Homework Solutions
Lecture Chapter 21
*18.47. Set Up: The capacitors between b and c are in parallel. This combination is in series with the 15 pF capacitor.
C1  15 pF, C2  90 pF and C3  11 pF
Solve: (a) For capacitors in parallel, Ceq  C1  C2  g
so C23  C2  C3  20 pF
(b) C1  15 pF is in series with C23  20 pF For capacitors in series,
1
1
1


g
Ceq C1 C2
so
CC
(15 pF)(20 pF)
1
1
1
 86 pF
and C123  1 23 


C123 C1 C23
C1  C23 15 pF  20 pF
Reflect: For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors. For capacitors
in series the equivalent capacitance is smaller than any of the individual capacitors.
*18.53. Set Up: C 
Q
 For two capacitors in parallel, Ceq  C1  C2  For two capacitors in series,
V
1
1
1
CC


and Ceq  1 2 
Ceq C1 C2
C1  C2
For capacitors in parallel, the voltages are the same and the charges add. For capacitors in series, the charges are the same
and the voltages add. Let C1  300 F, C2  500 F and C3  600 F
Solve: (a) The equivalent capacitance for C1 and C2 in parallel is C12  C1  C2  800 F This gives the circuit shown
in Figure (a) above. In that circuit the equivalent capacitance is
C
C12C3
(800 F)(600 F)

 343 F
C12  C3 800 F  600 F
This gives the circuit shown in Figure (b) above. In Figure (b), Q  CV  (343  106 F)(240 V)  823  105 C In
Figure (a) each capacitor therefore has charge 823  105 C The potential differences are
V3 
Q3 823  105 C
Q12 823  105 C
and


 103 V


13

7
V
V
12
C3 600  106 F
C12 800  106 F
Note that V3  V12  240 V Then in the original circuit, V1  V2  V12  103 V
Q1  V1C1  (103 V)(300  106 F)  309  105 C
Q2  V2C2  (103 V)(500  106 F)  515  105 C
Q1  309 C, Q2  515 C and Q3  823 C Note that Q1  Q2  Q3 
(b) V1  103 V, V2  103 V and V3  137 V
Reflect: Note that Q1  Q2  Q3 , V1  V2 and V1  V3  240 V
*18.63. Set Up: The two capacitors are in series. For capacitors in series the voltages add and the charges are the same.
1
1
1
Q


 g C   U  12 CV 2 
Ceq C1 C2
V
Solve: (a)
1
1
1
CC
(150 nF)(120 nF)


so Ceq  1 2 
 667 nF
Ceq C1 C2
C1  C2 150 nF  120 nF
Q  CV  (667 nF)(36 V)  24  106 C  24C
(b) Q  24C for each capacitor.
(c) U  12 CeqV 2  12 (667  10 9 F)(36 V) 2  432 J
(d) We know C and Q for each capacitor so rewrite U in terms of these quantities.
U  12 CV 2  12 C (Q /C ) 2  Q 2 /2C
150 nF: U 
(24  106 C) 2
9
 192J; 120 nF: U 
(24  106 C) 2
2(150  10 F)
2(120  109 F)
Note that 192J  240J  432J, the total stored energy calculated in part (c).
 240 J
Q 24  106 C
Q 24  106 C
120
nF:
V


16
V;


 20 V
C 150  109 F
C 120  109 F
Note that these two voltages sum to 36 V, the voltage applied across the network.
150 nF: V 
Reflect: Since Q is the same the capacitor with smaller C stores more energy (U  Q 2 /2C ) and has a larger voltage
(V  Q /C )
19.1. Set Up: 1A  1 C/s An electron has charge of magnitude e  160  1019 C The current tells us how much
charge flows in a given time. We can find the number of electrons that correspond to a certain amount of charge.
1


18
Solve: (a) 1.50 A  (1.50 C/s) 
  9.38  10 electrons/s
19

1.60
10
C/electron


(b) Q  I t  (150 A)(300 s)  450 C
Q 750 C

 500 s
I
150 A
Reflect: The amount of charge associated with typical currents is quite large. In Chapter 17 we found that the force
between objects that have net charge 1.0 C is immense. The same amount of charge enters and leaves any part of the
circuit each second and each circuit element remains neutral.
(c) t 
19.11. Set Up: R 
Solve: (a) R 
(b) R 
al L
A
L

A
 For aluminum, al  263  108  m For copper, c  172  108  m
(263  108   m)(380 m)
(100  102 m)(500  102 m)
 200  104 
c L
R r 2 (200  104 ) (0750  103 m) 2
L


 00205 m  205 cm

c
r2
172  108 
19.40. Set Up: P  I 2 R 
V2
 VI  V  IR The heater consumes 540 W when V  120 V Energy  Pt 
R
Solve: (a) P 
V2
V 2 (120 V) 2
so R 

 267
R
540 W
P
(b) P  VI so I 
P 540 W

 450 A
V 120 V
(c) Energy  (0540 kW)(1 h)  0540 kWh The cost is (0540 kWh)(72 cents/kWh)  39 cents
(d) Assuming that R remains 267, P 
V 2 (110 V)2

 453 W P is smaller by a factor of (110/120) 2 
267
R
*19.47. Set Up: The resistors can be placed in any combination of series and parallel. The equivalent resistance of two
resistors in series is Req  R1  R2 , which is larger than either resistance by itself. However, the equivalent resistance of
two resistors in parallel is given by Req 
1
, which is always less than the smaller of the two resistances. Thus, to
1
1

R1 R2
obtain the maximum equivalent resistance the three resistors should all be in series and to obtain the minimum equivalent
resistance they should all be in parallel.
Solve: (a) For three resistors in series we have Req  R1  R2  R3  36  47  51  134.
1
 15.
1
1
1


36 47 51
Reflect: In addition to these two extreme cases, we can combine the three resistors in a variety of other combinations: for
example, we can form a series combination of a single resistor and a pair of resistors that are connected in parallel. Can
you find all possible combinations of these three resistors?
(b) For three resistors in parallel we have Req 
19.54. Set Up: We can use the power rating in P  V 2 /R to find the resistance of a bulb. Then we can find the voltage
across each bulb when they are connected in series and in parallel.
V 2 (120 V) 2
Solve: R 

 144
P
100 W
V
120 V

 0417 A The voltage across each bulb is V1  IR  60 V
(a) Req  2 R  288 I 
Req 288
V 2 (60 V) 2

 25 W
R
144
(b) In parallel, the voltage across each bulb is 120 V and the power dissipated by each is 100 W.
P