ENGR-2150 Chap-24 Classroom Problems
Prob. 24.15
C1  C2  C3  C4  400 F
Vab  28.0 V
Figure 24.15a
(a) Calculate the charge on each capacitor?
(b) Calculate the potential difference across each capacitor?
(c) Calculate the potential difference between points a and b?
IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents. In
each equivalent network apply the rules for Q and V for capacitors in series and
parallel; start with the simplest network and work back to the original circuit.
SETUP: SIMPLIFY CIRCUIT
EXECUTE:
Simplify the circuit by replacing the capacitor combinations by their equivalents:
C1 and C2 are in series and are equivalent to C12 (Figure 24.15b).
1
1
1
 
C12 C1 C2
Figure 24.15b
C12 
 4.00 106 F 4.00 106 F  2.00 106 F
C1C2

C1  C2
4.00  106 F  4.00 106 F
C12 and C3 are in parallel and are equivalent to C123 (Figure 24.15c).
C123  C12  C3
C123  2.00 106 F  4.00 106 F
C123  6.00 106 F
Figure 24.15c
C123 and C4 are in series and are equivalent to C1234 (Figure 24.15d).
1
C1234

1
1

C123 C4
Figure 24.15d
C1234 
 6.00 106 F 4.00 106 F  2.40 106 F
C123C4

C123  C4
6.00 106 F  4.00 106 F
The circuit is equivalent to the circuit shown in Figure 24.15e.
V1234  V  28.0 V
Q1234  C1234V   2.40 106 F  28.0 V   67.2 C
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ENGR-2150 Chap-24 Classroom Problems
Now build back up the original circuit, step by step. C1234 represents C123 and C4 in
series (Figure 24.15f).
Q123  Q4  Q1234  67.2 C
(charge same for capacitors in
series)
Figure
24.15f
Q123 67.2 C

 11.2 V
C123 6.00  F
Q 67.2 C
V4  4 
 16.8 V
C4 4.00  F
Then V123 
Note that V4  V123  16.8 V  11.2 V  28.0 V, as it should.
Next consider the circuit as written in Figure 24.15g.
V3  V12  28.0 V  V4
V3  11.2 V
Q3  C3V3   4.00 F11.2 V 
Q3  44.8 C
Q12  C12V12   2.00 F11.2 V 
Q12  22.4 C
Figure 24.15g
Finally, consider the original circuit, as shown in Figure 24.15h.
Q1  Q2  Q12  22.4 C
(charge same for capacitors in
series)
V1 
Q1 22.4 C

 5.6 V
C1 4.00  F
V2 
Q2 22.4 C

 5.6 V
C2 4.00  F
Figure 24.15h
Note that V1  V2  11.2 V, which equals V3 as it should.
Summary: Q1  22.4 C, V1  5.6 V
Q2  22.4 C, V2  5.6 V
Q3  44.8 C, V3  11.2 V
Q4  67.2 C, V4  16.8 V
(c) Vad  V3  11.2 V
EVALUATE: V1  V2  V4  V , or V3  V4  V . Q1  Q2 , Q1  Q3  Q4 and Q4  Q1234 .
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ENGR-2150 Chap-24 Classroom Problems
Prob. 24.33
Vab = 220 v
(a)
(b)
(c)
(d)
(e)
Total Network Charge = Q = ?
Charge on each capacitor?
The total energy stored in the network?
The energy stored in each capacitor?
The potential difference across each capacitor?
IDENTIFY:
The two capacitors are in parallel. Ceq  C1  C2 . C 
Q
. U  12 CV 2 .
V
SET UP: For capacitors in parallel, the voltages are the same and the charges add.
EXECUTE:
(a) Ceq  C1  C2  35 nF  75 nF  110 nF .
Qtot  CeqV  (110 109 F)(220 V)  24.2 C
(b) V  220 V for each capacitor.
35 nF: Q35  C35V  (35 109 F)(220 V)  7.7 C
75 nF: Q75  C75V  (75 109 F)(220 V)  16.5 C
Note that Q35  Q75  Qtot .
(c) U tot  12 CeqV 2  12 (110 109 F)(220 V)2  2.66 mJ
(d) Energy stored in each capacitor:
35 nF: U35  12 C35V 2  12 (35 109 F)(220 V)2  0.85 mJ ;
75 nF: U75  12 C75V 2  12 (75 109 F)(220 V)2  1.81 mJ
Since V is the same the capacitor with larger C stores more energy.
(f) Capacitors in parallel will have the same potential . Therefore 220 V for each
capacitor.
EVALUATE: The capacitor with the larger C has the larger Q.
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