Chapter 26B - Capacitor
Circuits
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
Objectives: After completing this
module, you should be able to:
• Calculate the equivalent capacitance of a
number of capacitors connected in series
or in parallel.
• Determine the charge and voltage across
any chosen capacitor in a network when
given capacitances and the externally
applied potential difference.
Electrical Circuit Symbols
Electrical circuits often contain two or more
capacitors grouped together and attached
to an energy source, such as a battery.
The following symbols are often used:
Ground
+ - + - + - + -
Battery
+
-
Capacitor
+
+
-
Series Circuits
Capacitors or other devices connected
along a single path are said to be
connected in series. See circuit below:
+
+
C1
- +
- +
-+
-+
C2
Battery
-
C3
Series connection
of capacitors.
“+ to – to + …”
Charge inside
dots is induced.
Charge on Capacitors in Series
Since inside charge is only induced, the
charge on each capacitor is the same.
Q1
+
+
C1
Q2
- +
- +
Q3
-+
-+
C2
Battery
-
C3
Charge is same:
series connection
of capacitors.
Q = Q1 = Q2 =Q3
Voltage on Capacitors in Series
Since the potential difference between
points A and B is independent of path, the
battery voltage V must equal the sum of
the voltages across each capacitor.
V1
+
+
C1
•A
V2
- +
- +
V3
-+
-+
C2
Battery
-
C3
•
B
Total voltage V
Series connection
Sum of voltages
V = V1 + V2 + V3
Equivalent Capacitance: Series
V1
+
+
V2
- +
- +
C1
C2
V3
-+
-+
-
Q
Q
C ; V 
V
C
C3
V = V1 + V2 + V3
Q1= Q2 = Q3
Q Q1 Q2 Q3
 

C C1 C2 C3
1
1
1
1
 

Ce C1 C2 C3
Equivalent Ce
for capacitors
in series:
n
1
1

Ce i 1 Ci
Example 1. Find the equivalent capacitance
of the three capacitors connected in series
with a 24-V battery.
Ce for
series:
n
1
1

Ce i 1 Ci
1
1
1
1



Ce 2m F 4m F 6m F
1
 0.500  0.250  0.167
Ce
1
1
 0.917 or Ce 
Ce
0.917
C1
C2
C3
+ - + -+ + - + -+ 2 mF 4 mF 6 mF
24 V
Ce = 1.09 mF
Example 1 (Cont.): The equivalent circuit can
be shown as follows with single Ce.
C1
C2
C3
+ - + -+ + - + -+ 2 mF 4 mF 6 mF
n
1
1

Ce i 1 Ci
Ce
1.09 mF
24 V
Ce = 1.09 mF
24 V
Note that the equivalent capacitance Ce
for capacitors in series is always less than
the least in the circuit. (1.09 mF < 2 mF)
Example 1 (Cont.): What is the total charge
and the charge on each capacitor?
C1
C2
C3
+ - + -+ + - + -+ 2 mF 4 mF 6 mF
Ce
1.09 mF
24 V
24 V
QT = CeV = (1.09 mF)(24 V);
For series circuits:
QT = Q1 = Q2 = Q3
Ce = 1.09 mF
Q
C
V
Q  CV
QT = 26.2 mC
Q1 = Q2 = Q3 = 26.2 mC
Example 1 (Cont.): What is the voltage across
each capacitor?
Q
Q
C ; V 
V
C
Q1 26.2 m C
V1 

 13.1 V
C1
2 mF
Q2 26.2 m C
V2 

 6.55 V
C2
4 mF
Q3 26.2 mC
V3 

 4.37 V
C3
6 mF
C1
C2
C3
+ - + -+ + - + -+ 2 mF 4 mF 6 mF
24 V
VT = 24 V
Note: VT = 13.1 V + 6.55 V + 4.37 V = 24.0 V
Short Cut: Two Series Capacitors
The equivalent capacitance Ce for two series
capacitors is the product divided by the sum.
1
1
1
  ;
Ce C1 C2
C1C2
Ce 
C1  C2
Example:
(3 m F)(6 m F)
Ce 
3 m F  6m F
C1
C2
+ - + + - + 3 mF 6 mF
Ce = 2 mF
Parallel Circuits
Capacitors which are all connected to the
same source of potential are said to be
connected in parallel. See below:
Parallel capacitors:
“+ to +; - to -”
- -
C3
+
+
- -
C2
+
+
+
+
C1
- -
Voltages:
VT = V1 = V2 = V3
Charges:
QT = Q1 + Q2 + Q3
Equivalent Capacitance: Parallel
Q
C  ; Q  CV
V
Parallel capacitors
in Parallel:
- -
Ce = C1 + C2 + C3
C3
+
+
- -
C2
+
+
+
+
C1
- -
Q = Q1 + Q2 + Q3
Equal Voltages:
CV = C1V1 + C2V2 + C3V3
Equivalent Ce
for capacitors
in parallel:
n
Ce   Ci
i 1
Example 2. Find the equivalent capacitance
of the three capacitors connected in parallel
with a 24-V battery.
Ce for
parallel:
n
Ce   Ci
VT = V1 = V2 = V3
Q = Q1 + Q2 + Q3
i 1
24 V
Ce = (2 + 4 + 6) mF
2 mF
C1 C2
C3
4 mF
6 mF
Ce = 12 mF
Note that the equivalent capacitance Ce for
capacitors in parallel is always greater than
the largest in the circuit. (12 mF > 6 mF)
Example 2 (Cont.) Find the total charge QT
and charge across each capacitor.
Q = Q1 + Q2 + Q3
24 V
2 mF
C1 C2
C3
4 mF
6 mF
Ce = 12 mF
V1 = V2 = V3 = 24 V
Q
C  ; Q  CV
V
QT = CeV
Q1 = (2 mF)(24 V) = 48 mC
QT = (12 mF)(24 V)
Q1 = (4 mF)(24 V) = 96 mC
QT = 288 mC
Q1 = (6 mF)(24 V) = 144 mC
Example 3. Find the equivalent capacitance
of the circuit drawn below.
24 V
C1
4 mF
24 V
4 mF
C1
C3,6
C2
3 mF
C3
6 mF
C3,6
(3m F)(6m F)

 2m F
3m F  6m F
Ce = 4 mF + 2 mF
Ce = 6 mF
24 V
2 mF
Ce
6 mF
Example 3 (Cont.) Find the total charge QT.
Ce = 6 mF
24 V
C1
4 mF
24 V
4 mF
C1
C2
3 mF
C3
6 mF
C3,6
Q = CV = (6 mF)(24 V)
QT = 144 mC
24 V
2 mF
Ce
6 mF
Example 3 (Cont.) Find the charge Q4 and
voltage V4 across the the 4-mF capacitor.
V4 = VT = 24 V
24 V
4 mF
C1
C2
3 mF
C3
6 mF
Q4 = (4 mF)(24 V)
Q4 = 96 mC
The remainder of the charge: (144 mC – 96 mC)
is on EACH of the other capacitors. (Series)
Q3 = Q6 = 48 mC
This can also be found from
Q = C3,6V3,6 = (2 mF)(24 V)
Example 3 (Cont.) Find the voltages across
the 3 and 6-mF capacitors.
Q3 = Q6 = 48 mC
24 V
4 mF
C1
C2
3 mF
C3
6 mF
48m C
V3 
 16.0V
3m F
48m C
V6 
 8.00V
6m F
Note: V3 + V6 = 16.0 V + 8.00 V = 24 V
Use these techniques to find voltage and
capacitance across each capacitor in a circuit.
Summary: Series Circuits
n
1
1

Ce i 1 Ci
Q = Q1 = Q2 = Q3
V = V1 + V2 + V3
For two capacitors at a time:
C1C2
Ce 
C1  C2
Summary: Parallel Circuits
n
Ce   Ci
i 1
Q = Q1 + Q2 + Q3
V = V1 = V2 =V3
For complex circuits, reduce the circuit in steps
using the rules for both series and parallel
connections until you are able to solve problem.
CONCLUSION: Chapter 26B
Capacitor Circuits