rules was more complicated than simply applying the rules for
combining resistors as we did in part (a). In some circuits, however, Kirchoff’s rules are an indispensable tool.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 21
62.
Picture the Problem
: Four resistors and a battery are connected as shown
in the circuit diagram at the right.
Strategy:
Apply the rules of combining resistors by adding
2 4
and R R
in
series, then adding that combination in parallel with
3
.R
The effective
resistance of those three are then added in series with
1
R
to get the
equivalent resistance of the entire circuit, from which the current
1
I
can be
determined using Ohm’s Law. Then Ohm’s Law can be applied to
1
R
in
order to find
B
V
and
C
V
.
Solution:
1.
Find
eq
R
of the circuit:
1
1
eq 1
3 2 4
1 1 1 1
11
7.5 6.2 12
16.3
R R
R R R
−
−
⎡ ⎤
⎡ ⎤
=
+ + = + + Ω
⎢ ⎥
⎢ ⎥
+ Ω + Ω
⎣ ⎦
⎣ ⎦
= Ω
2.
Find
1
I
by applying
Ohm’s Law to the circuit:
1
eq
15 V
0.92 A
16.3
V
I
R
= = =
Ω
3.
Subtract the voltage drop
across
1
R
from
A
V
to find
C
V
:
(
)
(
)
C A 1 1
0 0.92 A 11 10 VV V I R= − = − Ω = −
4.
Add 15 V to
C
V
to find
B
V
:
B C
10 V 15 V 5 VV V
ε
= + = − + =
Insight:
If point C were grounded instead,
B
V
would be +15 V and
A
V
would be +10 V.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 22
63.
Picture the Problem
: Four resistors and two batteries are connected as
shown in the circuit diagram at right.
Strategy:
The circuit can be analyzed by applying Kirchoff’s rules. First
apply the Junction Rule to point A in the circuit, then apply the Loop Rule
to two loops, the lefthand loop 1 and the outside loop 2 labeled in the
diagram. These three equations can be combined to algebraically find
1 2 3
, , and .
I
I I
From the currents we can find the potential difference
between the points A and B.
Solution:
1. (a)
Apply the
Junction Rule to point A:
1 2 3
(i)I I I
=
+
……
2.
Apply the Loop Rule to
loop 1, beginning in lower left
hand corner, and solve for
3
I
:
11 33 14
1 4
3 1 1
3 3
012 V
12 V 13.7
10 A (ii)
1.2
I R I R I R
R R
I I I
R R
= − − −
+
Ω
= − = −
Ω
……
3.
Apply the Loop Rule to
loop 2, beginning in lower left
hand corner, and solve for
2
I
:
11 22 14
1 4
2 1 1
2 2
012 V 9.0 V
12.09.0 V 3.0 V 13.7
(iii)
6.7 6.7
I R I R I R
R R
I I I
R R
=
− − − −
+
− Ω
⎛ ⎞ ⎛ ⎞
= − = −
⎜ ⎟ ⎜ ⎟
Ω Ω
⎝ ⎠ ⎝ ⎠
……
4.
Substitute equations (ii)
and (iii) into equation (i).
1 1 1
1
3
6.7
1
13.7 13.7
1.2 6.7
13.7 3.0 13.7
10 A A
1.2 6.7 6.7
13.7 13.7 3
1 10 A A
1.2 6.7 6.7
10 A A
0.72 A
1
I
I I
I
I
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= − + −
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞
+ + = +
⎜ ⎟
⎝ ⎠
+
= =
+ +
5.
Substitute
1
I
into
equations (ii) and (iii):
( )
( )
3
2
13.7
10 A 0.72 A 1.8 A
1.2
3.0 13.7
A 0.72 A 1.0 A
6.7 6.7
I
I
= − =
⎛ ⎞
= − = −
⎜ ⎟
⎝ ⎠
6.
The currents through each resistor are as follows: 3.9 Ω, 9.8 Ω: 0.72 A; 1.2 Ω: 1.8 A; 6.7 Ω: 1.0 A.
7.
The potential at point
A
is greater than that at point
B
because
3
I
flows in the direction shown in the diagram and
produces a potential drop across
3
.R
8.
Find the potential drop across
3
:R
(
)
(
)
A B 3 3
1.8 A 1.2 2.2 VV V I R− = = Ω =
Insight:
Because we obtained a negative value for
2
I
, it must be flowing in the direction opposite that indicated by the
arrow in the circuit diagram.
64.
Picture the Problem
: Three resistors, a switch, and two batteries
are connected as shown in the circuit diagram at the right.
Strategy:
The circuit can be analyzed by applying Kirchoff’s
rules. First apply the Junction Rule to the intersection above the
9.0V battery, then apply the Loop Rule to two loops, the left
hand loop 1 and the righthand loop 2 labeled in the diagram.
These three equations can be combined to algebraically find
1 2 3
, , and .
I
I I
Only Loop 1 is needed to find the current through
the batteries when the switch is open.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 23
Solution:
1. (a)
Apply the Junction Rule:
1 2 3
... (i)I I I
+
=
2.
Apply the Loop Rule to
loop 1, beginning in lower left
hand corner, and solve for
1
I
:
1 1 2 2
2 2
1 2 2
1
0 6.0 V 9.0 V
3.0 V
3.0 V 4.0
1.5 A 2.0 ... (ii)
2.0 2.0
I R I R
I R
I I I
R
= − + − +
− +
− Ω
= = + = − +
Ω Ω
3.
Apply the Loop Rule to
loop 2, beginning in lower left
hand corner, and solve for
3
I
:
22 33
22
3 2 2
3
0 9.0 V
9.0 V
9.0 V 4.0
1.8 A 0.80 ... (iii)
5.0 5.0
I R I R
I R
I I I
R
= − + −
−
Ω
= = − = −
Ω Ω
4.
Substitute (ii) and (iii)
into (i) and solve for
2
I
:
(
)
(
)
2 2 2
2
1.5 A 2.0 1.8 A 0.80
3.3 A
0.87 A
3.8
I
I I
I
− + + = −
= =
5.
Now apply equation (ii):
(
)
1
1.5 2.0 0.87 A 0.24 AI = − + =
6.
About 0.9 A flows through the 9.0V battery, and about 0.2 A flows through the 6.0V battery.
7. (b)
When the switch is open, no
current flows in loop 2, so there
remains only the series circuit of loop 1.
Apply the Loop Rule, beginning in the
lower lefthand corner, and solve for
I
:
1 2
1 2
0 6.0 V 9.0 V
3.0 V 3.0 V
0.50 A
2.0 4.0
IR IR
I
R R
= − + − +
− −
= = = −
+ + Ω
Insight:
When the switch is open, 0.50 A flows through both batteries, but the minus sign indicates that it flows in the
counterclockwise direction, opposite the direction that is shown in the diagram.
65.
Picture the Problem
: Two capacitors,
C
1
=
C
and
C
2
= 2
C
, are connected to a battery.
Strategy:
Recall the characteristics of circuits containing capacitors connected in series and in parallel.
Solution:
1. (a)
If the two capacitors are connected in series, they have the same charge on their plates. Therefore, we
can compare their energies by considering the relation
2
2.
U Q C
=
We conclude that capacitor
C
1
stores twice as much
energy as does
C
2
.
2.
(b)
If the two capacitors are connected in parallel, they have the same potential difference between their plates.
Therefore, we can compare their energies by considering the relation
2
1
2
.U CV=
We conclude that in this case
capacitor
C
2
stores twice as much energy as does
C
1
.
Insight:
The fact that the smaller capacitor stores more energy when connected in series seems counterintuitive at first,
until you realize that it takes more work to put a given amount of charge on a small capacitor than it does to put the
same charge on a large capacitor.
66.
Picture the Problem
: Two capacitors are connected in series, and then a third capacitor is connected in series with the
original two.
Strategy:
Recall the characteristics of circuits containing capacitors connected in series.
Solution:
1. (a)
The equivalent capacitance of a circuit will decrease each time an additional capacitor is connected in
series.
2.
(b)
The best explanation is
III
. Adding a capacitor in series decreases the equivalent capacitance since each capacitor
now has less voltage across it, and hence stores less charge. Statements I and II are each false.
Insight:
Another way to understand the effect is to realize that adding another capacitor in series increases the effective
separation
d
between the plates, so that the capacitance
0
C A d
ε
=
decreases.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 24
67.
Picture the Problem
: Two capacitors are connected in parallel, and then a third capacitor is connected in parallel with
the original two.
Strategy:
Recall the characteristics of circuits containing capacitors connected in parallel.
Solution:
1. (a)
The equivalent capacitance of a circuit will increase each time an additional capacitor is connected in
parallel.
2.
(b)
The best explanation is
II.
Adding a capacitor in parallel will increase the total amount of charge stored, and
hence increase the equivalent capacitance. Statement I is false and statement III is true only for capacitors in series.
Insight:
Another way to understand the effect is to realize that adding another capacitor in parallel increases the
effective plate area
A
, so that the capacitance
0
C A d
ε
=
increases.
68.
Picture the Problem
: Three capacitors are connected together as shown in the
diagram at the right.
Strategy:
Following the rules of combining capacitors, we can use equation
2117 to find the equivalent capacitance of
C
2
and
C
3
, and then add the result
to
C
1
according to equation 2114, because
C
1
is connected in parallel with
the pair
C
2
and
C
3
.
Solution:
Use equations 2114
and 2117 to find
C
eq
:
1
1
eq 1
2 3
1 1 1 1
15 F 21 F
8.2 F 22 F
C C
C C
μ
μ
μ μ
−
−
⎛ ⎞
⎛ ⎞
= + + = + + =
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
Insight:
The equivalent capacitance of
C
2
and
C
3
is only about 6.0
µ
F, much smaller than the algebraic sum
8.2 22 F 30 F,
μ
μ
+ =
because they are connected in series and combined according to equation 2117.
69.
Picture the Problem
: Three capacitors are connected in series as shown in the
diagram at the right.
Strategy:
First use equation 2117 to find the equivalent capacitance of the three
capacitors connected in series. Then use the equation
Q CV
=
(equation 209) to
find the amount of charge stored on each capacitor. Finally, solve equation 209 to
find the voltage drop across
C
3
.
Solution:
1.
Use equation 2117 to find
C
eq
:
1
eq
1 1 1
3.0 F
4.5 F 12 F 32 F
C
μ
μ μ μ
−
⎛ ⎞
= + + =
⎜ ⎟
⎝ ⎠
2.
Solve equation 209 for
Q
:
(
)
(
)
6
eq
3.0 10 F 12 V 36 C
Q C V
μ
−
= = × =
3.
Apply equation 209 again to
C
3
:
3
3
36 C
1.1 V
32 F
Q
V
C
μ
μ
Δ = = =
Insight:
If instead the three capacitors had been connected in parallel, there would be a potential drop of 12 V across
each, and the three capacitors together would have stored 580
µ
C of charge.
70.
Picture the Problem
: The capacitors
C
1
and
C
2
>
C
1
are connected to a battery in various ways.
Strategy:
Recall the rules regarding the calculation of equivalent capacitance for circuits containing capacitors
connected in series and in parallel.
Solution:
Two capacitors in series have an equivalent capacitance less than the smallest of the capacitors, and two
capacitors connected in parallel have an equivalent capacitance greater than the largest of the capacitors. We conclude
that the ranking of the circuits in order of increasing equivalent capacitance is C < A < B < D.
Insight:
In order to produce a very large capacitance for high
p
ower applications, a large number of capacitors are often
wired in parallel in a structure called a
capacitor bank
.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 25
71.
Picture the Problem
: Initially, the plates of six capacitors are
charged as shown in the figure. The switches are then closed,
allowing charge to move freely between the capacitors.
Strategy:
Use the expression
C Q V=
(equation 209) and the
fact that the potential difference across each capacitor will be the
same at equilibrium in order to determine the rankings of the
charges.
Solution: 1. (a)
For circuit A the
voltages are equal after the switch
is closed and the total charge is 3
q
:
upper
lower
2 lower upper
upper lower
2
2
3
C C
Q
Q
V V Q Q
C C
Q Q q
= ⇒ = ⇒ =
+ =
2.
Substitute the first equation into the second:
(
)
upper upper upper, A lower, A
2 3 ; 2
Q Q q Q q Q q
+ = ⇒ = =
3.
For circuit B the voltages are
equal after the switch is closed
and the total charge is −4
q
:
upper
lower
1
3 lower upper
3
upper lower
3
4
C C
Q
Q
V V Q Q
C C
Q Q q
= ⇒ = ⇒ =
+ = −
4.
Substitute the first equation into the second:
(
)
1
upper upper upper, B lower, B
3
4 3; Q Q q Q q Q q
+
= − ⇒ = − = −
5.
For circuit C the voltages are
equal after the switch is closed
and the total charge is 6
q
:
upper
lower
1
2 lower upper
2
upper lower
2
6
C C
Q
Q
V V Q Q
C C
Q Q q
= ⇒ = ⇒ =
+ =
6.
Substitute the first equation into the second:
(
)
1
upper upper upper, C lower, C
2
6 4; 2Q Q q Q q Q q+ = ⇒ = =
7.
The ranking in order of increasing charge on the left plate of the upper capacitor is B < A < C.
8.
(b)
The ranking in order of increasing charge on the left plate of the lower capacitor is B < A = C.
Insight:
Because the charges are equal and opposite on each capacitor, we can determine that the ranking in order of
increasing charge on the right plate of the upper capacitor is C < A < B and on the lower capacitor is A = C < B.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 26
72.
Picture the Problem
: Three capacitors are connected together as shown in the
diagram at the right, and a battery is connected across terminals A and B.
Strategy:
The voltage drop across
C
1
equals the battery emf, so we can find the
energy stored by
C
1
by using equation 2017. To find the energy stored by
C
2
and
C
3
, we must first find the equivalent capacitance
C
23
of the two capacitors using
equation 2117. Then use the battery emf, which is the voltage across
C
23
, to find
the charge
Q
stored on both
C
2
and
C
3
. Finally, use equation 2018 to find the
energy stored in
C
2
and
C
3
.
Solution:
1.
Apply equation 2017 to
C
1
:
(
)
( )
2
2 6 4
1 1
1
2 2
15 10 F 9.0 V 6.1 10 J 0.61 mJU CV
− −
= = × = × =
2.
(b)
Combine
C
2
and
C
3
using equation 2117:
1
6
eq
1 1
6.0 F 6.0 10 F
8.2 F 22 F
C
μ
μ μ
−
−
⎛ ⎞
= + = = ×
⎜ ⎟
⎝ ⎠
3.
Find the charge stored on
C
2
and
C
3
:
(
)
(
)
6 5
eq
6.0 10 F 9.0 V 5.4 10 CQ C V
− −
= = × = ×
4.
Find the energy stored in
C
2
:
(
)
( )
2
5
2
4
2
6
5.4 10 C
1.8 10 J 0.18 mJ
2
2 8.2 10 F
Q
U
C
−
−
−
×
= = = × =
×
5.
Find the energy stored in
C
3
:
(
)
( )
2
5
2
5
3
6
5.4 10 C
6.6 10 J 66 J
2
2 22 10 F
Q
U
C
μ
−
−
−
×
= = = × =
×
Insight:
The three capacitors store a total of 0.86 mJ. Verify for yourself that if all three capacitors were connected in
parallel and attached to a 9.0V battery, they would store a total of 1.8 mJ. The 8.2
µ
F capacitor connected in series
limits the amount of charge that is stored on the 22
µ
F capacitor, reducing its ability to store energy.
73.
Picture the Problem
: Two capacitors are connected in parallel with each other and with a 15V battery.
Strategy:
The equivalent capacitance of the two capacitors connected in parallel is simply the sum of the capacitances,
as indicated by equation 2114. Find the equivalent capacitance and then use it together with the voltage across the
plates in order to find the charge stored by each capacitor (equation 209).
Solution:
1. (a)
Add the capacitances:
eq 1 2
7.5 F 15 F 23 FC C C
μ μ μ
= + = + =
2.
(b)
Because each capacitor has the same voltage across its plates, and since
Q
=
CV
, the 15
µ
F capacitor stores more
charge.
3. (c)
Solve equation 209 for
Q
1
:
(
)
(
)
6
1 1
7.5 10 F 15 V 110 C
Q C V
μ
−
= = × =
4.
Repeat for
Q
2
:
(
)
(
)
6
2 2
15 10 F 15 V 230 CQ C V
μ
−
= = × =
Insight:
If the capacitors had been connected in series, with the battery connected across the pair, their equivalent
capacitance would have been only 5.0
µ
F and they would each have stored only 75
µ
C of charge. See problem 74.
74.
Picture the Problem
: Two capacitors are connected in series together, and a 15V battery is connected across the pair.
Strategy:
The equivalent capacitance of the two capacitors connected in series is given by equation 2117. Find the
equivalent capacitance and then use it together with the battery voltage in order to find the charge stored by each
capacitor (equation 209).
Solution:
1. (a)
Use equation 2117 to find
C
eq
:
1
eq
1 1
5.0 F
7.5 F 15 F
C
μ
μ μ
−
⎛ ⎞
= + =
⎜ ⎟
⎝ ⎠
2.
(b)
As discussed in the text, capacitors connected in series all have charge of the same magnitude on their plates.
Their charges are the same.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 27
3. (c)
Solve equation 209 for
Q
:
(
)
(
)
6
eq
5.0 10 F 15 V 75 C
Q C V
μ
−
= = × =
Insight:
If the capacitors had been connected in parallel with each other and with the battery, their equivalent
capacitance would have been 22.5
µ
F and they would have stored a total of 340
µ
C of charge. See problem 73.
75.
Picture the Problem
: Five capacitors are connected in a network as
indicated by the diagram at the right.
Strategy:
By applying equations 2114 and 2117, we can determine an
algebraic expression for the equivalent capacitance
C
eq
of the combination,
set it equal to the given value, and solve for the unknown capacitance
C
.
To find
C
eq
, we consider
C
3
and
C
4
to be connected in series with each
other and in parallel with
C
2
. That group of three capacitors is connected
in series with the unknown
C
, and the entire group of four is connected in
parallel with
C
1
.
Solution:
1.
Apply equations 2114
and 2117 to find an expression for
C
eq
:
1
1
1
eq 1 2
3 4
1 1 1
C C C
C C C
−
−
−
⎧
⎫
⎡ ⎤
⎛ ⎞
⎪
⎪
⎢ ⎥
= + + + +
⎨
⎬
⎜ ⎟
⎢ ⎥
⎝ ⎠
⎪
⎪
⎣ ⎦
⎩ ⎭
2.
Set
eq
9.22 FC
μ
=
and solve for
C
:
1
1
1
1
1 1 1
9.22 F 7.22 F 4.25 F
12.0 F 8.35 F
1 1 1
2.00 F 9.17 F
1 1
2.56 F
2.00 F 9.17 F
C
C
C
μ μ μ
μ μ
μ μ
μ
μ μ
−
−
−
−
⎧
⎫
⎡ ⎤
⎛ ⎞
⎪
⎪
= + + + +
⎢ ⎥
⎨
⎬
⎜ ⎟
⎢ ⎥
⎝ ⎠
⎪
⎪
⎣ ⎦
⎩ ⎭
= +
⎛ ⎞
= − =
⎜ ⎟
⎝ ⎠
Insight:
If
C
were increased from 2.56
µ
F to 12.0
µ
F,
C
e
q
would increase from 9.22
µ
F to 12.4
µ
F.
76.
Picture the Problem
: Two capacitors are connected in series and a battery is connected across the pair.
Strategy:
Use equation 2017 to find an algebraic expression for the sum of the energies
U
1
and
U
2
that are stored in the
two capacitors
C
1
and
C
2
, keeping in mind that each capacitor will store the same amount of charge because they are
connected in series. Then use equation 2117 to find an expression for the equivalent capacitance
C
eq
of the pair of
capacitors, and use equation 2017 again to find the amount of energy
C
eq
stores when a potential
V
is applied across it.
Solution:
1.
Combine
eqs. 2017 and 209 to find
an expression for
1 2
U U+
:
2 2
2 2 2
1 2 1 1 2 2 1 2
1 2 1 2
1 1 1 1 1 1 1
2 2 2 2 2
Q Q
U U CV C V C C Q
C C C C
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ = + = + = +
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2.
Substitute equation 2117
into equation 2017:
2
2 2 2
eq eq eq
eq eq 1 2
1 1 1 1 1 1 1
2 2 2 2
Q
U C V C Q Q
C C C C
⎛ ⎞ ⎛ ⎞
⎛ ⎞
= = = = +
⎜ ⎟ ⎜ ⎟
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠
⎝ ⎠ ⎝ ⎠
3.
The two expressions are equivalent, so we have shown that
1 2 eq
U U U
+
=
.
Insight:
If capacitor
C
1
were very much larger than
C
2
, say
1 2
10C C
=
, it would store only a small fraction of the total
energy. That is because the potential drop across
C
1
would only be
1
11
,V
so
( )
( )
2
2
5
1 1
1 2 2
2 11 121
10U C V C V= =
but
( )
2
2
10 50
1
2 2 2
2 11 121
U C V C V= =
. Note that in this case,
(
)
2 2
10 55
1 11
eq 2 2
2 11 11 121
.
U C V C V= × =
The smaller capacitor
C
2
stores ten
times the energy of the larger capacitor
C
1
.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 28
77.
Picture the Problem
: Two capacitors, a switch, and a battery are connected as
indicated in the diagram at the right.
Strategy:
When the switch is in position A, current flows until the potential
difference across the plates of
C
1
is 12.0 V. When the switch is moved to position
B, capacitor
C
1
charges up
C
2
until the two capacitors have the same voltage across
their plates. We can determine the amount of charge
Q
i
that is initially on
C
1
by
using equation 209. Then we can set the voltage across each capacitor equal to
each other and solve for the amount of charge
Q
2
on
C
2
. We can then use equation
209 one more time to find the voltage.
Solution:
1.
Find the initial charge
on the 11.2
µ
F capacitor.
(
)
( )
6
i i
11.2 10 F 12.0 V 134 CQ CV
μ
−
= = × =
2.
Set the voltage across each capacitor equal
after the switch is moved to position B:
1 i 2 2
1 1 2
Q Q Q Q
V
C C C
−
= = =
3.
Solve the expression from step 2 for
Q
2
:
1
2
1
2 i 2
2
i
2
11.2 F
9.50 F
134 C
61.5 C
1
1
C
C
C
Q Q Q
C
Q
Q
μ
μ
μ
μ
= −
= = =
+
+
4.
Find the voltage drop across
C
2
, which
must be the same as that across
C
1
:
2
2
61.5 C
6.47 V
9.50 F
Q
V
C
μ
μ
= = =
Insight:
The capacitor
C
2
obtains a charge when the switch is thrown that is proportional to the magnitude of its
capacitance. For instance, if
C
2
were very much smaller, say, 0.950
µ
F, the charge on its plates would decrease to
Q
2
= 10.5
µ
C and the voltage across each capacitor would increase to 11.0 V. As
C
2
approaches zero, it will store zero
charge and the potential across
C
1
would remain 12.0 V.
78.
Picture the Problem
: A resistor and a capacitor form a series
RC
circuit as
shown in the diagram at right.
Strategy:
Apply equation 2118 to find the charge
(
)
q t
on the plates of the
capacitor as a function of time, where
t
= 0 corresponds to the instant at which
the switch is closed.
Solution:
Apply equation 2118 directly:
(
)
(
)
( )
( )
( )
( )
( )
3
6
/
4.2 10 s
150 23 10 F
6
5
1
4.2 ms 23 10 F 9.0 V 1
1.5 10 C 150 C
t
q t C e
q e
τ
μ
ε
−
−
−
×
−
Ω ×
−
−
= −
⎡ ⎤
⎢ ⎥
= × −
⎢ ⎥
⎢ ⎥
⎣ ⎦
= × =
Insight:
At
t
= 0 there is no charge on the capacitor. There will be 210
µ
C on the capacitor when it is fully charged.
79.
Picture the Problem
: A resistor and a capacitor form a series
RC
circuit as
shown in the diagram at the right.
Strategy:
Apply equation 2118 to find the charge
(
)
q t
on the plates of the
capacitor as a function of time, where
t
= 0 corresponds to the instant at which
the switch is closed.
Solution:
1. (a)
Solve equation
2118 for the case when
t
τ
=
:
( )
(
)
( )
( )
( )
( )
6 1
4
1
45 10 F 9.0 V 1
2.6 10 C 260 C
q C e
e
q
ττ
τ
τ μ
ε
−
− −
−
= −
= × −
= × =
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker,
Physics
, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 29
2.
(b)
Solve equation 2119
for the case when
t
τ
=
:
( )
1
9.0 V
28 mA
120
I e e
R
ττ
τ
ε
− −
= = =
Ω
Insight:
The charge
( )
q t
starts out at zero and increases to 405
µ
C when the capacitor is fully charged, whereas
(
)
I
t
starts out at 75 mA and decreases to zero when the capacitor is fully charged.
80.
Picture the Problem
: Three
RC
circuits have the emf, resistance, and capacitance given in the accompanying table.
Initially, the switch on the circuit is open and the capacitor is uncharged.
Strategy:
The initial current in an RC circuit is given by
(
)
0.
I
R
ε
=
The time constant of a series
RC
circuit is given
by
RC
τ
=
and the time to acquire half its final charge is
(
)
(
)
1
2
1 ln 2.
t
q t C e C t
τ
τ
ε
ε
−
= − ≡ ⇒ =
Use the data
given in the table to calculate the initial currents and the time constants of each circuit to determine the rankings.
Solution:
1. (a)
Divide
R
ε
for each circuit to find
(
)
(
) ( )
A B C
0 3.0 A, 0 3.0 A, and 0 1.0 A.
I I I= = =
The ranking of
the initial currents is therefore C < A = B.
2. (b)
Mulitply
RC
for each circuit to find
A B C
12 s, 3.0 s, and 18 s.
τ μ
τ
μ
τ
μ
=
= =
The ranking of the times for the
capacitors to acquire half of their final charges is therefore B < A < C.
Insight:
Initially, the capacitor behaves like a short circuit, allowing current to flow at a rate given by
.
I
R
ε
=
As the
capacitor charges up it reduces the potential difference across the resistor until it reaches zero and no current flows.
81.
Picture the Problem
: A battery, a resistor, and a capacitor form a series
RC
circuit.
Strategy:
The time constant of a series
RC
circuit is given by
.RC
τ
=
The capacitor is fully charged at
,t
=
∞
at which
time equation 2118 reduces to
( )
max
q q C
ε
∞ = =
because
0.
e
−∞
=
The initial current at
t
= 0 is given by
( )
( )
0
I
R
ε
=
(equation 2119) because
0
1.e
−
=
Solution:
1. (a)
Calculate
:
τ
=
(
)
(
)
6
175 55.7 10 F 9.75 msRCτ
−
= = Ω × =
2.
(b)
Calculate
( )
max
q q∞ =
:
(
)
(
)
6 4
max
55.7 10 F 12.0 V 6.68 10 C 668 C
q C
μ
ε
− −
= = × = × =
3. (c)
Calculate
( )
0I t =
:
( )
12.0 V
0 68.6 mA
175
I
R
ε
= = =
Ω
Insight:
At a time equal to 3 time constants (9.75 ms × 3 = 29.3 ms), the charge has reached 95.0% of its maximum
value and the current has decreased to 4.98% of its initial value.
82.
Picture the Problem
: A battery, a resistor, and a capacitor form a series
RC
circuit.
Strategy:
The time constant of a series
RC
circuit is given by ,RC
τ
=
so we need only solve this expression for C to
find the correct capacitance to achieve the specified time constant. Then we can use equation 2119 to find the current
in the circuit at the specified time after the switch is closed.
Solution:
1. (a)
Calculate the required C:
3
5
3.5 10 s
2.4 10 F 24 F
145
C
R
τ
μ
−
−
×
= = = × =
Ω
2.
(b)
Solve equation 2119 at t = 7.0 ms:
( )
( ) ( )
7.0 ms 3.5 ms
9.0 V
7.0 ms 8.4 mA
145
I e
−
= =
Ω
Insight:
Two time constants have elapsed at 7.0 ms, at which time the current has decreased to 13.5% of its initial
value.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 30
83.
Picture the Problem
: A battery, a resistor, and a capacitor form a series RC circuit that operates a camera flash unit.
Strategy:
Equation 2118 determines the charge on a capacitor in a series RC circuit as a function of time. Solve
equation 2118 for the resistance required to charge the capacitor to 90% of its full charge in the specified time interval.
Solution:
Solve equation 2118 for R:
( )
( )
( )
max
max
6
max
1 1
ln 1
21 s
6.1 k
ln 1
1500 10 F ln 1 0.90
t RC
q q
e
C q
q t
q RC
t
R
C q q
ε
−
−
− = = −
⎛ ⎞
− = −
⎜ ⎟
⎝ ⎠
=
− = − = Ω
−
× −
Insight:
Doubling the resistance to 12.2 kΩ would double the time constant, and the charge on the capacitor would
reach only 68% of its full charge in 21 s.
84.
Picture the Problem
: The flash unit of a camera contains a battery,
resistor, capacitor, lamp, and two switches. The circuit is illustrated in the
diagram at right.
Strategy:
The capacitor will charge up if S
1
is closed and S
2
is open. In
this configuration we have a simple RC circuit. The time required to
charge the capacitor to 5.0 V can be found by combining equations 209
and 2118 and solving for t.
Solution:
1.
Substitute
q CV=
(equation 209) into equation 2118:
(
)
1
1
t RC
t RC
q C e
CV
e
C
ε
ε
−
−
= −
= −
2.
Rearrange the equation and solve for t:
( )
( )( )
3 6
1
ln 1
5.0 V
ln 1 50.0 10 140 10 F ln 1
9.0 V
5.7 s
t RC
e V
t
V
RC
V
t RC
ε
ε
ε
−
−
= −
− = −
⎛ ⎞ ⎛ ⎞
= − − = − × Ω × −
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
=
Insight:
The time required to charge the capacitor in the flash unit is responsible for most of the delay between pictures
that is required by many cameras. When it is fully charged, the capacitor can deliver its energy to the flashbulb much
more rapidly than could a battery.
85.
Picture the Problem
: Three resistors, a capacitor, and a battery are
connected in a circuit as depicted in the diagram at the right.
Strategy:
The circuit is a simple RC circuit if we replace the three
resistors with their equivalent resistance. The 13Ω and 6.5Ω resistors
are connected in series, and the pair is connected in parallel with the 24
Ω resistor. Use equations 217 and 2110 to find
eq
,R
then calculate the
time constant and the initial current according to equation 2119.
Solution:
1. (a)
Find
eq
:R
1
eq
1 1
10.8
24 13 6.5
R
−
⎛ ⎞
=
+ = Ω
⎜ ⎟
Ω Ω+ Ω
⎝ ⎠
2.
Calculate the time constant:
( )
(
)
6 4
eq
10.8 62 10 F 6.7 10 s 0.67 msR C
τ
− −
= = Ω × = × =
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 31
3.
(b)
Find the initial current:
( )
( )
0
eq
15 V
0 1.00 1.4 A
10.8
I e
R
τ
ε
−
= = =
Ω
4. (c)
The equivalent resistance of the three resistors must be increased in order to increase the time constant. The
resistance of the 6.5Ω resistor should be increased to increase the time constant, because increasing the resistance of
the 6.5Ω resistor increases
eq
.R
Insight:
We bent the rules for significant digits a little bit in step 1 in order to avoid rounding errors in steps 2 and 3.
The time constant could also be increased by choosing a larger capacitor.
86.
Picture the Problem
: A battery, a resistor, and a capacitor form a series RC circuit.
Strategy:
Equation 2118 determines the charge on a capacitor in a series RC circuit as a function of time. Solve
equation 2118 for the time required to charge the capacitor to 50% of its full charge. Then solve equation 2119 for the
time required for the current to drop to 10% of its initial value.
Solution:
1. (a)
Set
(
)
浡x
〮㔰 〮㔰
q t q C
ε
= =
in equation 2118 and solve for the time t:
(
)
0.50 1
0.50 1
1 0.50
ln0.50
ln0.50 ln 2.0
t RC
t RC
t RC
C C e
e
e
t
RC
t RC RC
ε
ε
−
−
−
= −
= −
= −
− =
= − =
2.
(b)
Set
( ) ( )
0.10 0 0.10I t I
R
ε
= =
and solve for t:
0.10
0.10
ln0.10 ln10
t RC
t RC
e
R R
e
t
t RC
RC
ε
ε
−
−
=
=
= − ⇒ =
Insight:
When a single time constant has passed, the charge has increased from zero to 63.2% of its final value, and the
current has dropped to 36.8% of its initial value.
87.
Picture the Problem
: A given car battery is rated as 250 amphours.
Strategy:
Examine the amphours unit to determine the quantity it measures.
Solution:
Recalling that 1 A = 1 C/s it follows that an amphour has the dimensions of charge, and hence 250 amp
hours is a measure of the amount of charge that passes through the battery before the battery’s energy is depleted.
Insight:
Charge is conserved in a circuit, so that just as much charge flows into the battery as flows out, and you don’t
actually “use up” a battery’s charge. Amphours are simply a measure of the amount of charge that flows out of one
terminal of the battery and back into the other terminal. However, chemical reactions in the battery give the charges
energy (voltage) that they can deliver to the circuit. When the chemicals become depleted there is no more energy to
give to the charges. When we multiply amphours (charge) by the voltage difference (energy per charge) between the
terminals, the result is the energy delivered by the battery to the circuit. The amphour rating of a battery is therefore an
indirect way to indicate the total amount of energy that the battery can deliver to a device.
88.
Picture the Problem
: The resistivity of a tungsten filament increases as it heats up.
Strategy:
Assume that the voltage across the filament remains constant but that its resistance increases.
Solution: 1. (a)
Assuming constant voltage, the appropriate expression for power consumption is
2
P V R= (equation
216). We conclude that the power consumption of the light bulb will decrease as it heats up.
2. (b)
The best explanation is
I
. The voltage is unchanged, and therefore an increase in resistance implies a reduced
power, as we can see from P = V
2
/R. Statements II and III each ignore the decrease in current as resistance increases.
Insight:
If the resistance decreased with temperature, there would be the danger of runaway current, where more current
produces more heating which lowers the resistance and allows more current to flow. Fluorescent light bulbs exhibit this
kind of behavior and must be accompanied by a ballast that controls the current through the bulb.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 32
89.
Picture the Problem
: A cylindrical wire is to be doubled in length, but it is desired that its resistance remain the same.
Strategy:
Use the expression for resistance,
R L A
ρ
=
(equation 213) to answer the question.
Solution: 1. (a)
The resistance of a wire depends on
.L A
Therefore, to keep the resistance the same when the length
L
is doubled means that the area A must be doubled as well. This implies that the radius must be increased.
2. (b)
The radius must be increased by a factor of
2
because A depends on the radius squared.
Insight:
We can also answer part (b) with a ratio:
new
new new new old
old old old old
old
2
2
A
r A L R L
r A L R L
A
π
ρ
ρ
π
= = = = =
90.
Picture the Problem
: Two electric space heaters, each with a power rating of 500 W when connected to a voltage V,
are connected in series to the same voltage.
Strategy:
Use the expression
2
P I R=
(equation 215) to answer the conceptual question.
Solution: 1. (a)
If two heaters, each with resistance R, are connected in series, the equivalent resistance is now 2R. The
heaters are connected to the same potential difference, however, and therefore they draw half the original current. From
P
= I
2
R we can see that each heater now dissipates onequarter the power of the single heater, for a total power
consumption of onehalf the original value. We conclude that the power consumed by the two heaters connected in
series is less than 1000 W.
2. (b)
The best explanation is
II
. The voltage is the same, but the resistance is doubled by connecting the heaters in
series. Therefore, the power consumed (P = V
2
/R) is less than 1000 W. Statement I ignores the change in resistance
and statement III erroneously claims the power depends upon the resistance squared.
Insight:
Connecting the heaters in series also reduces by half the potential difference across each heater. Using
2
P V R=
we can see that the power dissipated by each heater is onefourth the original value, or 125 W, for a total
power consumption of 250 W.
91.
Picture the Problem
: Two resistors, R
1
= R and R
2
= 2R, are connected to a battery.
Strategy:
Recall the characteristics of circuits containing resistors connected in series and in parallel.
Solution:
1. (a)
If the two resistors are connected in series, they have the same current flowing through them.
Therefore, we can compare the power they dissipate by considering the relation
2
.P I R=
We conclude that resistor R
2
dissipates twice as much power as does R
1
.
2.
(b)
If the two resistors are connected in parallel, they have the same potential difference across them. Therefore, we
can compare the power they dissipate by considering the relation
2
.P V R=
We conclude that in this case resistor R
1
dissipates twice as much power as does R
2
.
Insight:
The fact that in the parallel circuit the smaller resistor dissipates more power than the larger one seems
counterintuitive at first, until you realize that the smaller resistor allows more current to flow through its branch.
92.
Picture the Problem
: Consider the circuit with three lights shown at the right.
Strategy:
Recall the characteristics of circuits containing resistors connected in
series.
Solution:
1. (a)
Light 2 goes out when the switch is closed, so its intensity will
decrease when the switch is closed.
2.
(b)
The current in the circuit increases when the switch is closed, due to the
decrease in equivalent resistance from 3R to 2R. It follows that the power
2
P I R
=
(equation 215) dissipated in light bulbs 1 and 3 increases, resulting in an increase
in the intensity of light for these bulbs.
Insight:
Light bulb 2 turns off when the switch is closed because the switch “shorts out” the lamp. From a physical
perspective, the closed switch forces both ends of the light bulb filament to be at the same potential so no current flows.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 33
93.
Picture the Problem
: Consider the circuit with three lights shown at the right.
Strategy:
Recall the characteristics of circuits with resistors connected in series.
Solution:
1. (a)
When the switch is closed, light 2 is shorted out and the equivalent
resistance of the circuit drops from 3R to 2R. It follows that the current supplied by
the battery will increase when the switch is closed.
2.
(b)
The best explanation is
III
. Closing the switch shorts out the second resistor,
decreases the total resistance of the circuit, and increases the current. Statements I
and II are each false.
Insight:
Regarding statement II, while it is true that the light bulb remains connected to the battery as before, the new
connection created by the switch makes all the difference; it effectively removes light 2 from the circuit.
94.
Picture the Problem
: Consider the circuit with three lights shown at the right.
Strategy:
Recall the characteristics of circuits with resistors connected in series.
Solution:
1. (a)
The battery in this circuit maintains a constant voltage
ε
=牯獳⁴桥=
汩杨琠l×汢献sTh攠捯牲敳p潮摩dg=灯睥爠摩獳楰a瑥搠楳t
2
敱
.P R
ε
=
When the switch is
closed, the equivalent resistance of the circuit decreases from 3R to 2R. As a result,
the total power dissipated in the circuit will increase when the switch is closed.
2.
(b)
The best explanation is
II
. The equivalent resistance of the circuit is reduced by closing the switch, but the
voltage remains the same. Therefore, from P = V
2
/R we see that the power dissipated increases. Statements I and III
are each false because they ignore the change in current when the switch is closed.
Insight:
You might witness an effect like this in a string of holiday lights. Each bulb has a spring that will short out the
bulb when the filament breaks. This will increase the current through the other bulbs and increase the power disspated
by the string of lights. It will also hasten the failure of the other bulbs because each filament will now become hotter.
95.
Picture the Problem
: Consider the circuit with three lights shown at the right.
Strategy:
Recall the characteristics of circuits with resistors connected in parallel.
Solution:
1. (a)
Light 3 turns on when the switch is closed, so its intensity will
increase when the switch is closed.
2.
(b)
The intensities of lights 1 and 2 stay the same when the switch is closed. After all, they still have the same
potential difference and the same resistance. Therefore, it is clear from
2
P V R= that they will continue to dissipate
the same amount of power.
Insight:
Household circuits are arranged in parallel like this one because any device can be turned off or on without
affecting the other devices connected to the same circuit.
96.
Picture the Problem
: Consider the circuit with three lights shown at the right.
Strategy:
Recall the characteristics of circuits with resistors connected in parallel.
Solution:
1. (a)
When the switch is closed in figure 2129, the battery has an
additional path through which it can send current, in addition to the two paths that
were already available to it. Therefore, the current supplied by the battery to this
circuit will increase when the switch is closed.
2.
(b)
The best explanation is
I
. The current increases because three resistors are drawing current from the battery when
the switch is closed, rather than just two. Statement II ignores the new current flow, and statement III erroneously
concludes that adding a resistor will increase the resistance of the circuit, when it will actually decrease.
Insight:
Adding a light bulb to the circuit by closing the switch will decrease the equivalent resistance of the circuit.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 34
97.
Picture the Problem
: Consider the circuit with three lights shown at the right.
Strategy:
Recall the characteristics of circuits with resistors connected in parallel.
Solution:
1. (a)
The total power dissipated in the circuit will increase when the
switch is closed. Before the switch was closed, light 1 dissipated the power
2
P R
ε
=
as did light 2. Now all three lights dissipate the power
2
.P R
ε
=
2.
(b)
The best explanation is
II
. The equivalent resistance of the circuit is reduced by closing the switch, but the
voltage remains the same. Therefore, from P = V
2
/R we see that the power dissipated increases. Statement I
erroneously concludes that adding a resistor will increase the resistance of the circuit, when it will actually decrease.
Statement III ignores the change in current that results with the addition of light bulb 3.
Insight:
The constant voltage across each light bulb in a parallel circuit ensures that each bulb dissipates the same
amount of power.
98.
Picture the Problem
: A 12V battery is connected across terminals
A and B that are depicted in the diagram at the right. The battery
voltage drives electric current through all three resistors.
Strategy:
First find the effective resistance of the three resistors
using equations 217 and 2110. Ohm’s Law (equation 212) then
gives the current through the entire circuit, which is also the current
through the 12Ω resistor. That current can then be used to find the
potential drop across R
3
, which when subtracted from the battery
emf gives the potential drop across R
1
and R.
Solution:
1.
Find
eq
:R
1
1
eq 3
1
1 1 1 1
12 41.8
55 65
R R
R R
−
−
⎛ ⎞
⎛ ⎞
=
+ + = Ω+ + = Ω
⎜ ⎟
⎜ ⎟
Ω Ω
⎝ ⎠
⎝ ⎠
2.
Find the current through the 12Ω resistor:
3
eq
12 V
0.287 A
41.8
I
R
ε
= = =
Ω
3.
Find the voltage across
1
R
and R:
(
)( )
1 3 3
12.0 V 0.287 A 12 8.56 VV I R
ε
= − = − Ω =
4.
Use equation 216 to find
1
P
:
( )
2
2
1
1
1
8.56 V
1.3 W
55
V
P
R
= = =
Ω
(55Ω resistor)
5.
Repeat find
2
P
:
( )
2
2
1
2
8.56 V
1.1 W
65
V
P
R
= = =
Ω
(65Ω resistor)
6.
Use equation 215 to find
3
P
:
( ) ( )
2
2
3 3 3
0.287 A 12 0.99 WP I R= = Ω = (12 Ω resistor)
Insight:
If all three resistors were to be connected in parallel with the battery, the equivalent resistance would drop to
8.55 Ω and the dissipated powers would be 2.6 W in the 55Ω resistor, 2.2 W in the 65Ω resistor, and 12 W in the 12Ω
resistor.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 35
99.
Picture the Problem
: You are asked to connect three resistors of known values in
order to produce a certain effective resistance.
Strategy:
The configuration of resistors can be determined with logical reasoning.
Connecting all three of the resistors in series would make
eq
1080 R
=
Ω
, higher
than the 255 Ω desired effective resistance. Connecting the three in parallel would
make
eq
89.4 R = Ω
, smaller than the desired resistance. We could connect two
resistors in parallel and then place that pair in series with the third resistor. The
three possible arrangements yield
eq
R
values of 376 Ω, 527 Ω, and 629 Ω, all too
large. Therefore, we must connect two resistors in series, then connect the pair in
parallel with the third resistor. The three possible arrangements produce
eq
R
values
of 126 Ω, 255 Ω, and 270 Ω.
Solution: 1.
Connect the 146 Ω and 521 Ω resistors in series. Then connect this pair in parallel with the 413 Ω resistor.
2.
Check the solution produces the correct
eq
R
:
1
eq
1 1
255
146 521 413
R
−
⎛ ⎞
=
+ = Ω
⎜ ⎟
+ Ω Ω
⎝ ⎠
Insight:
One of the three possible configurations, the one with the two larger resistances in series and connected in
parallel with the 146Ω resistor, can be ruled out because the effective resistance would always be less than 146 Ω.
100.
Picture the Problem
: You are asked to connect three capacitors of known values in
order to produce a certain effective capacitance.
Strategy:
The configuration of capacitors can be determined with logical
reasoning. Connecting all three of the capacitors in series would make
eq
3.3 FC
μ
=
, smaller than the 22 µF desired effective capacitance. Connecting the
three in parallel would make
eq
34 F,C
μ
=
larger than the desired capacitance. We
could connect two capacitors in parallel and then place that pair in series with the
third capacitor. The three possible arrangements yield
eq
C
values of 5.7 µF, 6.6 µF,
and 8.5 µF, all too small. Therefore, we must connect two capacitors in series, then
connect the pair in parallel with the third capacitor. The three possible
arrangements produce
eq
C
values of 13 µF, 14 µF, and 22 µF.
Solution: 1.
Connect the 7.2µF and 9.0µF capacitors in series. Then connect this pair in parallel with the 18µF
capacitor.
2.
Check the solution produces the correct
eq
C
:
1
eq
1 1
18 F 22 F
7.2 F 9.0 F
C
μ μ
μ μ
−
⎛ ⎞
= + + =
⎜ ⎟
⎝ ⎠
Insight:
One of the three possible configurations, the two that connect the 18µF capacitor in series with a smaller
capacitor can be ruled out because the effective capacitance would always be less than 16.2 µF, the value of
eq
C
if the
7.2µF and 9.0µF capacitors were connected in parallel. Adding the 18µF capacitor in series only makes the
capacitance value of that branch even smaller, and reduces
eq
C
below 16.2 µF.
101.
Picture the Problem
: Your car carries an electric charge, and constitutes an electric current as long as it is moving.
Strategy:
Divide the charge on your car by the time it takes to make the journey in order to find the current.
Solution:
Solve equation 211:
6
8
85 10 C 1 h
3.1 10 A 31 nA
0.75 h 3600 s
Q
I
t
−
−
Δ ×
⎛ ⎞
= = = × =
⎜ ⎟
Δ
⎝ ⎠
Insight:
This approach is a little artificial, because the calculated current changes depending on how much time the trip
takes, not how fast the car is moving. The definition of electric current presupposes a continuous flow of charge, so that
the amount of current created by a single moving point charge is illdefined.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 36
102.
Picture the Problem
: Initially, the switch in the circuit shown at the right is open
and both capacitors are uncharged. The switch is closed and a long time has passed
after the current has ceased to flow.
Strategy:
The ammeter represents essentially a short circuit between the two
capacitors, so that they can be considered as connected in series. Use the
characteristics of capacitors that are connected in series to answer the questions.
Solution: 1.
(a)
The two capacitors have the same charge, and hence the smaller capacitor will have the larger voltage.
Setting
1 2
Q Q=
we find that
(
)
1 1 2 2 1 2 2
2 2.C V C V V C C V V= ⇒ = =
Once current has ceased to flow, there is no
voltage drop across the resistor, so
1 2
V V V+ =
and we conclude that
2
1
3
.V V=
2.
(b)
The equivalent capacitance of C
1
and C
2
in series is
[
]
2
敱
3
ㄱ 1 2.
C C C C= + =
Therefore, the charge on the
equivalent capacitance is
2
eq
3
,Q C V CV= =
and the charge on the right plate of C
2
is
2
3
.CV−
3. (c)
Because the capacitors were initially uncharged, and because they both have the same charge after a long time, we
conclude that the total charge that flowed through the ammeter is equal to the final charge on the capacitors, or
2
3
.CV
Insight:
More specifically, in terms of direction and sign, the charge
2
3
CV−
flowed to the left through the ammeter.
This is because it is the electrons that actually move in an electric circuit, and because the left plate of C
2
will end up
being positively charged and the right plate of C
1
will be negatively charged.
103.
Picture the Problem
: Initially, the switches in the circuits shown at the
right are open and the capacitors are uncharged. The switches are then
closed and the capacitors charge in a manner similar to a simple RC circuit.
Strategy:
First, note that after a long time the potential difference across
each of the three capacitors is equal to the potential difference across the
battery. Use that fact to determine the final charge on each capacitor.
Second, use the rules for finding the equivalent resistance of series and
parallel circuits to find the resistance of the branch of the circuit that contains the capacitor. The equivalent resistance of
that branch can then be used to determine the effective time constant that describes the charging of the capacitor.
Solution: 1.
(a)
In each of the three circuits, the capacitor has the same potential difference as the battery after a long
time, when the flow of current has ceased. Therefore, they each have the same charge Q = C
V. The ranking of the final
charges is thus A = B = C.
2.
(b)
The time required is proportional to the time constant, R
eq
C; therefore, the ranking follows the equivalent
resistance in the branch of the circuit where the current will come to a stop. Recall that the equivalent resistance of two
resistors in series is 2R. Therefore, the equivalent resistance in series with the capacitor in circuit A is a resistor R in
parallel with 2R, making
2
eq, A
3
.
R R=
The equivalent resistance in series with the capacitor in circuit B is 2R, and that
in circuit C is R. We conclude that the ranking of the time constants R
eq
C, which is also the ranking of the times for the
current to drop to 90% of its initial value, is A < C < B.
Insight:
Current continues to flow in the branches that do not contain a capacitor. This means that the current flowing
through the battery is
nc c
,
t
I I e
τ
−
+
where
nc
I
is the current in the branch that does not contain a capacitor,
c
I
is the
current in the capacitor’s branch, and
eq
R C
τ
=
is the time constant for current decay in the capacitor’s branch. A
careful analysis reveals that the exact times for the current through the battery to fall to 90% of its initial value are:
( )
(
)
(
)
2
A B C
3
ln 0.90 0.070, 2 ln 0.70 0.71, and ln 0.85 0.16.
t RC RC t RC RC t RC RC= − = = − = = − =
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 37
104.
Picture the Problem
: A conducting wire of circular crosssection is fashioned out of tungsten.
Strategy:
Write equation 213 for a wire of circular crosssection, and solve the resulting expression for the diameter d.
Use the resistivity of tungsten as listed in Table 211.
Solution:
1.
Write equation 213
for a wire of circular crosssection:
2
1
4
L L
R
A d
ρ ρ
π
= =
2.
Solve for d:
(
)
( )
( )
8
4 5.6 10 m 1.2 m
4
0.13 mm
5.0
L
d
R
ρ
π π
−
× Ω⋅
= = =
Ω
Insight:
If the diameter of the wire were doubled, it would reduce R by a factor of four, to 1.25 Ω.
105.
Picture the Problem
: A voltage drop occurs across each of the two wires in an extension cord because of the intrinsic
resistance of the copper wires.
Strategy:
Combine Ohm’s Law (equation 212) with the definition of resistance (equation 213) to write an expression
for the voltage drop across the two wires of the extension cord. The length of wire involved is twice the length of the
extension cord, or 2×12 ft×0.305 m/ft = 7.32 m.
Solution:
Combine equations 212 and 213:
( )
( )
( )
8
6 2
7.32 m
5.0 A 1.68 10 m 0.53 V
1.17 10 m
L
V IR I
A
ρ
−
−
= = = × Ω⋅ =
×
Insight:
Although this voltage drop is small in comparison to the 120 V required to operate the tool or appliance, the
power dissipated in the cord is even more important because it is responsible for the cord heating that could produce a
fire. In this case the power dissipated in the cord is a fairly modest
( ) ( )
2
2
5.0 A 0.105 2.6 W.P I R= = Ω =
106.
Picture the Problem
: Three terminals of a threeway light bulb are connected to the two
filaments in the manner indicated by the diagram at the right.
Strategy:
Use equation 216 together with the given information to find the resistances R
1
and R
2
and the power dissipated when 120 V is connected across terminals B and C.
Solution:
1. (a)
Solve equation 216 for R
1
:
( )
2
2
1
120 V
190
75.0 W
V
R
P
=
= = Ω
2.
(b)
Find the equivalent resistance of R
1
and R
2
:
2
eq 1 2
V
R R R
P
=
= +
3.
Solve the expression from step 2 for R
2
:
( )
( )
2 2
2
2 1
120 V 120 V
96
50.0 W 75.0 W
V
R R
P
=
− = − = Ω
4.
(c)
Find the power dissipated by R
2
:
( )
2
2
2
120 V
150 W
96
V
P
R
= = =
Ω
Insight:
Note that when 120 V is connected across the smallest resistance (96 Ω), the largest amount of power is
dissipated (150 W).
107.
Picture the Problem
: A portable CD player consumes electrical power as it operates.
Strategy:
The power consumed by the CD player equals the amount of current it draws multiplied by the electrical
potential at which the current is delivered (equation 214). The result can then be used to determine the energy
consumed and the time it could operate if it converted gravitational potential energy to electrical energy.
Solution:
1. (a)
Find P from equation 214:
(
)
(
)
3
7.5 10 A 3.5 V 0.026 WP I V
−
= = × =
2.
Calculate the energy consumed in 35 s:
(
)
(
)
0.026 W 35 s 0.91 JE P t= Δ = =
3. (b)
Find the time required to lift 1.0 m:
(
)
(
)
( )
( )( )
2
0.65 kg 9.81 m/s 1.0 m
240 s 4.0 min
0.0075 A 3.5 V
E mgh
t
P I V
Δ = = = = =
Insight:
Size AA alkaline batteries rated at 2800 mA∙h would operate this player for about 370 hours before needing to
be replaced.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 38
108.
Picture the Problem
: An electrical heating coil raises the temperature of the water in which it is immersed.
Strategy:
Use equation 1613 to find the amount of energy required to raise the temperature of the known mass of water
by the specified amount. Divide this energy by the time required to heat the water in order to find the electrical power
required to operate the heater. Finally, solve equation 216 for the voltage at which the heater operates.
Solution:
1.
Solve
equation 1613 for Q:
w w
Q m c T
=
Δ
2.
Divide Q by
tΔ
to find P:
w w
m c T
Q
P
t t
Δ
= =
Δ Δ
3.
Solve equation 216 for V,
substituting the expression
for P from step 2:
( )( )( )( )
( )
w w
4.6 kg 4186J/kg K 32 22 C 250
15 min 60 s/min
230 V 0.23 kV
m c T
V PR R
t
⋅ − ° Ω
Δ
⎛ ⎞
= = =
⎜ ⎟
Δ ×
⎝ ⎠
= =
Insight:
If the heating element operated at 120 V but had the same resistance, it would dissipate only 58 W and it would
require 56 minutes to increase the temperature of the same amount of water (4.2 L or 1.1 gallons) by 10 C°.
109.
Picture the Problem
: Two resistors, a capacitor, a switch, and a battery are
connected as indicated in the diagram at the right.
Strategy:
The capacitor acts like a short circuit immediately after the switch is
thrown because it is uncharged and the potential difference across its plates is zero.
Long after the switch is thrown, the capacitor has charged up to its maximum value
and no current flows through it. All of the current then remains in the perimeter of
the circuit, flowing through the 11Ω and 5.6Ω resistors. Use these principles
together with Ohm’s Law (equation 212) to find the current immediately after the
switch is thrown and long after the switch is thrown.
Solution:
1. (a)
The current flows through only the 11Ω resistor immediately after the switch is thrown, but through
both the 11Ω and the 5.6Ω resistors long after the switch is thrown. Therefore, the initial current is greater than the
final current.
2.
(b)
Find I immediately after the switch is thrown:
9.0 V
0.82 A
11
V
I
R
= = =
Ω
3. (c)
Find I long after the switch is thrown:
9.0 V
0.54 A
11 5.6
V
I
R
= = =
Ω+ Ω
Insight:
Verify for yourself that the voltage across the capacitor is 3.0 V long after the switch is thrown.
110.
Picture the Problem
: A silver wire and a copper wire each have the same volume and the same resistance, but different
radii.
Strategy:
Use equation 213 and the volume of a cylinder to simplify the ratio of the radii of the two wires.
Solution:
1.
Calculate the ratio of the radii,
substituting
2
A
r
π
=
and
R L A
ρ
=
:
silver
silver silver silver silver silver silver
copper copper copper copper copper copper
copper
A
r A L R L
r A L R L
A
π
ρ ρ
ρ ρ
π
= = = =
2.
Square both sides and substitute
2
:V r L
π
=
(
)
( )
2
2
2
silver silver
silver copper
silver
2
2
copper copper silver
copper copper
V r
r
r
r r
V r
ρ π
ρ
ρ
ρ π
⎛ ⎞
= =
⎜ ⎟
⎜ ⎟
⎝ ⎠
3.
Simplify and insert numerical values:
4
silver silver
copper copper
1
1
8
4
4
silver silver
8
copper copper
1.59 10 m
0.986
1.68 10 m
r
r
r
r
ρ
ρ
ρ
ρ
−
−
⎛ ⎞
=
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎛ ⎞
⎛ ⎞
× Ω⋅
= = =
⎜ ⎟
⎜ ⎟
⎜ ⎟
× Ω⋅
⎝ ⎠
⎝ ⎠
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 39
Insight:
The copper wire must have a slightly larger radius to compensate for its larger resistivity in order for each wire
to have the same resistance. The larger radius also shortens the wire if the volume is to remain constant, which in turn
helps to decrease the resistance of the copper wire.
111.
Picture the Problem
: Two resistors are connected in series with each other and with a
battery.
Strategy:
The same current must pass through both resistors because they are connected in
series. We can use the given voltage and current together with Ohm’s Law (equation 212) in
order to find the value of R
1
. The voltage across R
2
must be the difference between the emf
and the voltage across R
1
, allowing us to use Ohm’s Law again to calculate R
2
.
Solution:
1.
Apply Ohm’s Law to R
1
:
1
1
2.7 V
18
0.15 A
V
R
I
=
= = Ω
2.
Apply Ohm’s Law to R
2
:
2 1
2
12.0 2.7 V
62
0.15 A
V V
R
I I
ε
−
−
=
= = = Ω
Insight:
The larger resistor has a much larger voltage drop across it (9.3 V). This arrangement of resistors and a battery
is often called a voltage divider, because an adjustment of the relative values of R
1
and R
2
can produce a voltage
anywhere between zero and
ε
慴⁴桥a灯p湴i摷慹整w敥e ⁴桥⁴睯=牥獩獴潲献†䡥湣攠
ε
猠鍤楶楤敤鐠慴⁴桡琠灯p湴⸠
=
=
=
ㄱ2.†
Picture the Problem
: A heart pacemaker is constructed from a simple RC circuit.
Strategy:
Solve equation 2118 for the voltage across the capacitor, set it equal to the trigger voltage, and solve the
resulting expression for the resistance R. Set the time t equal to
60 s 75 0.80 s
=
楮牤i爠景爠瑨攠灡捥p慫敲⁴漠晩牥‷㔠
瑩te猠敡捨i湵瑥⸠
=
Solution:
1.
Divide both sides
of equation 2118 by C:
( )
( )
1 1
t t RC
q
V e e
C
τ
ε ε
− −
= = − = −
2.
(b)
Solve the expression for R:
( )
( )
( ) ( )
6
5
1 1
ln 1
0.80 s
ln 1
110 10 F ln 1 0.25 V 9.0V
2.6 10 0.26 M
t RC t RC
V V
e e
t V
RC
t
R
C V
ε
ε ε
ε
− −
−
= − ⇒ = −
⎛ ⎞
− = −
⎜ ⎟
⎝ ⎠
= =
−
× −⎡ ⎤
⎣ ⎦
= × Ω= Ω
Insight:
A larger resistor would increase the time constant and fire the heart fewer than 75 times per minute.
113.
Picture the Problem
: A long, thin wire of resistance R is cut into three equal lengths. The three lengths are then
connected as resistors in parallel.
Strategy:
Since resistance is directly proportional to length (equation 213), each of the three pieces has resistance
1
3
.
R
Use equation 2110 to find the equivalent resistance of the three resistors connected in parallel.
Solution:
Solve equation 2110 for
eq
R
:
1
eq
1 1 1
3 3 3
1 1 1
9
R
R
R R R
−
⎛ ⎞
= + + =
⎜ ⎟
⎝ ⎠
Insight:
If two of the three pieces were connected in series, and the pair of them connected in parallel with the third, the
equivalent resistance would be
2
9
.
R
If two pieces were connected in parallel, and the pair connected in series with the
third, the equivalent resistance would be
1
2
.R
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 40
114.
Picture the Problem
: Three resistors of values R,
1
2
,R
and
2R
are connected in series with each other and with a
battery. They are then connected in parallel with each other and with a battery.
Strategy:
Use equations 215 and 216 to determine which resistor dissipates the most power in the series and parallel
configurations. The current through each resistor is the same in the series configuration, and the voltage across the
resistor is the same in the parallel configuration.
Solution:
1. (a)
Since
2
P I R=
(equation 215), the largest resistance 2R dissipates power at the greatest rate.
2.
(b)
Since
2
P V R= (equation 216), the smallest resistance
1
2
R
dissipates power at the greatest rate.
Insight:
In the series circuit the 2R resistor also has the greatest voltage drop across it, and in the parallel circuit the
1
2
R
resistor has the greatest current through it.
115.
Picture the Problem
: Three resistors are connected together in the manner
indicated by the diagram at the right, and a 12.0V battery is connected across
terminals A and B.
Strategy:
The voltage across each of the two branches is 12.0 V, so we can use
Ohm’s Law (equation 212) to determine the current through each branch.
Solution:
1. (a)
The equivalent resistance
eq
R
for the upper two resistors in series is
35 82 117 ,Ω+ Ω= Ω
which is
greater than
45 .Ω
Therefore, the current in the
45
Ω
e獩獴潲猠gr敡瑥e⁴桡n⁴h攠e×rre湴渠nh攠×灰p爠扲b湣n,湣=×d楮朠
瑨攠
㌵Ω
resistor.
2.
(b)
Apply Ohm’s Law to the lower branch:
45
12.0 V
0.27 A
45
V
I
R
= = =
Ω
3.
Apply Ohm’s Law to the upper branch:
35 82
eq
12.0 V
0.103 A 103 mA
117
V
I I
R
= = = = =
Ω
Insight:
If the three resistors were connected in series, the current through all three resistors would be 74 mA.
116.
Picture the Problem
: Two resistors, a capacitor, a switch, and a battery are
connected as indicated in the diagram at the right.
Strategy:
The capacitor acts like a short circuit immediately after the switch is
thrown because it is uncharged and the potential difference across its plates is zero.
Long after the switch is thrown, the capacitor has charged up to its maximum value
and no current flows through it. All of the current then remains in the perimeter of
the circuit, flowing through the 0.73Ω, 11Ω, and 5.6Ω resistors. Use these
principles together with Ohm’s Law (equation 212) to find the current and terminal
voltage long after the switch is thrown.
Solution 1. (a)
With no current flowing, the current through the internal resistance is zero, and the voltage across the
battery terminals is 9.0 V.
2. (b)
When the switch is closed, current flows through the battery’s internal resistance, producing a voltage drop. The
voltage across the terminals will therefore decrease when the switch is closed.
3.
(c)
For the three resistances in series,
eq
11 5.6 0.73 17.33 .R = Ω+ Ω+ Ω= Ω
Apply
Ohm’s Law to find the voltage across the
terminals:
( )
i i
eq
i
eq
0.73
1 9.0 V 1 8.6 V
17.33
V IR R
R
R
R
ε
ε ε
ε
⎛ ⎞
= − = −
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎛ ⎞
Ω
⎛ ⎞
= − = − =
⎜ ⎟
⎜ ⎟
⎜ ⎟
Ω
⎝ ⎠
⎝ ⎠
Insight:
Reducing the circuit resistance will allow more current to flow, which in turn will increase the voltage drop
across the internal resistance of the battery, and the terminal voltage will decrease.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 41
117.
Picture the Problem
: The National Electric Code specifies the maximum current that can flow through wires of
standard sizes, which in effect limits the amount of power per length that can be dissipated by the wires.
Strategy:
Use equation 213 to find the resistance per length of the wires, then equation 215 to find the power
dissipated per length when the maximum specified current is flowing in the wires.
Solution:
1. (a)
Combine
equations 215 and 213:
2 2 2 2
2 2
1
4
P R
I I I I
L L A r d
ρ ρ ρ
π π
= = = =
2.
Find
P L
for 8gauge wire:
( )
( )
8
2
2
3
1
4
1.68 10 m
35 A 2.4 W/m
129 10 in 0.0254 m/in
P
L
π
−
−
× Ω⋅
= =
× ×
3. (b)
Find
P L
for 10gauge wire:
( )
( )
8
2
2
3
1
4
1.68 10 m
25 A 2.0 W/m
102 10 in 0.0254 m/in
P
L
π
−
−
× Ω⋅
= =
× ×
Insight:
If the wires were to dissipate powers greater than these values, sufficient heat could be produced to start a fire.
118.
Picture the Problem
: Five capacitors are connected as shown in the diagram
at right, and a 15.0V battery is connected to terminals A and B.
Strategy:
The charge on each capacitor can be determined by applying the
rules of combinations of capacitors (equations 2114 and 2117) to the
network of capacitors, and using
Q CV=
(equation 209) to find the charge
on each capacitor. The total energy stored in the system can be calculated
from
2
1
eq
2
U C V=
(equation 2017) where
eq
C
is the equivalent capacitance
of all five capacitors.
Solution:
1. (a)
Use equation 209 to find
1
Q
:
(
)
( )
6
1 1
7.22 10 F 15.0 V 108 CQ CV
μ
−
= = × =
2.
Find the equivalent capacitance of
2 3 4
, , and :C C C
1
234
1 1
4.25 F 9.17 F
12.0 F 8.35 F
C
μ μ
μ μ
−
⎛ ⎞
= + + =
⎜ ⎟
⎝ ⎠
3.
Put
234
C
in series with
15.0 FC
μ
=
:
1
C234
1 1
5.69 F
9.17 F 15.0 F
C
μ
μ μ
−
⎛ ⎞
= + =
⎜ ⎟
⎝ ⎠
4.
The charge on C is the same as the charge on
C234
:C
(
)
( )
6
C C234
5.69 10 F 15.0 V 85.4 CQ C V
μ
−
= = × =
5.
Find the equivalent capacitance of
3 4
and :C C
1
34
1 1
4.923 F
12.0 F 8.35 F
C
μ
μ μ
−
⎛ ⎞
= + =
⎜ ⎟
⎝ ⎠
6.
The charge
234
Q
is the same as
C
Q
and is shared by
2
C
and
34
C
in proportion to their capacitances:
( )
2
2 C
2 34
4.25 F
85.4 C
4.25 F 4.923 F
39.6 C
C
Q Q
C C
μ
μ
μ μ
μ
⎛ ⎞
⎛ ⎞
= =
⎜ ⎟
⎜ ⎟
+ +
⎝ ⎠
⎝ ⎠
=
7.
3 4
and Q Q
are equal because
3 4
and C C
are connected in series:
3 4 C 2
85.4 C 39.6 C 45.8 CQ Q Q Q
μ μ μ
= = − = − =
8. (b)
Apply equation 2017 to the entire system:
(
)
( )
( )
2 2
1 1
eq 1 C234
2 2
2
6 6
1
2
3
7.22 10 F 5.69 10 F 15.0 V
1.45 10 J 1.45 mJ
U C V C C V
U
− −
−
= = +
= × + ×
= × =
9.
(c)
Because
1
C
is in parallel with the other capacitors, increasing its value will increase the overall capacitance
eq
,C
so that the total energy
2
1
eq
2
U C V=
will increase.
Insight:
The total energy stored by the system can also be found by summing
2
2U Q C=
(equation 2018) for each of
the five capacitors, because the charge Q on each capacitor is known from part (a).
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 42
119.
Picture the Problem
: Three resistors, a capacitor, and a battery are
connected in a circuit as depicted in the diagram at right.
Strategy:
The circuit is a simple RC circuit if we replace the three resistors
with their equivalent resistance. Initially, the capacitor carries no charge and
acts like a closed switch. Therefore 15 V will be across the 24 Ω resistor and
the branch containing the 13 Ω and 6.5 Ω resistors. The power dissipated in
each resistor just after t = 0 can then be found using equations 215 and 216.
We can then use equations 217 and 2110 to find
eq
,R
calculate the time
constant, and find the charge on the capacitor as a function of time (equation
2118). The current after a very long time will be zero because the voltage
across the capacitor would have increased to 15 V.
Solution:
1. (a)
Use equation 216 to find the power
dissipated by the 24Ω resistor:
( )
2
24
15 V
9.4 W
24
P = =
Ω
2.
Find the current in the left branch of the circuit:
left
left
15 V
0.769 A
13 6.5
V
I
R
= = =
+ Ω
3.
Use equation 215 to find the power dissipated
by the 13Ω resistor:
( ) ( )
2
2
13 left
0.769 A 13 7.7 WP I R= = Ω =
4.
Repeat for the 6.5Ω resistor:
( ) ( )
2
2
6.5 left
0.769 A 6.5 3.8 WP I R= = Ω =
5.
As
,t →∞
the capacitor acts like an open switch and
0,I →
so that no power is dissipated by the resistors:
24 13 6.5
0P P P
=
= =
6. (b)
Find
eq
R
of the three resistors:
1
eq
1 1
10.8
24 13 6.5
R
−
⎛ ⎞
=
+ = Ω
⎜ ⎟
Ω Ω+ Ω
⎝ ⎠
7.
Calculate the time constant:
(
)
( )
6
eq
10.8 62 10 F 0.67 msR C
τ
−
= = Ω × =
8.
Use equation 2118 to find q at t
=
0.35 ms:
(
)
( )
( )
/
6 (0.35 ms)/(0.67 ms)
4
1
62 10 F 15 V 1
3.8 10 C 0.38 mC
t
q C e
e
q
τ
ε
−
− −
−
= −
⎡ ⎤
= × −
⎣ ⎦
= × =
9. (c)
As
,t →∞
the capacitor is fully charged to 15 V:
(
)
( )
2
2 6
1 1
2 2
62 10 F 15 V 7.0 mJU C
ε
−
= = × =
10.
(d)
Doubling
ε
慵獥猠
2
1
2
U C
ε
=
to increase by a factor of four. It quadruples.
Insight:
If the capacitor were doubled in size, the time constant would also double, but it would actually take less time
(0.31 ms instead of 0.35 ms) for the capacitor to accumulate 0.38 mC of charge because the larger capacitor can store
more charge. The energy stored in the capacitor a long time after the switch is closed would also double.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 43
120.
Picture the Problem
: Two resistors are connected in parallel with each other and with a battery.
Strategy:
Use equation 216 to find an algebraic expression for the sum of the powers P
1
and P
2
that are dissipated by
the two resistors R
1
and R
2
, keeping in mind that each resistor has the same voltage across it because they are connected
in parallel. Then use equation 2110 to find an expression for the equivalent resistance R
eq
of the pair of resistors, and
use equation 216 again to find the amount of power dissipated by R
eq
when a potential V is applied across it.
Solution:
1.
Combine equations 216 and
2110 to find an expression for
1 2
:P P+
2 2
2
1 2
1 2 1 2
1 1V V
P P V
R R R R
⎛ ⎞
+ = + = +
⎜ ⎟
⎝ ⎠
2.
Substitute equation 2110 into equation 216:
2
2 2
total
eq eq 1 2
1 1 1V
P V V
R R R R
⎛ ⎞
⎛ ⎞
= = = +
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
3.
The two expressions are equivalent, so we have shown that
1 2 total
.P P P
+
=
Insight:
If one resistor were much larger than the other, the one with the smallest resistance would dissipate the greatest
amount of power.
121.
Picture the Problem
: A battery with emf
ε
慮搠楮瑥牮慬r牥獩rt慮捥a r is connected to two known resistors, and the
current passing through the battery is measured in each case.
Strategy:
Write an equation for each resistor and algebraically solve the resulting two equations for the two unknowns
ε
湤= r. In each case the emf of the battery equals the voltage drop across the internal resistance plus the voltage drop
across the external resistor. The current through the internal resistance is the same as the current through the external
resistor because they are connected in series.
Solution:
1.
Write Ohm’s law
for each of the two cases:
1 1 1
2 2 2
.................(i)
................(ii)
I r I R
I r I R
ε
ε
= +
= +
2.
Subtract (ii) from (i) and solve for r:
(
)
( )( ) ( )( )
1 2 1 1 2 2
2 2 1 1
1 2
0
0.45 A 55 0.65 A 25
43
0.65 0.45 A
I I r I R I R
I R I R
r
I I
= − + −
Ω − Ω
−
=
= = Ω
− −
3.
Substitute r into (i) and solve for
:
ε
=
(
)
(
) ( )( )
1 1 1
0.65 A 25 0.65 A 43 44 VI R I r
ε
= + = Ω + Ω =
Insight:
A procedure like the one described in the problem statement can be used to characterize a battery’s condition.
A battery with a reduced emf and a large internal resistance is likely either starting to fail or is already ruined.
122.
Picture the Problem
: Two resistors are connected in series with each other and with a battery, and the current passing
through them is measured. Then the same two resistors are connected in parallel with each other and with a battery, and
the current passing through one of them is measured.
Strategy:
Use Ohm’s Law (equation 212) to write two equations, one for the series configuration and one for the
parallel configuration. Combine the two equations algebraically in order to find the two resistances
1 2
and .R R
Solution:
1.
Write Ohm’s Law for the series case:
1 2
2
2
2.0 V
6.0 V 4.0 V
IR IR
IR I
R
ε
= +
= + ⇒ =
2.
Write Ohm’s Law for
2
R
when it is connected in
parallel with the battery:
2 2
2
2
6.0 V
6.0 V 6.0 V 40
13
0.45 A 3
I R
R
I
ε
= =
=
= = Ω= Ω
3.
Substitute the expression for I from step 1 into the
equation for the voltage drop across
1
R
:
( )
1 1
1
1 2
2
4.0 V
4.0 V 80
2.0 27
2.0 V 3
V IR
V
R R
I R
Δ
= =
Δ
=
= = = Ω= Ω
Insight:
While the method described here can be used to determine an unknown resistance, it depends upon the ability
to accurately measure current. The Wheatstone Bridge described in problem 123 may be more accurate because it
determines an unknown resistance by measuring a potential difference, which can be easier than measuring a current.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 44
123.
Picture the Problem
: Five resistors are connected together in the
Wheatstone Bridge arrangement depicted in the diagram at the right.
Strategy:
The current I
3
is zero when the potential difference across the
12.5Ω resistor is the same as the potential difference across R. When
that condition is true, both ends of the 85.0Ω resistor are at the same
potential and no current will flow through it (I
3
= 0). Furthermore, if I
3
is
zero then
1 4
I
I=
and
2 5
,
I
I=
and we’d have only two parallel branches
of current flowing through the circuit. Use these principles together with
Ohm’s Law (equation 212) to find the unknown resistance R.
Solution:
1.
Set
4 5
V VΔ = Δ
:
4 4 5
I
R I R
=
2.
Find the currents using Ohm’s Law for
each parallel branch:
4
1 4 2
R R
R R R R
ε
ε
⎛ ⎞ ⎛ ⎞
=
⎜ ⎟ ⎜ ⎟
+ +
⎝ ⎠ ⎝ ⎠
3.
Solve the expression for R by multiplying
both sides by
( )
2
R R+
and dividing by
:
ε
=
( )
( ) ( )
( )( )
( )
4 2
1 4
4 2 1 4
4 2
1
12.5 15.0
7.50
25.0
R R R R
R R
R R R R R R
R R
R
R
ε
ε
⎛ ⎞
+ =
⎜ ⎟
+
⎝ ⎠
+ = +
Ω Ω
=
= = Ω
Ω
Insight:
This can be a very precise way of measuring an unknown resistance R because a galvanometer can be
connected in series with R
3
to ensure
3
0I =
with great sensitivity. The accuracy of the determination of R then depends
only on how accurately we know the values of the other three resistors.
124.
Picture the Problem
: A simple circuit is used to find the electrical
resistance of a person and his or her footwear.
Strategy:
The same electrical current that flows through R also flows through
the person and his shoes because they are connected in series. The current
creates a voltage drop across R according to Ohm’s Law.
Solution:
Solve Ohm’s
Law for the current
flowing through R:
6
6
3.70 V
1.00 10
3.70 10 A
V
I
R
−
= =
×
Ω
= ×
Insight:
This apparatus will allow a current of 50.0 µA to pass through the
person if the person had zero resistance, so it is designed to ensure the current
never exceeds 150 µA.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 45
125.
Picture the Problem
: A simple circuit is used to find the electrical
resistance of a person and his or her footwear.
Strategy:
The electrical current that flows through R, the person, and his shoes
creates a voltage drop across R according to Ohm’s Law. The resistance R
ps
of
the person and his shoes is connected in series with R. Use these principles to
find the resistance of the person and his shoes.
Solution:
1.
Write Ohm’s
Law for the circuit:
(
)
ps
I
R R
ε
= +
2.
Solve for R
ps
, inserting
the current determined in
problem 124:
ps
6
6
7
ps
50.0 V
1.00 10
3.70 10 A
1.25 10 12.5 M
R R
I
R
ε
−
= −
=
− × Ω
×
= × Ω = Ω
Insight:
The person’s shoes are well within the ANSI specifications of 0.1×10
7
Ω to 100×10
7
Ω.
126.
Picture the Problem
: A simple circuit is used to find the electrical
resistance of a person and his or her footwear.
Strategy:
The electrical current that flows through R, the person, and his shoes
creates a voltage drop across R according to Ohm’s Law. The resistance R
ps
of
the person and his shoes is connected in series with R. Use these principles to
find the current through the person and the reading on the voltmeter.
Solution:
1.
Solve Ohm’s
Law for the current:
ps
6 7
6
50.0 V
1.00 10 4.00 10
1.22 10 A
I
R R
I
ε
−
=
+
=
×
+ × Ω
= ×
2.
Use Ohm’s Law again to
find the voltage drop across R:
(
)
(
)
6 6
1.22 10 A 1.00 10 1.22 VV I R
−
= = × × Ω =
Insight:
The higher resistance of this person’s shoes reduces the current and the reading on the voltmeter when
compared to the previous two questions.
127.
Picture the Problem
: A simple circuit is used to find the electrical
resistance of a person and his or her footwear. In this case the shoes are wet.
Strategy:
The wet shoes will produce a lower resistance R
ps
for the person and
his shoes. Use Ohm’s Law to answer the question.
Solution:
The moisture on the person’s shoes will reduce their electrical
resistance and thereby increase the current flowing through the circuit. The
larger current I will produce a larger voltage drop IR across the resistor, so we
expect the reading on the voltmeter to increase.
Insight:
The voltmeter is essentially reading the current through the person, not
the resistance of his shoes. The lower the current, the higher the shoe’s
resistance, and the better the protection against the potentially lethal effects of
an electrical shock.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 46
128.
Picture the Problem
: Three resistors are connected to a battery in the manner
indicated by the diagram at the right.
Strategy:
The potential difference across
1
R
is equal to the battery emf. The
potential difference across
2 3
and R R
is given by Ohm’s Law (equation 212) and
is determined by the amount of current that flows through the two resistors. The
current can also be found using Ohm’s Law applied to the entire lower branch.
Solution:
1. (a)
1
V
ε
Δ =
because the
battery is connected across
1
:R
1
12.0 VV
ε
Δ = =
2.
Use Ohm’s Law to find
2
:V
Δ
( )
( )( )
2 lower 2 2
2 3
12.0 V
200.0
200.0 300.0
0.0240 A 200.0 4.80 V
V I R R
R R
ε
⎛ ⎞
⎛ ⎞
Δ
= = = Ω
⎜ ⎟
⎜ ⎟
+ + Ω
⎝ ⎠
⎝ ⎠
= Ω =
3.
Repeat to find
3
:VΔ
(
)
(
)
3 lower 3
0.0240 A 300.0 7.20 VV I RΔ = = Ω =
4.
(b)
Apply Ohm’s Law to
1
R
to find
1
:
I
1
1
12.0 V
0.120 A 120 mA
100.0
I
R
ε
= = = =
Ω
5.
The current through the lower branch
was found earlier:
2 3 lower
0.0240 A 24.0 mAI I I
= = = =
Insight:
The current through the lower branch is onefifth the current through the upper branch because the resistance
of the lower branch is five times greater than that of the upper branch.
129.
Picture the Problem
: Three resistors are connected to a battery in the manner
indicated by the diagram at the right.
Strategy:
Write an expression for the equivalent resistance of three resistors
using equations 217 and 2110. Then use Ohm’s Law (equation 212) to find
the value of R such that the current supplied by the battery is 0.0750 A. Repeat
the same procedure to find the value of R such that the potential difference
2
V
Δ
across
2
R
is 2.65 V.
Solution:
1. (a)
Find
eq
R
of the 3 resistors:
1
1
eq
1 2 3 1 2
1 1 1 1
R
R R R R R R
−
−
⎛ ⎞
⎛ ⎞
= + = +
⎜ ⎟
⎜ ⎟
+ +
⎝ ⎠
⎝ ⎠
2.
Use Ohm’s Law to set
0.0750 A:I =
eq 1 2
1 1
I
R R R R
ε
ε
⎛ ⎞
= = +
⎜ ⎟
+
⎝ ⎠
3.
Solve the expression for R:
1 2
1
1
2
1
1 1
1 0.0750 A 1
225
12.0 V 225
329
I
R R R
I
R R
R
ε
ε
−
−
− =
+
⎛ ⎞
⎛ ⎞
=
− − = − − Ω
⎜ ⎟
⎜ ⎟
Ω
⎝ ⎠
⎝ ⎠
= Ω
4.
(b)
Write Ohm’s Law for
2
R
and solve for R:
( )( )
2 lower 2 2
2
2
2
2
12.0 V 225
225 794
2.65 V
V I R R
R R
R
R R
V
ε
ε
⎛ ⎞
Δ = =
⎜ ⎟
+
⎝ ⎠
Ω
=
− = − Ω= Ω
Δ
Insight:
Decreasing R will create a larger potential difference across
2
R
because more current will flow through the
lower branch.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 47
130.
Picture the Problem
: The circuit described in the problem statement is
shown with the switch in the open position. Once the switch is closed at
t = 0, current will flow in the circuit and charge will begin to accumulate
on the capacitor plates.
Strategy:
The two resistors combine as resistors in series to form a simple
RC
circuit with the capacitor. Decreas ing the resistance decreases the time
constant and decreases the amount of time required to charge the capacitor
plates.
Solution:
1. (a)
The final value of the charge on the capacitor will stay the same if the first resistor is changed, because
the voltage across the capacitor and therefore the charge on the capacitor plates depends only upon the emf of the
battery.
2.
(b)
The time required for the capacitor to charge to 80% of its final value will decrease
b
ecause the time constant will
decrease when the resistance in the RC circuit is decreased.
3. (c)
Use the expression from
step 3 of example 219:
(
)
(
)
(
)
( )
6
ln 0.200 63 275 182 10 F ln 0.200 99.0 mst RC
−
= − = − + Ω × =
Insight:
As expected, the time required to charge the capacitor to 80% of its final value decreased by
19 ms / 118 ms = 16% because the resistance of the circuit decreased by 63 Ω / 401 Ω = 16%.
131.
Picture the Problem
: The circuit described in the problem statement is
shown with the switch in the open position. Once the switch is closed at
t = 0, current will flow in the circuit and charge will begin to accumulate
on the capacitor plates.
Strategy:
The final potential difference across the plates of the capacitor
equals the emf of the battery. We can therefore use equation 209 to find
the charge on the plates of the capacitor. Decreasing the capacitance
decreases the time constant and decreases the amount of time required to
charge the capacitor plates.
Solution:
1. (a)
Use equation 209 to find
Q when the capacitor is fully charged:
(
)
(
)
6
91.0 10 F 3.00 V 273 CQ C
μ
ε
−
= = × =
2.
(b)
The time required for the capacitor to charge to 80% of its final value will decrease
b
ecause the time constant will
decrease when the capacitance in the RC circuit is decreased.
3. (c)
Use the expression from
step 3 of example 219:
(
)
(
)
(
)
( )
6
ln 0.200 126 275 91.0 10 F ln 0.200 58.7 mst RC
−
= − = − + Ω × =
Insight:
As expected, the time required to charge the capacitor to 80% of its final value was cut in half because the
capacitance and therefore the time constant of the circuit was cut in half.
Chapter 21
: Electric Current and DirectCurrent Circuits James S. Walker, Physics, 4
th
Edition
Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21 – 48
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