Chapter 18
Direct-Current Circuits
Quick Quizzes
1.
(a), (d). Bulb R1 becomes brighter. Connecting a wire from b to c provides a nearly zero
resistance path from b to c and decreases the total resistance of the circuit from R1 + R2 to
just R1. Ignoring internal resistance, the potential difference maintained by the battery is
unchanged while the resistance of the circuit has decreased. The current passing through
bulb R1 increases, causing this bulb to glow brighter. Bulb R2 goes out because essentially
all of the current now passes through the wire connecting b and c and bypasses the
filament of Bulb R2.
2.
(b). When the switch is opened, resistors R1 and R2 are in series, so that the total circuit
resistance is larger than when the switch was closed. As a result, the current decreases.
3.
(a). When the switch is closed, resistors R1 and R2 are in parallel, so that the total circuit
resistance is smaller than when the switch was open. As a result, the total current
increases.
4.
(a) unchanged; (b) unchanged; (c) increase; (d) decrease. Neglecting internal resistance,
the terminal potential difference of the battery remains constant (equal to the emf) as
bulbs are added. Since each bulb is added in parallel with the battery, the current in each
bulb, and hence the brightness of the bulb, is unchanged as new bulbs are added. The total
current supplied by the battery increases as bulbs are added in parallel. Thus, the power
delivered by the battery increases and its lifetime decreases.
5.
((a) decrease; (b) decrease; (c) decrease; (d) increase. Adding bulbs in series increases the
total resistance of the circuit and results in a decrease in the current in the bulbs. Thus, the
brightness of individual bulbs deceases as bulbs are added. Neglecting internal resistance,
the terminal potential difference of the battery remains constant (equal to the emf) and the
power supplied by the battery decreases as bulbs are added. This means that the battery
will supply energy at a lower rate, and its lifetime will be increased.
6.
(c). After the capacitor is fully charged, current flows only around the outer loop of the
circuit. This path has a total resistance of 3 Ω , so the 6-V battery will supply a current of
2 Amperes.
99
100
CHAPTER 18
Answers to Even Numbered Conceptual Questions
2.
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
4.
(a) ii (b) i
6.
(a) ii (b) i
8.
A short circuit can develop when the last bit of insulation frays away between the two
conductors in a lamp cord. Then the two conductors touch each other, opening a low
resistance branch in parallel with the lamp. The lamp will immediately go out, carrying no
current and presenting no danger. A very large current will be produced in the power
source, the house wiring, and the wire in the lamp cord up to and through the short. The
circuit breaker will interrupt the circuit quickly but not before considerable heating and
sparking is produced in the short-circuit path.
10.
A wire or cable in a transmission line is thick and made of material with very low
resistivity. Only when its length is very large does its resistance become significant. To
transmit power over a long distance it is most efficient to use low current at high voltage,
minimizing the I 2 R power loss in the transmission line.
12.
The bulbs of set A are wired in parallel. The bulbs of set B are wired in series, so removing
one bulb produces an open circuit with infinite resistance and zero current.
14.
(a) ii. The power delivered may be expressed as P = I 2 R , and while resistors connected in
series have the same current in each, they may have different values of resistance.
2
(b) ii. The power delivered may also be expressed as P = ( ∆V ) R , and while resistors
connected in parallel have the same potential difference across them, they may have
different values of resistance.
16.
Compare two runs in series to two resistors connected in series. Compare three runs in
parallel to three resistors connected in parallel. Compare one chairlift followed by two
runs in parallel to a battery followed immediately by two resistors in parallel.
The junction rule for ski resorts says that the number of skiers coming into a junction must
be equal to the number of skiers leaving. The loop rule would be stated as the total change
in altitude must be zero for any skier completing a closed path.
18.
Because water is a good conductor, if you should become part of a short circuit when
fumbling with any electrical circuit while in a bathtub, the current would follow a
pathway through you, the water, and to ground. Electrocution would be the obvious
result.
Direct-Current Circuits
101
20.
Even if the fuse were to burn out or the circuit breaker to trip, there would still be a
current path through the device, and it would not be protected.
22.
When connected in series, all bulbs carry the same current. Thus, the one with the lowest
resistance dissipates the least power  P = I 2 R  and glows the dimmest. This is seen to be
the bulb that is labeled “200 W”. If they were connected in parallel, all bulbs would have
the same potential difference across them. Then, the one with the lowest resistance will
2
dissipate the most power  P = ( ∆V ) R  and glow the brightest.
24.
The break is closer to A.
102
CHAPTER 18
Answers to Even Numbered Problems
(c)
2.18 Ω , I 4 = 6.0 A, I 8 = 3.0 A, I12 = 2.0 A
30 V
(b)
2.3 V
5.13 Ω
(b)
4.53 V
(b)
25.0 W, 6.25 W, 6.25 W; 37.5 W
2.
(a)
24 Ω
4.
(a)
6.
15 Ω
8.
(a)
(b)
1.0 A
10.
RA = 6.0 Ω, RB = 3.0 Ω
12.
(a)
14.
14.2 W, 28.4 W, 1.33 W, 4.00 W
16.
0.714 A, 1.29 A, 12.6 V
18.
5.4 V with a at a higher potential than b
20.
I 2 = − 2.0 A (flows from b toward a), I1 = 1.0 A
22.
0.50 W
24.
(a)
26.
0.28 A (dead battery), 1.7 × 10 2 A (starter)
28.
(a) 9.0 V with b at a higher potential than a
(b) 0.42 A directed from b to a
32.
(a)
34.
47 µ F
36.
(a)
10.0 µ F
38.
(a)
8.0 A
40.
No. Their combined power requirements exceed the 1 800 W available.
42.
(a)
44.
0.865 or 86.5%
46.
(a)
48.
(a)
75.0 V
4.59 Ω
(b)
2.00 ms
(b)
0.0816 or 8.16%
(b)
180 µ C
(b)
415 µ C
120 V
(c)
(c)
0.80 A
(b)
0.56 µ A
14 Ω
(b)
56 W
40.0 W
(b)
80.0 V, 40.0 V, 40.0 V
4.1 × 10 −11 J
114 µ C
(d)
(c)
2.0 A
5.8 × 10 2 W
Direct-Current Circuits
50.
(a) 0 in the 3-kΩ resistor and 333 µ A in the others
(b) 50.0 µ C
52.
14 Ω
54.
14 s
56.
R = 20 Ω or R = 98 Ω
60.
Q1 = ( 240 µ C ) ( 1 − e −1 000 t 6.0 s ) , Q2 = ( 360 µ C ) ( 1 − e −1 000 t 6.0 s )
62.
(a) 0.099 9 Ω , I R1 = 50.0 A, I R2 = I R3 = I100 = 45.0 mA
103
(b) 1.09 Ω , I R1 = I R2 = 4.55 A, I R3 = I100 = 45.5 mA
(c) 9.99 Ω, I R1 = I R2 = I R3 = 0.450 A, I100 = 50.0 mA
64.
(a) The two bulbs have the same current, power delivered, and hence same brightness.
(b) Again, the two bulbs have the same current, power delivered, and brightness.
(c) The bulbs in case 2 are brighter than those in case 1 because each bulb carries twice the
current it had in case 1.
(d) In case 1, both bulbs go out if either bulb fails because the current in the circuit would
be zero. In case 2, if one bulb fails, the other bulb is unaffected and maintains its
original current and brightness.
104
CHAPTER 18
Problem Solutions
18.1
From ∆V = I ( R + r ) , the internal resistance is
r=
18.2
(a)
∆V
9.00 V
−R=
− 72.0 Ω = 4.92 Ω
I
0.117 A
Req = R1 + R2 + R3 = +4.0 Ω + 8.0 Ω + 12 Ω = 24 Ω
(b) The same current exists in all resistors in a series combination.
I=
∆V 24 V
=
= 1.0 A
Req 24 Ω
(c) If the three resistors were connected in parallel,
−1
−1
 1
1
1 
1
1 
 1
Req = 
+
+
+
+
 =
 = 2.18 Ω
 4.0 Ω 8.0 Ω 12 Ω 
 R1 R2 R3 
Resistors in parallel have the same potential difference across them, so
I4 =
18.3
∆V 24 V
24 V
24 V
=
= 6.0 A , I 8 =
= 3.0 A , and I12 =
= 2.0 A
R4 4.0 Ω
8.0 Ω
12 Ω
For the bulb in use as intended,
0.800 W
Rbulb =
( ∆V )
P
2
=
( 120 V )
2
75.0 W
= 192 Ω
Now, presuming the bulb resistance is unchanged,
the current in the circuit shown is
I=
∆V
120 V
=
= 0.620 A
Req 0.800 Ω + 192 Ω + 0.800 Ω
and the actual power dissipated in the bulb is
P = I 2 Rbulb = ( 0.620 A ) ( 192 Ω ) = 73.8 W
2
120 V
192 W
0.800 W
Direct-Current Circuits
18.4
a
(a) The current through this series combination is
I=
( ∆V )bc
Rbc
=
b
9.0 W
12 V
= 2.0 A
6.0 Ω
Therefore, the terminal potential difference of the
power supply is
105
c
6.0 W
DV
∆V = I Req = ( 2.0 A ) ( 9.0 Ω + 6.0 Ω ) = 30 V
(b) When connected in parallel, the potential difference across either resistor is the
voltage setting of the power supply. Thus,
∆V = I 9 R9 = ( 0.25 A )( 9.0 Ω ) = 2.3 V
18.5
7.00 W
(a) The equivalent resistance of the two parallel
resistors is
4.00 W
9.00 W
−1
1 
 1
Rp = 
+
 = 4.12 Ω
Ω
Ω
7.00
10.0

10.0 W
a
Thus,
Rab = R4 + Rp + R9 = ( 4.00 + 4.12 + 9.00 ) Ω = 17.1 Ω
(b) I ab =
( ∆V ) ab
Rab
=
34.0 V
= 1.99 A , so I 4 = I 9 = 1.99 A
17.1 Ω
Also,
( ∆V ) p = I ab Rp = ( 1.99 A )( 4.12 Ω ) = 8.18 V
Then,
I7 =
and
I10 =
( ∆V ) p
R7
( ∆V ) p
R10
=
8.18 V
= 1.17 A
7.00 Ω
=
8.18 V
= 0.818 A
10.0 Ω
b
106
CHAPTER 18
18.6
The equivalent resistance of the parallel combination of three resistors is
−1
1
1 
 1
Rp = 
+
+
 = 3.0 Ω
 18 Ω 9.0 Ω 6.0 Ω 
Hence, the equivalent resistance of the circuit connected to the 30 V source is
Req = R12 + Rp = 12 Ω + 3.0 Ω = 15 Ω
18.7
If a potential difference is applied between points a and b, the vertical resistor with a free
end is not part of any closed current path. Hence, it has no effect on the circuit and can
be ignored. The remaining four resistors between a and b reduce to a single equivalent
resistor, Req = 2.5R , as shown below:
R
a
18.8
R
R
R
b
a
R
R
R/2
b
a
2.5R
b
(a) The rules for combining resistors in series and parallel are used to reduce the circuit
to an equivalent resistor in the stages shown below. The result is Req = 5.13 Ω .
10.0 W
3.00 W
4.00 W
10
—
3
3.00 W
Figure 1
Figure 2
W
3.00 W
3.00 W
4.00 W
3.00 W
5.00 W
3.00 W
22
—
3
W
66
—
31
W
Req = 5.13 W
3.00 W
Figure 3
Figure 4
(b) From P = ( ∆V ) Req , the emf of the power source is
2
∆V = P⋅Req =
( 4.00 W )( 5.13 Ω ) = 4.53 V
Figure 5
Direct-Current Circuits
18.9
Turn the circuit given in Figure P18.9 90° counterclockwise to observe that it is
equivalent to that shown in Figure 1 below. This reduces, in stages, as shown in the
following figures.
b
25.0
V
10.0
W
b
I20
5.00
W
5.00
c
W
20.0
W
10.0
W
25.0
V
10.0
W
I20
5.00
W
10.0
W
a
Figure 1
25.0
W
a
Figure 2
b
I
25.0
V
2.94
W
10.0
W
I
25.0
V
a
Figure 3
12.9
W
Figure 4
From Figure 4,
I=
∆V 25.0 V
=
= 1.93 A
R
12.9 Ω
(b) From Figure 3,
( ∆V )ba = I Rba
= ( 1.93 A ) ( 2.94 Ω ) = 5.68 V
(a) From Figures 1 and 2, the current through the 20.0 Ω resistor is
I 20 =
( ∆V )ba
Rbca
=
5.68 V
= 0.227 A
25.0 Ω
107
108
CHAPTER 18
18.10
First, consider the parallel case. The resistance of resistor B is
RB =
( ∆V )B
IB
=
6.0 V
= 3.0 Ω
2.0 A
In the series combination, the potential difference across B is given by
( ∆V )B = ( ∆V )battery − ( ∆V ) A = 6.0 V-4.0 V = 2.0 V
The current through the series combination is then
Is =
( ∆V ) B
RB
=
2.0 V 2
= A
3.0 Ω 3
and the resistance of resistor A is RA =
18.11
( ∆V ) A
Is
=
4.0 V
= 6.0 Ω
23A
The equivalent resistance is Req = R + Rp , where Rp is the total resistance of the three
parallel branches;
1
1
1
( 30 Ω )( R + 5.0 Ω )
 1

 1

Rp = 
+
+
+
 =
 =
R + 35 Ω
 120 Ω 40 Ω R + 5.0 Ω 
 30 Ω R + 5.0 Ω 
−1
Thus, 75 Ω = R +
−1
( 30 Ω ) ( R + 5.0 Ω ) R2 + ( 65 Ω ) R + 150 Ω 2
R + 35 Ω
=
R + 35 Ω
which reduces to R2 − ( 10 Ω ) R − 2 475 Ω 2 = 0 or ( R − 55 Ω )( R + 45 Ω ) = 0 .
Only the positive solution is physically acceptable, so R = 55 Ω
Direct-Current Circuits
18.12
(a) The total current from a to b is equal to the maximum
current allowed in the 100 Ω series resistor adjacent
to point a. This current has a value of
I max =
109
100 W
a
100 W
100 W
Pmax
25.0 W
=
= 0.500 A
R
100 Ω
The total resistance is
−1
1 
 1
Rab = 100 Ω + 
+
 = 100 Ω + 50.0 Ω = 150 Ω
 100 Ω 100 Ω 
Thus, ( ∆V )max = I max Rab = ( 0.500 A )( 150 Ω ) = 75.0 V
2
(b) The power dissipated in the series resistor is P1 = I max
R = 25.0 W , and the power
dissipated in each of the identical parallel resistors is
P2 = P3 = ( I max 2 ) R = ( 0.250 A ) ( 100 Ω ) = 6.25 W
2
2
The total power delivered is
P = P1 + P2 + P3 = ( 25.0 + 6.25 + 6.25 ) W = 37.5 W
b
110
CHAPTER 18
18.13
The resistors in the circuit can be combined in the stages shown below to yield an
equivalent resistance of Rad = ( 63 11) Ω .
6.0 W
3.0 W
a 3.0 W
b
I1
c
d
3.0 W
6.0 W
4.0 W
I2
3.0 W
a
I2
2.0 W
I
e
12 W
I12
I1
3.0 W
c
d
b
e
3.0 W
2.0 W
I
18 V
Figure 1
Figure 2
18 V
6.0 W
3.0 W I1
a
I
I2
b
3W
a
d
W
d
( ∆V )ad
Rad
=
18 V
Figure 4
Figure 5
18 V
= 3.14 A
( 63 11) Ω
( ∆V )bd = I Rbd = ( 3.14 A ) ( 30 11 Ω ) = 8.57 V
so
( ∆V )bd
3.0 Ω + 2.0 Ω
=
8.57 V
= 1.71 A
5.0 Ω
( ∆V )be = I 2 Rbe = ( 1.71 A )( 3.0 Ω ) = 5.14 V
Finally, from Figure 1,
I12 =
( ∆V )be
R12
W
18 V
Now, look at Figure 2 and observe that
I2 =
63
—
11
a
I
18 V
Figure 3
Then, from Figure 4,
30
—
11
I
5.0 W
From Figure 5, I =
b
=
5.14 V
= 0.43 A
12 Ω
d
Direct-Current Circuits
18.14
2.00 W
The resistance of the parallel combination of the
3.00 Ω and 1.00 Ω resistors is
−1
1 
 1
Rp = 
+
 = 0.750 Ω
 3.00 Ω 1.00 Ω 
111
3.00
W
18.0
V
1.00
W
4.00 W
The equivalent resistance of the circuit connected to
the battery is
2.00 W
Req = 2.00 Ω + Rp + 4.00 Ω = 6.75 Ω
and the current supplied by the battery is
I=
∆V 18.0 V
=
= 2.67 A
Req 6.75 Ω
18.0
V
0.750
W
4.00 W
The power dissipated in the 2.00-Ω resistor is
P2 = I 2 R2 = ( 2.67 A ) ( 2.00 Ω ) = 14.2 W
2
and that dissipated in the 4.00-Ω resistor is
P4 = I 2 R4 = ( 2.67 A ) ( 4.00 Ω ) = 28.4 W
2
The potential difference across the parallel combination of the 3.00 Ω and 1.00 Ω
resistors is
( ∆V ) p = I Rp = ( 2.67 A ) ( 0.750 Ω ) = 2.00 V
Thus, the power dissipation in these resistors is given by
( ∆V ) p
2
P3 =
P1 =
( ∆V ) p
R1
2
3.00 Ω
R3
2
and
( 2.00 V )
=
( 2.00 V )
=
1.00 Ω
= 1.33 W
2
= 4.00 W
112
CHAPTER 18
18.15
(a) Connect two 50-Ω resistors in parallel to get 25 Ω . Then connect that parallel
combination in series with a 20-Ω resistor for a total resistance of 45 Ω .
(b) Connect two 50-Ω resistors in parallel to get 25 Ω .
Also, connect two 20-Ω resistors in parallel to get 10 Ω .
Then, connect these two parallel combinations in series to obtain 35 Ω .
18.16
Going counterclockwise around the upper loop,
applying Kirchhoff’s loop rule, gives
+15.0 V − ( 7.00 ) I1 − ( 5.00 ) ( 2.00 A ) = 0
or
I1 =
15.0 V − 10.0 V
= 0.714 A
7.00 Ω
7.00 W
I1
I2
From Kirchhoff’s junction rule, I1 + I 2 − 2.00 A = 0
so
I 2 = 2.00 A − I1 = 2.00 A − 0.714 A = 1.29 A
Going around the lower loop in a clockwise direction gives
+ E− ( 2.00 ) I 2 − ( 5.00 ) ( 2.00 A ) = 0
or
E = ( 2.00 Ω ) ( 1.29 A ) + ( 5.00 Ω ) ( 2.00 A ) = 12.6 V
15.0 V
5.00 W
2.00 W
A
e
Direct-Current Circuits
18.17
3.00 W
We name the currents I1 , I 2 , and I 3 as shown.
Using Kirchhoff’s loop rule on the rightmost
loop gives
I1
I2
5.00
W
+12.0 V- ( 1.00+3.00 ) I 3
− ( 5.00 + 1.00 ) I 2 − 4.00 V = 0
( 2.00 ) I 3 + ( 3.00 ) I 2 = 4.00 V
or
113
(1)
1.00
W
I3
1.00
W
8.00
W
4.00
V
12.0
V
Applying the loop rule to the leftmost loop yields
+4.00 V+ ( 1.00+5.00 ) I 2 − ( 8.00 ) I1 = 0
( 4.00 ) I1 − ( 3.00 ) I 2 = 2.00 V
or
From Kirchhoff’s junction rule, I1 +I 2 =I 3
(2)
(3)
Solving equations (1), (2) and (3) simultaneously gives
I1 =0.846 A, I 2 =0.462 A, and I 3 = 1.31 A
All currents are in the directions indicated by the arrows in the circuit diagram.
18.18
Observe that the center branch of this circuit, that is the branch containing points a and
b, is not a continuous conducting path, so no current can flow in this branch. The only
current in the circuit flows counterclockwise around the perimeter of this circuit. Going
counterclockwise around the this outer loop and applying Kirchhoff’s loop rule gives
−8.0 V − ( 2.0 Ω ) I − ( 3.0 Ω ) I +12 V − ( 10 Ω ) I − ( 5.0 Ω ) I = 0
or
I=
12 V − 8.0 V
= 0.20 A
20 Ω
Now, we start at point b and go around the upper panel of the circuit to point a, keeping
track of changes in potential as they occur. This gives
∆Vab = Va − Vb = −4.0 V+ ( 6.0 Ω )( 0 ) − ( 3.0 Ω )( 0.20 A ) + 12 V − ( 10 Ω )( 0.20 A ) = +5.4 V
Since ∆Vab > 0 , point a is 5.4 V higher in potential than point b
114
CHAPTER 18
18.19
(a) Applying Kirchhoff’s loop rule, as you go clockwise around the loop, gives
+ 20.0 V − ( 2 000 ) I − 30.0 V − ( 2 500 ) I + 25.0 V − ( 500 ) I = 0 ,
or I = 3.00 × 10 −3 A = 3.00 mA
(b) Start at the grounded point and move up the left side, recording changes in
potential as you go, to obtain
VA = + 20.0 V − ( 2 000 Ω ) ( 3.00 × 10 −3 A ) − 30.0 V − ( 1 000 Ω ) ( 3.00 × 10 −3 A )
or VA = − 19.0 V
(c)
( ∆V )1 500 = ( 1 500 Ω ) ( 3.00 × 10 −3 A ) = 4.50 V
(The upper end is at the higher potential.)
18.20
Following the path of I1 from a to b, and recording
changes in potential gives
Vb − Va = + 24 V − ( 6.0 Ω )( 3.0 A ) = + 6.0 V
e
I3
R
I2
3.0 W
Now, following the path of I 2 from a to b, and
recording changes in potential gives
Vb − Va = − ( 3.0 Ω ) I 2 = + 6.0 V , or I 2 = − 2.0 A
Thus, I 2 is directed from from b toward a and has magnitude of 2.0 A.
Applying Kirchhoff’s junction rule at point a gives
I 3 = I1 + I 2 = 3.0 A + ( −2.0 A ) = 1.0 A
24 V
a
I1
6.0 W
b
115
Direct-Current Circuits
18.21
First simplify the circuit
by combining the series
resistors. Then, apply
Kirchhoff’s junction rule
at point a to find
I1 + I 2 = 2.00 A
3.00 W
1.00 W
4.00 W
1.00 W
e1
I1
5.00
W
8.00 V
12.0 W
6.00 W
e1
I = 2.00 A
8.00 V
12.0 W
I2
I = 2.00 A a
Next, we apply Kirchhoff’s loop rule to the rightmost loop to obtain
−8.00 V + ( 6.00 ) I1 − ( 12.0 ) I 2 = 0
or
−8.00 V + ( 6.00 ) I1 − ( 12.0 ) ( 2.00 A − I1 ) = 0
This yields
I1 = 1.78 A
Finally, apply Kirchhoff’s loop rule to the leftmost loop to obtain
+ ε 1 − ( 4.00 )( 2.00 A ) − ( 6.00 ) I1 + 8.00 V = 0
or
18.22
ε 1 = ( 4.00 )( 2.00 A ) + ( 6.00 )( 1.78 A ) − 8.00 V = 10.7 V
From Kirchhoff’s point rule, note that I1 = I − I 2 in the
circuit shown at the right. Going counterclockwise around
the upper pane of the circuit, Kirchhoff’s loop rule gives
30 W
 170 Ω 
I =
(1)
 I2
 90 Ω 
Now, going counterclockwise around the outer perimeter
of the circuit, Kirchhoff’s loop rule gives
I2
90 W
− ( 50 Ω ) I 2 − ( 30 Ω ) I 2 + ( 90 Ω ) ( I − I 2 ) = 0
or
50 W
I1
–
20 W
I
+
12 V
+12 V − ( 50 Ω ) I 2 − ( 30 Ω ) I 2 − ( 20 Ω ) I = 0 or, using Equation (1)

 170 Ω  
 50 Ω + 30 Ω + ( 20 Ω )  90 Ω   I 2 = 12 V which gives I 2 = 0.10 A



The power delivered to the 50-Ω resistor is P = I 22 ( 50 Ω ) = ( 0.10 A ) ( 50 Ω ) = 0.50 W
2
116
CHAPTER 18
18.23
(a) We name the currents I1 , I 2 , and I 3 as shown.
Applying Kirchhoff’s loop rule to loop abcfa ,
gives + ε 1 − ε 2 − R2 I 2 − R1 I1 = 0
or 3I 2 + 2I1 = 10.0 mA
b
I1
I2
e1
(1)
R3
e2
70.0 V
d
I3
e3
80.0 V
60.0 V
R2
Applying the loop rule to loop edcfe yields
3.00 kW
2.00 kW
a
+ ε 3 − R3 I 3 − ε 2 − R2 I 2 = 0
or 3I 2 + 4 I 3 = 20.0 mA
4.00 kW
c
R1
f
e
(2)
Finally, applying Kirchhoff’s junction rule at junction c gives
I 2 = I1 + I 3
(3)
Solving equations (1), (2), and (3) simultaneously yields
I1 = 0.385 mA, I 2 = 3.08 mA, and I 3 = 2.69 mA
(b) Start at point c and go to point f, recording changes in potential to obtain
Vf − Vc = − ε 2 − R2 I 2 = −60.0 V − ( 3.00 × 10 3 Ω )( 3.08 × 10 −3 A ) = −69.2 V
or ∆V
18.24
cf
= 69.2 V and point c is at the higher potential
1.50 V
(a) Applying Kirchhoff’s loop rule to the circuit gives
+ 3.00 V − ( 0.255 Ω + 0.153 Ω + R )( 0.600 A ) = 0
or R =
3.00 V
− ( 0.255 Ω + 0.153 Ω ) = 4.59 Ω
0.600 A
0.255 W
1.50 V
– +
0.153 W
R
– +
I = 0.600 A
117
Direct-Current Circuits
(b) The total power input to the circuit is
Pinput = ( ε 1 + ε 2 ) I = ( 1.50 V + 1.50 V )( 0.600 A ) = 1.80 W
Ploss = I 2 ( r1 + r2 ) = ( 0.600 A ) ( 0.255 Ω + 0.153 Ω ) = 0.147 W
2
Thus, the fraction of the power input that is dissipated internally is
Ploss 0.147 W
=
= 0.081 6 or 8.16%
Pinput
1.80 W
18.25
2.0 W
Applying Kirchhoff’s junction rule at junction a gives
I1 = I 2 + I 3
(1)
24 V
4.0 W
+
+ 24 V − ( 2 .0 + 4.0 ) I1 − ( 3.0 ) I 3 = 0
or
2 I1 + I 3 = 8.0 A
12 V
(2)
and for the lower loop,
+ 12 V + ( 3.0 ) I 3 − ( 1.0 + 5.0 ) I 2 = 0
or
2 I 2 − I 3 = 4.0 A
(3)
Solving equations (1), (2), and (3) simultaneously gives
I1 = 3.5 A, I 2 = 2.5 A, and I 3 = 1.0 A
I1
I3
Using Kirchhoff’s loop rule on the upper loop yields
+
3.0 W
5.0 W
a
I2
1.0 W
118
CHAPTER 18
18.26
Using Kirchhoff’s loop rule on the outer
perimeter of the circuit gives
I1
or
I1 + 6 I 3 = 1.2 × 10 A
(1)
For the rightmost loop, the loop rule gives
I3
0.01 W
1.00 W
12.0 V
10.0 V
+ 12 V − ( 0.01) I1 − ( 0.06 ) I 3 = 0
3
I2
Live
battery
0.06 W
Starter
Dead
battery
+10 V + ( 1.00 ) I 2 − ( 0.06 ) I 3 = 0
or
I 2 − 0.06 I 3 = −10 A
(2)
Applying Kirchhoff’s junction rule at either junction gives
I1 = I 2 + I 3
(3)
Solving equations (1), (2), and (3) simultaneously yields
I 2 = 0.28 A ( in dead battery ) and I 3 = 1.7 × 10 2 A (in starter)
18.27
Assume currents I1 , I 2 , and I 3 in the directions shown.
Then, using Kirchhoff’s junction rule at junction a gives
I 3 = I1 + I 2
(1)
Applying Kirchhoff’s loop rule on the upper loop,
20.0 V
30.0 W
5.00 W
10.0 V 20.0 W
+ 20.0 V − ( 30.0 ) I1 + ( 5.00 ) I 2 − 10.0 V = 0
or
6 I1 − I 2 = 2.00 A
(2)
and for the lower loop, + 10.0 V − ( 5.00 ) I 2 − ( 20.0 ) I 3 = 0
or
I 2 + 4 I 3 = 2 .00 A
Solving equations (1), (2), and (3) simultaneously yields
I1 = 0.353 A, I 2 = 0.118 A, and I 3 = 0.471 A
(3)
I2
I1
I3
a
119
Direct-Current Circuits
18.28
(a) Since there is not a continuous path in the
center branch, no current exists in that part
of the circuit. Then, applying Kirchhoff’s loop
rule to the outer perimeter gives
+18 V + 36 V − ( 1.0 + 4.0 + 3.0 + 8.0 + 2.0 ) Ω  I = 0
or I =
18 V
+ –
2.0 W
8.0 W
I
zero current
5.0 W
– +
a b
6.0 W
12 V
1.0 W 4.0 W
– +
I
I
54 V
= 3.0 A
18 Ω
3.0 W
I
36 V
Now, start at point b and go around the lower loop to point a, recording changes in
potential to obtain
Va − Vb = −36 V + ( 4.0 Ω + 1.0 Ω )( 3.0 A ) + ( 6.0 Ω + 5.0 Ω )( 0 ) + 12 V = −9.0 V
or ∆V ab = 9.0 V with point b at a higher potential than a
18 V
+ –
(b) Assume currents as shown in the modified
circuit. Applying Kirchhoff’s loop rule to the
upper loop gives
I1
I
− ( 11) I + 12 V − ( 7.0 ) I − ( 13 ) I1 + 18 V=0
or 18 I + 13 I1 = 30 A
13 W
(1)
I1 – I
–
+
11 W 12 V
− ( 5.0 ) ( I1 − I ) + 36 V + ( 7.0 ) I − 12 V + ( 11) I =0
or 23 I − 5 I1 = −24 A
b
a
–
5.0 W
For the lower loop, the loop rule yields
7.0 W
(2)
Solving equations (1) and (2) simultaneously gives I1 = 2.9 A , and
I = − 0.42 A
Thus, the current in the 7.0-Ω resistor is 0.42 A flowing from b to a .
+
36 V
I1
I1 – I
120
CHAPTER 18
18.29
Applying Kirchhoff’s junction rule at
junction a gives
I 3 = I1 + I 2
I1
3.00 V
12.0 V
(1)
I2
a
I3
4.00 W
3.00 W
5.00 W
Using Kirchhoff’s loop rule on the leftmost
loop yields
−3.00 V − ( 4.00 ) I 3 − ( 5.00 ) I1 + 12.0 V = 0
or
5 I1 + 4 I 3 = 9.00 A
(2)
and for the rightmost loop,
−3.00 V − ( 4.00 ) I 3 − ( 3.00 + 2.00 ) I 2 + 18.0 V = 0
or
5 I 2 + 4 I 3 = 15.0 A
(3)
Solving equations (1), (2), and (3) simultaneously gives
I1 = 0.323 A, I 2 = 1.523 A, and I 3 = 1.846 A
Therefore, the potential differences across the resistors are
∆V2 = I 2 ( 2 .00 Ω ) = 3.05 V , ∆V3 = I 2 ( 3.00 Ω ) = 4.57 V
∆V4 = I 3 ( 4 .00 Ω ) = 7.38 V , and ∆V5 = I1 ( 5 .00 Ω ) = 1.62 V
18.30
The time constant is τ = RC . Considering units, we find
 Volts   Coulombs   Coulombs 
RC → ( Ohms )( Farads ) = 


=
 Amperes   Volts   Amperes 


Coulombs
=
 = Second
 Coulombs Second 
or τ = RC has units of time.
18.31
(a) τ = RC = ( 2.0 × 106 Ω )( 6.0 × 10 −6 F ) = 12 s
(b) Qmax = C ε = ( 6.0 × 10 −6 F ) ( 20 V ) = 1.2 × 10 − 4 C
18.0 V
2.00 W
Direct-Current Circuits
18.32
(a) τ = RC = ( 100 Ω ) ( 20.0 × 10 −6 F ) = 2.00 × 10 −3 s = 2.00 ms
(b) Qmax = C ε = ( 20.0 × 10 −6 F ) ( 9.00 V ) = 1.80 × 10 −4 C = 180 µ C
(c)
18.33
1

Q = Qmax ( 1 − e −t τ ) = Qmax ( 1 − e −τ τ ) = Qmax  1 −  = 114 µ C
e

Qmax = C ε = ( 5.0 × 10 −6 F ) ( 30 V ) = 1.5 × 10 −4 C , and
τ = RC = ( 1.0 × 10 6 Ω )( 5.0 × 10 −6 F ) = 5.0 s
Thus, at t = 10 s = 2τ
Q = Qmax ( 1 − e −t τ ) = ( 1.5 × 10 −4 C )( 1 − e −2 ) = 1.3 × 10 −4 C
18.34
The charge on the capacitor at time t is Q = Qmax ( 1 − e −t τ ) , where
Q = C ( ∆V ) and Qmax = C ε . Thus, ∆V = ε ( 1 − e −t τ ) or e −t τ = 1 − ( ∆V )
We are given that,
ε = 12 V , and at t = 1.0 s , ∆V = 10 V
e −1.0 s τ = 1 −
Therefore,
10 12 − 10
1
=
=
or e +1.0 s τ = 6.0
12
12
6.0
Taking the natural logarithm of each side of the equation gives
1.0 s
τ
= ln ( 6.0 )
or
τ=
1.0 s
= 0.56 s
ln ( 6.0 )
Since the time constant is τ = RC , we have
C=
τ
R
=
0.56 s
= 4.7 × 10 −5 F = 47 µ F
12 × 10 3 Ω
ε
121
122
CHAPTER 18
18.35
From Q = Qmax ( 1 − e −t τ ) , we have at t = 0.900 s ,
Q
= 1 − e − 0.900 s τ = 0.600
Qmax
Thus, e − 0.900 s τ = 0.400 , or −
0.900 s
τ
giving the time constant as τ = −
18.36
(a)
I max =
ε
R=
= ln ( 0.400 )
0.900 s
= 0.982 s
ln ( 0.400 )
, so the resistance is
R
ε
I max
=
48.0 V
= 9.60 × 10 4 Ω
0.500 × 10 -3 A
The time constant is τ = RC , so the capacitance is found to be
C=
τ
R
=
0.960 s
= 1.00 × 10 −5 F = 10.0 µ F
9.60 × 10 4 Ω
(b) Qmax = C ε = ( 10.0 µ F )( 48.0 V ) = 480 µ C , so the charge stored in the capacitor at
t = 1.92 s is
1.92 s
−


Q = Qmax ( 1 − e −t τ ) = ( 480 µ C )  1 − e 0.960 s  = ( 480 µ C ) ( 1 − e −2 ) = 415 µ C


18.37
(a) The current drawn by each appliance is
Heater:
I=
P 1 300 W
=
= 10.8 A
∆V
120 V
Toaster:
I=
P 1 000 W
=
= 8.33 A
∆V
120 V
Grill:
I=
P 1 500 W
=
= 12.5 A
∆V
120 V
(b) If the three appliances are operated simultaneously, they will draw a total current of
I total = ( 10.8 + 8.33 + 12.5 ) A = 31.7 A . Therefore, a 30 ampere circuit breaker is
insufficient to handle the load .
Direct-Current Circuits
18.38
123
(a) The equivalent resistance of the parallel combination is
−1
−1
 1
1
1 
1
1 
 1
+
+
+
+
Req = 
 =
 = 15 Ω
 150 Ω 25 Ω 50 Ω 
 R1 R2 R3 
so the total current supplied to the circuit is
I total =
∆V 120 V
=
= 8.0 A
Req
15 Ω
(b) Since the appliances are connected in parallel,
the voltage across each one is ∆V = 120 V .
(c)
I lamp =
(d) Pheater =
18.39
∆V 120 V
=
= 0.80 A
Rlamp 150 Ω
( ∆V )
2
Rheater
=
( 120 V )
2
= 5.8 × 10 2 W
25 Ω
From P= ( ∆V ) R , the resistance of the element is
2
R=
( ∆V )
P
2
=
( 240 V )
2
3 000 W
= 19.2 Ω
When the element is connected to a 120-V source, we find that
(a)
I=
∆V 120 V
=
= 6.25 A , and
R 19.2 Ω
(b) P = ( ∆V ) I = ( 120 V )( 6.25 A ) = 750 W
18.40
The maximum power available from this line is
Pmax = ( ∆V ) I max = ( 120 V )( 15 A ) = 1 800 W
Thus, the combined power requirements (2 400 W) exceeds the available power, and you
cannot operate the two appliances together .
124
CHAPTER 18
18.41
(a) The area of each surface of this axon membrane is
A = A ( 2π r ) = ( 0.10 m )  2π ( 10 × 10 −6 m )  = 2π × 10 −6 m 2
and the capacitance is
C = κ ∈0
 2π × 10 −6 m 2 
A
−8
= 3.0 ( 8.85 × 10 −12 C 2 N ⋅ m 2 ) 
 = 1.67 × 10 F
-8
d
 1.0 × 10 m 
In the resting state, the charge on the outer surface of the membrane is
Qi = C ( ∆V )i = ( 1.67 × 10 −8 F ) ( 70 × 10 −3 V ) = 1.17 × 10 −9 C → 1.2 × 10 −9 C
The number of potassium ions required to produce this charge is
NK+ =
Qi 1.17 × 10 −9 C
=
= 7.3 × 10 9 K + ions
1.6 × 10 -19 C
e
and the charge per unit area on this surface is
σ=
−20
2
Qi 1.17 × 10 −9 C 
1e
  10 m
=


A 2π × 10 -6 m 2  1.6 × 10 -19 C   1 Å 2

1e
=
=
4
2
 8.6 × 10 Å
1e
( 290 Å )
2
This corresponds to a low charge density of one electronic charge per square of side
290 Å, compared to a normal atomic spacing of one atom per several Å 2 .
(b) In the resting state, the net charge on the inner surface of the membrane is
− Qi = − 1.17 × 10 −9 C , and the net positive charge on this surface in the excited state
is
Q f = C ( ∆V ) f = ( 1.67 × 10 −8 F )( +30 × 10 −3 V ) = + 5.0 × 10 −10 C
The total positive charge which must pass through the membrane to produce the
excited state is therefore
∆Q = Q f − Qi
= + 5.0 × 10 −10 C − ( − 1.17 × 10 −9 C ) = 1.67 × 10 −9 C → 1.7 × 10 −9 C
corresponding to
N Na+ =
∆Q
1.67 × 10 −9 C
=
= 1.0 × 1010 Na+ ions
-19
+
1.6 × 10 C Na ion
e
Direct-Current Circuits
125
(c) If the sodium ions enter the axon in a time of ∆t = 2.0 ms , the average current is
I=
∆Q 1.67 × 10 −9 C
=
= 8.3 × 10 −7 A = 0.83 µ A
−3
∆t
2.0 × 10 s
(d) When the membrane becomes permeable to sodium ions, the initial influx of
sodium ions neutralizes the capacitor with no required energy input. The energy
input required to charge the now neutral capacitor to the potential difference of the
excited state is
2
1
1
2
W = C ( ∆V ) f = ( 1.67 × 10 −8 F )( 30 × 10 −3 V ) = 7.5 × 10 −12 J
2
2
18.42
The capacitance of the 10 cm length of axon was found to be C = 1.67 × 10 −8 F in the
solution of Problem 18.41.
(a) When the membrane becomes permeable to potassium ions, these ions flow out of
the axon with no energy input required until the capacitor is neutralized. To
maintain this outflow of potassium ions and charge the now neutral capacitor to the
resting action potential requires an energy input of
2
1
1
2
W = C ( ∆V ) = ( 1.67 × 10 −8 F )( 70 × 10 −3 V ) = 4.1 × 10 −11 J .
2
2
(b) As found in the solution of Problem 18.41, the charge on the inner surface of the
membrane in the resting state is − 1.17 × 10 −9 C and the charge on this surface in the
excited state is + 5.0 × 10 −10 C . Thus, the positive charge which must flow out of the
axon as it goes from the excited state to the resting state is
∆Q = 5.0 × 10 −10 C + 1.17 × 10 −9 C = 1.67 × 10 −9 C ,
and the average current during the 3.0 ms required to return to the resting state is
I=
18.43
∆Q 1.67 × 10 −9 C
=
= 5.6 × 10 −7 A = 0.56 µ A
−3
∆t
3.0 × 10 s
From Figure 18.28, the duration of an action potential pulse is 4.5 ms. From the solution
Problem 18.41, the energy input required to reach the excited state is W1 = 7.5 × 10 −12 J .
The energy input required during the return to the resting state is found in Problem
18.42 to be W2 = 4.1 × 10 −11 J . Therefore, the average power input required during an
action potential pulse is
P=
Wtotal W1 + W2 7.5 × 10 −12 J+4.1 × 10 −11 J
=
=
= 1.1 × 10 −8 W = 11 nW
4.5 × 10 −3 s
∆t
∆t
126
CHAPTER 18
18.44
From Q = Qmax ( 1 − e −t τ ) , the ratio Q Qmax at t = 2τ is found to be
Q
1
= 1 − e −2τ τ = 1 − 2 = 0.865 , or Q is 86.5% of Qmax at t = 2τ
Qmax
e
18.45
The resistive network between a an b reduces, in the stages shown below, to an
equivalent resistance of Req = 7.5 Ω .
2.4 W
5.1 W
a
18.46
a
a
1.8 W
b
2.4 W
2.4 W
3.6 W
8.6 W
1.8 W
3.5 W
1.5 W
3.6 W
b
a
3.6 W
b
b
(a) The circuit reduces as shown below to an equivalent resistance of Req = 14 Ω .
5.0 W
28 V
3.0 W
10 W
3.0 W
10 W
4.0 W
5.0 W
4.0 W
2.0 W
28 V
3.0 W
10 W
3.0 W
10 W
4.0 W
5.0 W
3.0 W
10 W
5.0 W
4.0 W
2.0 W
28
V
10 W
2.0 W
Figure 1
28
V
7.5 W
Figure 2
5.0 W
10 W
28
V
10 W
Figure 3
Figure 5
Figure 4
(b) The power dissipated in the circuit is P =
5.0 W
4.0 W
4.0 W
( ∆V )
Req
2
=
I
5.0 W
( 28 V )
14 Ω
28
V
14
W
Figure 6
2
= 56 W
(c) The current in the original 5.0-Ω resistor (in Figure 1) is the total current supplied
by the battery. From Figure 6, this is
I=
∆V 28 V
=
= 2.0 A
Req 14 Ω
127
Direct-Current Circuits
18.47
(a) The resistors combine to an equivalent resistance of Req = 15 Ω as shown.
6.0 W
a
I1
15 V
I3
I2
+
–
2.4 W
c
e
I4
6.0 W
15 V
+
–
f
b
15 V
+
–
I2
6.0 W
a
I3
c
I1
+
15 V –
6.0
W
a
I1
6.0 W
+
15 V –
3.0 W
6.0 W
b
3.6 W
Figure 2
c
6.0 W
I3
d
Figure 1
I1
e
6.0 W
d
a
I2
6.0 W
9.0 W
b
2.4 W
c
I1
I5
6.0 W
6.0 W
6.0 W
a
15 W
6.0 W
d
Figure 3
b
d
Figure 4
(b) From Figure 5, I1 =
b
Figure 5
∆Vab 15 V
=
= 1.0 A
Req
15 Ω
Then, from Figure 4,
∆Vac = ∆Vdb = I1 ( 6.0 Ω ) = 6.0 V and ∆Vcd = I1 ( 3.0 Ω ) = 3.0 V
∆Vcd
3.0 V
=
= 0.50 A
6.0 Ω 6.0 Ω
From Figure 3,
I2 = I3 =
From Figure 2,
∆Ved = I 3 ( 3.6 Ω ) = 1.8 V
I4 =
Then, from Figure 1,
and
I5 =
∆Vfd
9.0 Ω
=
∆Ved 1.8 V
=
= 0.30 A
6.0 Ω 6.0 Ω
∆Ved 1.8 V
=
= 0.20 A
9.0 Ω 9.0 Ω
(c) From Figure 2, ∆Vce = I 3 ( 2.4 Ω ) = 1.2 V . All the other needed potential differences
were calculated above in part (b). The results were
∆Vac = ∆Vdb = 6.0 V ; ∆Vcd = 3.0 V ; and ∆Vfd = ∆Ved = 1.8 V
128
CHAPTER 18
(d) The power dissipated in each resistor is found from P = ( ∆V ) R with the
following results:
2
( ∆V )ac
2
Pac =
Rac
( ∆V )ed
=
2
Ped =
Red
( ∆V )cd
=
2
Pcd =
18.48
Rcd
=
( 6.0 V )
= 6.0 W
6.0 Ω
( 1.8 V )
( 3.0 V )
Pce =
= 0.54 W
Pfd =
= 1.5 W
Pdb =
( ∆V ) fd
R fd
( ∆V )db
=
2
2
6.0 Ω
Rce
=
2
2
6.0 Ω
( ∆V )ce
2
2
Rdb
=
( 1.2 V )
2
= 0.60 W
2.4 Ω
( 1.8 V )
2
= 0.36 W
9.0 Ω
( 6.0 V )
2
= 6.0 W
6.0 Ω
(a) From P = ( ∆V ) R , the resistance of each of the
three bulbs is given by
2
R=
( ∆V )
2
P
=
( 120 V )
R1
2
60.0 W
= 240 Ω
120 V
R2
R3
As connected, the parallel combination of
R2 and R3 is in series with R1 . Thus, the
equivalent resistance of the circuit is
−1
−1
 1
1 
1 
 1
+
+
Req = R1 + 
 = 240 Ω + 
 = 360 Ω
 240 Ω 240 Ω 
 R2 R3 
The total power delivered to the circuit is
P=
( ∆V )
Req
2
=
( 120 V )
2
360 Ω
= 40.0 W
(b) The current supplied by the source is I =
∆V 120 V 1
=
= A . Thus, the potential
Req 360 Ω 3
difference across R1 is
( ∆V )1 = I R1 =  A  ( 240 Ω ) = 80.0 V
3
1


The potential difference across the parallel combination of R2 and R3 is then
( ∆V )2 = ( ∆V )3 = ( ∆V )source − ( ∆V ) 1= 120 V − 80.0 V = 40.0 V
Direct-Current Circuits
18.49
(a) From
I=
129
ε = I ( r + Rload ) , the current supplied when the headlights are the entire load is
ε
r + Rload
=
12.6 V
= 2.48 A
( 0.080+5.00 ) Ω
The potential difference across the headlights is then
∆V = I Rload = ( 2.48 A )( 5.00 Ω ) = 12.4 V
(b) The starter motor connects in parallel with the headlights. If I hl is the current
supplied to the headlights, the total current delivered by the battery is
I = I hl + 35.0 A
The terminal potential difference of the battery is ∆V = ε − I r , so the total current is
I = (ε − ∆V ) r while the current to the headlights is I hl = ∆V 5.00 Ω . Thus,
I = I hl + 35.0 A becomes
ε − ∆V =
r
∆V
+ 35.0 A
5.00 Ω
which yields
∆V =
18.50
ε − ( 35.0 A ) r = 12.6 V − ( 35.0 A )( 0.080 Ω ) =
1 + r ( 5.00 Ω )
1+ ( 0.080 Ω ) ( 5.00 Ω )
(a) After steady-state conditions have been
reached, there is no current in the branch
containing the capacitor.
I R3 = 0 ( steady-state )
Thus, for R3 :
9.65 V
12.0 W
R1
e
9.00 V
15.0
kW
For the other two resistors, the steady-state
current is simply determined by the 9.00-V emf
across the 12.0-k Ω and 15.0-k Ω resistors in series:
For R1 and R2 :
I ( R1 + R2 ) =
ε
R1 + R2
=
9.00 V
= 333 µ A
(12.0 k Ω + 15.0 k Ω)
( steady-state )
R2
C
10.0
mF
R3
3.00
kW
130
CHAPTER 18
(b) When the steady-state has been reached, the potential difference across C is the
same as the potential difference across R2 because there is no change in potential
across R3 . Therefore, the charge on the capacitor is
Q = C ( ∆V )R
2
= C ( I R2 ) = ( 10.0 µ F ) ( 333 × 10 −6 A )( 15.0 × 10 3 Ω ) = 50.0 µ C
18.51
Applying Kirchhoff’s junction rule at junction a
gives
I 2 = I1 + I 3
a
5.0 W
I1
I2
(1)
Applying Kirchhoff’s loop rule on the leftmost
loop yields
I1 + 2 I 2 = 1.0 A
9.0 V
10 W
(2)
For the rightmost loop,
+ ( 10 ) I 2 + 4.0 V + ( 10 ) I 3 − 14 V = 0
or
I 2 + I 3 = 1.0 A
Solving equations (1), (2) and (3) simultaneously gives
I1 = 0, I 2 = I 3 = 0.50 A
10 W
4.0 V
+ 9.0 V − ( 5.0 ) I1 − 4.0 V − ( 10 ) I 2 = 0
or
I3
(3)
14 V
Direct-Current Circuits
18.52
131
With the switch open, the circuit may be reduced as follows:
R
10 W
90 W
a
b
90 W
100 W
R
a
R
b
10 W
50 W
R + 50 W
a
b
a
b
100 W
With the switch closed, the circuit reduces as shown below:
R
10 W
90 W
R
a
b
90 W
9W
9W
a
R + 18 W
b
a
b
10 W
Since the equivalent resistance with the switch closed is one-half that when the switch is
open, we have
R + 18 Ω =
18.53
1
( R + 50 Ω ) , which yields R = 14 Ω
2
When a generator with emf ε and internal resistance r supplies current I, its terminal
voltage is ∆V = ε − I r .
If ∆V = 110 V when I = 10.0 A , then
110 V = ε − ( 10.0 A ) r
(1)
Given that ∆V = 106 V when I = 30.0 A , yields
106 V = ε − ( 30.0 A ) r
(2)
Solving equations (1) and (2) simultaneously gives
ε = 112 V
18.54
and r = 0.200 Ω
At time t, the charge on the capacitor will be Q = Qmax ( 1 − e −t τ ) where
τ = RC = ( 2.0 × 10 6 Ω )( 3.0 × 10 −6 F ) = 6.0 s
When Q = 0.90 Qmax , this gives 0.90 = 1 − e −t τ
or
e −t τ = 0.10
t
Thus, − = ln ( 0.10 )
τ
giving t = − ( 6.0 s ) ln ( 0.10 ) = 14 s
132
CHAPTER 18
18.55
(a) For the first measurement, the equivalent circuit is
as shown in Figure 1. From this,
1
R1
2
1
Ry + Rx
2
Ry
Rx
Figure 1
(1)
a
For the second measurement, the equivalent circuit
is shown in Figure 2. This gives
Rac = R2 =
b
c
Ry
Rab = R1 = Ry + Ry = 2 Ry
so Ry =
R1
a
Ry
(2)
R2
Ry
c
Rx
Figure 2
Substitute (1) into (2) to obtain
1 1
1

R2 =  R 1  + Rx , or Rx = R2 − R1
2 2
4

(b) If R1 = 13 Ω and R2 = 6.0 Ω , then Rx = 2.8 Ω
Since this exceeds the limit of 2.0 Ω , the antenna is inadequately grounded .
18.56
Assume a set of currents as shown in the circuit
diagram at the right. Applying Kirchhoff’s loop
rule to the leftmost loop gives
I
75 V
+ 75 − ( 5.0 ) I − ( 30 ) ( I − I1 ) = 0
or
7 I − 6 I1 = 15
5.0 W
30 W
I – I1
I1
40 W
R
(1)
For the rightmost loop, the loop rule gives
7 R 
− ( 40 + R ) I1 + ( 30 ) ( I − I1 ) = 0 , or I =  +
 I1
 3 30 
(2)
Substituting equation (2) into (1) and simplifying gives
310 I1 + 7 ( I1 R ) = 450
Also, it is known that PR = I12 R = 20 W , so I1 R =
(3)
20 W
I1
(4)
Direct-Current Circuits
133
Substitution of equation (4) into (3) yields
310 I1 +
140
= 450 or 310 I12 − 450 I1 + 140 = 0
I1
Solving this quadratic equation gives two possible values for the current I1 . These are
20 W
I1 = 1.0 A and I1 = 0.452 A . Then, from R= 2 , we find two possible values for the
I1
resistance R. These are
R = 20 Ω or R = 98 Ω
18.57
When connected in series, the equivalent resistance is Req = R1 + R2 + ⋅⋅⋅ + Rn = n R . Thus,
the current is I s = ( ∆V ) Req = ( ∆V ) n R , and the power consumed by the series
configuration is
Ps = I s2 Req =
( ∆V )
( n R)
2
2
( n R) =
( ∆V )
2
nR
For the parallel connection, the power consumed by each individual resistor is
P1 =
( ∆V )
2
R
, and the total power consumption is
n ( ∆V )
Pp = n P1 =
R
Ps ( ∆V )
1
R
1
=
⋅
= 2 or Ps = 2 Pp
2
n
n R n ( ∆V )
n
Pp
2
Therefore,
2
134
18.58
CHAPTER 18
Consider a battery of emf ε connected between
points a and b as shown. Applying Kirchhoff’s
loop rule to loop acbea gives
I
I2
− ( 1.0 ) I1 − ( 1.0 ) ( I1 − I 3 ) + ε = 0
or
2 I1 − I 3 = ε
e
I1
a
(1)
e
1.0 W
I1 – I3
c
I3
1.0
W
3.0 W
5.0 W
I2 + I3
d
Applying the loop rule to loop adbea gives
b
1.0 W
− ( 3.0 ) I 2 − ( 5.0 ) ( I 2 + I 3 ) + ε = 0
or
8 I2 + 5 I3 = ε
(2)
For loop adca , the loop rule yields
− ( 3.0 ) I 2 + ( 1.0 ) I 3 + ( 1.0 ) I1 = 0 or I1 + I 3 = 3I 2
(3)
Solving equations (1), (2) and (3) simultaneously gives
I1 =
13
ε , I 2 = 4 ε , and I 3 = − 1 ε
27
27
27
Then, applying Kirchhoff’s junction rule at junction a gives
I = I1 + I 2 =
18.59
13
ε + 4 ε = 17 ε . Therefore, Rab = ε = ε = 27 Ω
27
27
27
17
I ( 17 ε 27 )
r = 10.0 W
(a) and (b) - With R the value of the load resistor, the
current in a series circuit composed of a 12.0 V battery,
an internal resistance of 10.0 Ω , and a load resistor is
12.0 V
12.0 V
I=
R + 10.0 Ω
I
and the power delivered to the load resistor is
(144 V ) R
2
PL = I 2 R =
R
( R + 10.0 Ω )
2
135
Direct-Current Circuits
Some typical data values for the graph
are
R (Ω)
1.00
5.00
10.0
15.0
20.0
25.0
30.0
3.60 W
ÃL
PL (W)
1.19
3.20
3.60
3.46
3.20
2.94
2.70
10.0 W
R
The curve peaks at PL = 3.60 W at a load resistance of R = 10.0 Ω .
18.60
The total resistance in the circuit is
−1
−1
 1
 1
1 
1 
R=
+
+
 = 1.2 kΩ
 =
 R1 R2 
 2.0 kΩ 3.0 kΩ 
and the total capacitance is C = C1 + C2 = 2.0 µ F+3.0 µ F=5.0 µ F
Thus, Qmax = C ε = ( 5.0 µ F )( 120 V ) = 600 µ C
and
τ = RC = ( 1.2 × 10 3 Ω )( 5.0 × 10 −6 F ) = 6.0 × 10 −3 s =
6.0 s
1 000
The total stored charge at any time t is then
Q = Q1 + Q2 = Qmax ( 1 − e −t τ ) or Q1 + Q2 = ( 600 µ C ) ( 1 − e −1 000 t 6.0 s )
(1)
Since the capacitors are in parallel with each other, the same potential difference exists
across both at any time.
Therefore,
( ∆V ) C =
C 
Q1 Q2
, or Q2 =  2  Q1 = 1.5 Q1
=
C1 C2
 C1 
Solving equations (1) and (2) simultaneously gives
Q1 = ( 240 µ C ) ( 1 − e −1 000 t 6.0 s )
and
Q2 = ( 360 µ C ) ( 1 − e −1 000 t 6.0 s )
(2)
136
CHAPTER 18
18.61
(a) Using the rules for combining resistors in series and parallel, the circuit reduces as
shown below:
10.0 W 25.0 V
– +
10.0 W 25.0 V
– +
I
a
10.0 W
5.00 W
5.00
W
10.0 W 25.0 V
– +
I
b
I1
10.0 W
a
I
b
5.00 W
20.0
W
I1
a
25.0 W
Step 2
Step 1
b
2.94 W
Step 3
From the figure of Step 3, observe that
I=
25.0 V
= 1.93 A
12.94 A
∆Vab = I ( 2.94 Ω ) = ( 1.93 A )( 2.94 Ω ) = 5.68 V
and
(b) From the figure of Step 1, observe that I1 =
18.62
∆Vab
5.68 V
=
= 0.227 A
25.0 Ω 25.0 Ω
(a) When the power supply is connected to points A and B, the circuit reduces as
shown below to an equivalent resistance of Req = 0.099 9 Ω .
100 W
R2 + R3 + 100 W = 111 W
Req = 0.0999 W
A
R1
B
R2
C
R3
D
A
R1 = 0.100 W
B
I1
+ –
+ –
+ –
5.00 V
5.00 V
From the center figure above, observe that I R1 = I1 =
and
I R2 = I R3 = I100 =
5.00 V
5.00 V
= 50.0 A
0.100 Ω
5.00 V
= 0.045 0 A = 45.0 mA
111 Ω
Direct-Current Circuits
137
(b) When the power supply is connected to points A and C, the circuit reduces as
shown below to an equivalent resistance of Req = 1.09 Ω .
100 W
R2 + 100 W = 110 W
Req = 1.09 W
A
R1
B
R2
C
R3
D
A
R1 + R2 = 1.10 W
C
I1
+ –
+ –
5.00 V
5.00 V
+ –
5.00 V
From the center figure above, observe that I R1 = I R2 = I1 =
I R3 = I100 =
and
5.00 V
= 4.55 A
1.10 Ω
5.00 V
= 0.045 5 A = 45.5 mA
110 Ω
(c) When the power supply is connected to points A and D, the circuit reduces as
shown below to an equivalent resistance of Req = 9.99 Ω .
100 W
100 W
Req = 9.99 W
A
R1
B
R2
C
R3
D
A
R1 + R2 + R3 = 11.1 W
D
I1
+ –
+ –
5.00 V
5.00 V
+ –
5.00 V
From the center figure above, observe that I R1 = I R2 = I R3 = I1 =
and
18.63
I100 =
5.00 V
= 0.450 A
11.1 Ω
5.00 V
= 0.050 0 A = 50.0 mA
100 Ω
In the circuit diagram at the right, note that all points
labeled a are at the same potential and equivalent
to each other. Also, all points labeled c are equivalent.
To determine the voltmeter reading, go from point e
to point d along the path ecd, keeping track of all
changes in potential to find:
∆Ved = Vd − Ve = −4.50 V + 6.00 V = + 1.50 V
a
a
a
a
I
b
A
10.0 W
I2
5.00 W
d
V
e
5.00 W
I1
I3
6.00 W
f
c
6.00 V
+ – c
+ –
c
4.50 V
c
138
CHAPTER 18
Apply Kirchhoff’s loop rule around loop abcfa to find
− ( 6.00 Ω ) I + ( 6.00 Ω ) I 3 = 0
or
I3 = I
(1)
Apply Kirchhoff’s loop rule around loop abcda to find
− ( 6.00 Ω ) I + 6.00 V − ( 10.0 Ω ) I 2 = 0
or
I 2 = 0.600 A − 0.600 I
(2)
Apply Kirchhoff’s loop rule around loop abcea to find
− ( 6.00 Ω ) I + 4.50 V − ( 5.00 Ω ) I1 = 0
or
I1 = 0.900 A − 1.20 I
(3)
Finally, apply Kirchhoff’s junction rule at either point a or point c to obtain
I + I 3 = I1 + I 2
(4)
Substitute equations (1), (2), and (3) into equation (4) to obtain the current through the
ammeter. This gives
I + I = 0.900 A − 1.20 I + 0.600 A − 0.600I
or
18.64
3.80 I = 1.50 A
and
I = 1.50 A 3.80 = 0.395 A
In the figure given below, note that all bulbs have the same resistance, R .
A
DV
A
DV
B
Case 1
B
Case 2
(a) In the series situation, Case 1, the same current I1 flows through both bulbs. Thus,
the same power, P1 = I12 R , is supplied to each bulb. Since the brightness of a bulb is
proportional to the power supplied to it, they will have the same brightness. We
conclude that the bulbs have the same current, power supplied, and brightness .
(b) In the parallel case, Case 2, the same potential difference ∆V is maintained across
each of the bulbs. Thus, the same current I 2 = ∆V R will flow in each branch of
this parallel circuit. This means that, again, the same power
to each bulb, and the two bulbs will have equal brightness .
P2 = I 22 R is supplied
Direct-Current Circuits
139
(c) The total resistance of the single branch of the series circuit (Case 1) is 2R. Thus, the
current in this case is I1 = ∆V 2R . Note that this is one half of the current I 2 that
flows through each bulb in the parallel circuit (Case 2). Since the power supplied is
proportional to the square of the current, the power supplied to each bulb in Case 2
is four times that supplied to each bulb in Case 1. Thus, the bulbs in Case 2 are
much brighter than those in Case 1.
(d) If either bulb goes out in Case 1, the only conducting path of the circuit is broken
and all current ceases. Thus, in the series case, the other bulb must also go out . If
one bulb goes out in Case 2, there is still a continuous conducting path through the
other bulb. Neglecting any internal resistance of the battery, the battery continues to
maintain the same potential difference ∆V across this bulb as was present when
both bulbs were lit. Thus, in the parallel case, the second bulb remains lit with
unchanged current and brightness when one bulb fails.
140
CHAPTER 18