Chapter 21:
Problems: 2, 6, 8, 19, 23, 36, 44, 54, 58, 62
Solutions to Problems : Chapter 21
Problems appeared on the end of chapter 21 of the Textbook
2.
Picture the Problem: A current flows through the filament of a flashlight bulb.
Strategy: Use equation 21-1 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 21-1 for
ΔQ :
ΔQ = I Δt = ( 0.18 A )( 78 s ) = 14 C
2. (b) Divide by e to find the number of
electrons:
Ne =
14 C
= 8.8 × 1019 electrons
−19
1.60 × 10 C/electron
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 20-2: U = qΔV = (14 C )( 3.0 V ) = 42 J.
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 21-4. For part (b) we can use the definition of current
(equation 21-1) and the charge on an electron to find the required time for 10 million electrons to pass
through the circuit.
P 85 W
= 0.71 A
Solution: 1. (a) Solve equation I = =
ε 120 V
21-4 for I:
2. (b) Solve equation 21-1 for
Δt :
Δt =
7
−
−19
−
ΔQ N e e (1× 10 e )(1.6 × 10 C/e ))
=
=
= 2 × 10−12 s = 2 ps
I
I
0.71 A
Insight: Ten million electrons in two picoseconds! It’s a good thing electric companies don’t charge
by the electron!
8. Picture the Problem: A silver wire of known dimensions has an intrinsic resistance.
Strategy: Use table 21-1 together with equation 21-3 to determine the resistance of the wire.
Solution: Apply equation 21-3 ,
using ρ from table 21-1 and
A = π r 2 = 14 π D 2 :
R=ρ
L
4.9 m
= (1.59 × 10−8 Ω ⋅ m )
= 0.41 Ω
2
−3
π
A
×
0.49
10
m
(
)
4
Insight: Metals typically have a very low resistance. In this case, a wire only half a millimeter thick
and 16 ft long has a resistance of less than half an ohm!
19. 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 21-4). Solve this equation for the current produced.
Solution: Solve equation 21-4 for I:
I=
P 3.8 ×103 W
=
= 58 A
V
65 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.
23. 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 21-4. First find the power
P consumed and then determine the energy ΔU consumed by multiplying the power by the time over
which the charger is operated (120 minutes or 2.0 hours). The cost per kilowatt-hour is then the total
cost divided by the energy consumed.
Solution: 1. Calculate the power delivered to the
battery:
P = IV = (15 A )(12 V ) = 180 W
2. Multiply P by Δt to find ΔU :
ΔU = P Δt = ( 0.18 kW )( 2.0 h ) = 0.36 kWh
3. Divide the total cost by ΔU in kilowatt-hours:
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.
36. 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 55-Ω resistor connected in
parallel with R, then add the equivalent resistance of the pair to 12
Ω to find Req for the entire circuit.
−1
Solution: 1. Use
equations 21-7 and
21-10 to find an
expression for Req :
2. Solve the expression for R:
1 ⎞
⎛1
Req = 12 Ω + ⎜ +
⎟ = 26 Ω
⎝ R 55 Ω ⎠
−1
1 ⎞
⎛1
⎜ +
⎟ = 14 Ω ⇒
⎝ R 55 Ω ⎠
1
1
1
+
=
R 55 Ω 14 Ω
1
1
1
=
−
R 14 Ω 55 Ω
−1
1 ⎞
⎛ 1
−
R=⎜
⎟ = 19 Ω
⎝ 14 Ω 55 Ω ⎠
Insight: The 19-Ω and 55-Ω 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.
44. Picture the Problem: Six resistors are connected to
a battery in the manner indicated in the diagram.
Strategy: The current I 3 through the 13.8-Ω resistor
determines the voltage ΔVAB because of Ohm’s
Law. That potential difference can then be used to
find I 2 and I 4 , and the current I1 is the sum of the
other three currents.
Solution: 1. (a) Find ΔVAB using Ohm’s
Law:
ΔVAB = I 3 R3 = ( 0.775 A )(13.8 Ω ) = 10.7 V
2. Apply Ohm’s Law to R4 to find I 4 :
I4 =
3. Apply Ohm’s Law to R5 and R6 to find I 2 =
I2 :
4. Add the currents to find I1 :
ΔVAB 10.7 V
=
= 0.622 A
R4
17.2 Ω
ΔVAB
10.7 V
=
= 0.852 A
R5 + R6 8.45 + 4.11 Ω
I1 = I 2 + I 3 + I 4 = 0.852 + 0.775 + 0.622 A = 2.248 A
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.
54. Picture the Problem: Three capacitors are connected in series as
shown in the diagram at the right.
Strategy: First use equation 21-17 to find the equivalent
capacitance of the three capacitors connected in series. Then use
the equation Q = C V (equation 20-9) to find the amount of charge
stored on each capacitor. Finally, solve equation 20-9 to find the
voltage drop across C3.
−1
⎛ 1
1
1 ⎞
+
+
Solution: 1. Use equation 21-17 to find Ceq: Ceq = ⎜
⎟ = 3.0 μ F
4.5
F
12
F
32
μ
μ
μF ⎠
⎝
2. Solve equation 20-9 for Q:
Q = CeqV = ( 3.0 × 10 − 6 F ) (18 V ) = 53 μ C
3. Apply equation 20-9 again to C3:
ΔV3 =
Q 53 μ C
=
= 1.7 V
C3 32 μ F
Insight: If instead the three capacitors had been connected in parallel, there would be a potential drop
of 18 V across each, and the three capacitors together would have stored 870 µC of charge.
58. Picture the Problem: Five capacitors are connected in a
network as indicated by the diagram at the right.
Strategy: By applying equations 21-14 and 21-17, we can
determine an algebraic expression for the equivalent
capacitance Ceq of the combination, set it equal to the
given value, and solve for the unknown capacitance C. To
find Ceq, we consider C3 and C4 to be connected in series
with each other and in parallel with C2. That group of
three capacitors is connected in series with the unknown
C, and the entire group of four is connected in parallel
with C1.
−1 −1 ⎫
⎧
⎛ 1
1 ⎞ ⎤ ⎪
⎪1 ⎡
Ceq = C1 + ⎨ + ⎢C2 + ⎜ +
⎟ ⎥ ⎬
⎝ C3 C4 ⎠ ⎥⎦ ⎪
⎪⎩ C ⎢⎣
⎭
Solution: 1. Apply
equations 21-14
and 21-17 to find an
expression for Ceq:
2. Set Ceq = 9.22 μ F and solve
for C:
−1
−1 −1 ⎫
⎧
⎛
⎞ ⎤ ⎪
1
1
⎪1 ⎡
9.22 μ F = 7.22 μ F + ⎨ + ⎢ 4.25 μ F + ⎜
+
⎟ ⎥ ⎬
⎝ 12.0 μ F 8.35 μ F ⎠ ⎦⎥ ⎪
⎪⎩ C ⎢⎣
⎭
1
1
1
= +
2.00 μ F C 9.17 μ F
−1
−1
⎛
⎞
1
1
−
C =⎜
⎟ = 2.56 μ F
2.00
F
9.17
F
μ
μ
⎝
⎠
Insight: If C were increased from 2.56 µF to 12.0 µF, Ceq would increase from 9.22 µF to 12.4 µF.
62. Picture the Problem: A resistor and a capacitor form a series
RC circuit as shown in the diagram at the right.
Strategy: Apply equation 21-18 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 :
2. (b) Solve
equation 21-19
for the case when
t =τ :
q (τ ) = Cε (1 − e −τ τ )
= ( 45 × 10− 6 F ) ( 9.0 V ) (1 − e −1 )
q (τ ) = 2.6 × 10 −4 C = 260 μ C
I (τ ) =
ε e ττ
−
R
=
9.0 V −1
e = 28 mA
120 Ω
Insight: The charge q ( t ) starts out at zero and increases to 405 µC when the capacitor is fully charged,
whereas I ( t ) starts out at 75 mA and decreases to zero when the capacitor is fully charged.