p. 48
Capacitors in Series and Parallel
In a parallel combination of capacitances C1 and C2 , the voltage
V across each capacitor is the same, but the charges q1 and q2 on
each capacitor are different.
Even though two capacitors are connected in parallel across it, the
seat of emf “sees” a single equivalent capacitance CP:
q1  q2
or
V
q q
CP  1  2 or
V V
CP  C1  C2
CP 
Example
A 3.0 F and a 5.0 F capacitor are connected in parallel across a 12 V battery.
a. Find the charge on each capacitor.
q1  C1V1; q1   3.0 F 12 V  ;
q1  36 C
q2  C2V2 ; q2   5.0 F 12 V  ; q2  60 C
b. Find the equivalent capacitance.
CP = C1 + C2; CP = 3.0 F + 5.0 F; CP = 8.0 F
In a series combination of capacitances C1 and C2 ,
the same amount of charge q is on the plates of
each capacitor, but the voltages V1 and V2 across
each capacitor are different.
Even though two capacitors are connected in series
with it, the seat of emf “sees” a single equivalent
capacitance CP, which may be determined as
follows.
V  V1  V2
q
q
q

 ;
CS C1 C2
An example follows on the next page.
1
1
1


CS C1 C2
p. 49
Example
A 3.0 F and a 5.0 F capacitor are connected in series with a 12 V battery.
a. Find the equivalent capacitance.
1
1
1


;
CS 3.0 F 5.0 F
1
53
15 F

; CS 
; CS  1.9 F
CS 15 F
8
b. Find the charge on each capacitor.
q = CSV: q = (1.9 F)(12 V); q = 23 C
c. Find the potential drop (or voltage) across each capacitor.
V1 
q
23 C
: V1 
;
C1
3.0 F
V2 
q
23 C
: V2 
; V1  4.6 V
C2
5.0 F
V1  7.7 V
Capacitance Networks
It is possible to use the formulas for capacitors connected in series and parallel to find the equivalent
capacitance of more elaborate capacitance networks.
Example (Chapter 20 problem 87)
Determine the equivalent capacitance between A and B for
the group of capacitors in the drawing.
87.
REASONING Our approach to this problem is to deal with the arrangement in parts. We will combine
separately those parts that involve a series connection and those that involve a parallel connection.
SOLUTION The 24, 12, and 8.0-F capacitors are in series. Using Equation 20.19, we can find the
equivalent capacitance for the three capacitors:
1
1
1
1



Cs 24  F 12  F 8.0  F
or
Cs  4.0  F
This 4.0-F capacitance is in parallel with the 4.0-F capacitance already shown in the text diagram.
Using Equation 20.18, we find that the equivalent capacitance for the parallel group is
p. 50
Cp  4.0  F + 4.0  F = 8.0  F
This 8.0-F capacitance is between the 5.0 and the 6.0-F capacitances and in series with them.
Equation 20.19 can be used, then, to determine the equivalent capacitance between A and B in the text diagram:
1
1
1
1



Cs 5.0  F 8.0  F 6.0  F
or
Cs  2.0  F