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and substituting for C T:
1/C 1
V 1 � � � E (10.32)
���
1/C1 � 1/C2 � 1/C3 A similar equation will result for each capacitor of the network.
For capacitors in parallel, as shown in Fig. 10.62, the voltage is the
same across each capacitor, and the total charge is the sum of that on
each capacitor:
Q T � Q 1 � Q 2 � Q 3
However, Q � CV
Therefore, CT E � C1V1 � C2V2 � C3V3 but E � V 1 � V 2 � V 3
CAPACITORS IN SERIES AND PARALLEL ⏐⏐⏐ 411
(10.33)
Thus, CT � C1 � C2 � C3 (10.34)
which is similar to the manner in which the total resistance of a series
circuit is found.
EXAMPLE 10.14 For the circuit of Fig. 10.63:
a. Find the total capacitance.
b. Determine the charge on each plate.
c. Find the voltage across each capacitor.
Solutions:
a. � � �
� � �
� 0.005 � 10 6 � 0.02 � 10 6 � 0.1 � 10 6
� 0.125 � 10 6
and CT � �8 mF
b. QT � Q1 � Q2 � Q3 QT � CTE � (8 � 10 �6 F)(60 V) � 480 mC
c. V1 � � �2.4 V
V2 � � �9.6 V
480 � 10
V3 � � �48.0 V
and E � V1 � V2 � V3 � 2.4 V � 9.6 V � 48 V
� 60 V (checks)
�6 480 � 10
Q3 C
� ���6
C3 10 � 10 F
�6 480 � 10
Q2 C
� ���6
C2 50 � 10 F
�6 1 1 1 1
� � � �
CT C1 C2 C3
1
1
1
���6
���6
���6
200 � 10 F 50 � 10 F 10 � 10 F
1
��6
0.125 � 10
Q1 C
� ���6
C1 200 � 10 F
+
E
–
+
E
–
Q T
+
–
Q1 V1 +
–
Q2 V2 FIG. 10.62
Parallel capacitors.
C 1
C 2
C 3
CT 60 V
200 mF 50 mF 10 mF
FIG. 10.63
Example 10.14.
+
–
Q3 V3