University of Alabama
Department of Physics and Astronomy
PH 106 / LeClair
Spring 2008
Exercise: Electrical Energy & Capacitance: Solutions
1.
Find the equivalent capacitance of the capacitors in the figure below.
+

12V
3
μ
F
6
μ
F
8
μ
F
8
μ
F 4
μ
F
Before we start, it is useful to remember that one farad times one volt gives one coulomb:
1 [
F
]
·
1 [
V
] = 1 [
C
]
, and that
capacitance times voltage gives stored charge:
Q
=
CV
. Knowing this now will save some confusion on units later on. For
that matter, it is also good to remember that the prefix
μ
means
10

6
.
In order to find a single equivalent capacitor that could replace all five in the diagram above, we need to look for purely
series and parallel combinations that can be replaced by a single capacitor. The uppermost
8
μ
F and
4
μ
F capacitors are
purely in series, so they can be replaced by a single equivalent we will call
C
84
, as shown below:
Using our rule for combining series capacitors, we can find the value of
C
84
easily:
1
C
84
=
1
8
μ
F
+
1
8
μ
F
=
⇒
C
84
=
8
3
μ
F
≈
2
.
67
μ
F
Now we have this equivalent capacitance purely in
parallel
with the second
8
μ
F capacitor. We can replace
C
84
and the
second
8
μ
F capacitors with a single equivalent, which we will call
C
884
:
Using our addition rule for parallel capacitors, we can find its value:
C
884
=
C
84
+ 8
μ
F
≈
10
.
67
μ
F
This leaves us with three capacitors in series, as shown below:
Adding together these three in series, we have the overall equivalent capacitance,
C
eq
:
1
C
eq
=
1
C
884
+
1
3
μ
F
+
1
6
μ
F
=
⇒
C
eq
≈
1
.
68
μ
F
Since we now have one single capacitor connected to a single voltage source, we can find the total charge stored in the
equivalent capacitor,
Q
eq
:
Q
eq
=
C
eq
V
= (1
.
68
μ
F
) (12
V
)
≈
20
.
16
μ
C
Now what if we wanted to get the charge and voltage on each single capacitor? In that case, we have to work backwards
and rebuild our original circuit. We know that the
C
eq
capacitor is really three capacitors in series  the
6
μ
F, the
3
μ
F,
and
C
884
. Series capacitors always have the same charge, and one must have the same charge as the equivalent capacitor:
Q
6
μ
F
=
Q
3
μ
F
=
Q
884
Since we know the charge and capacitance for all three of these capacitors, we can now find the voltage
on each, since
V
=
Q/C
:
V
6
μ
F
=
Q
6
μ
F
6
μ
F
=
20
.
16
μ
C
6
μ
F
≈
3
.
4
V
V
3
μ
F
=
Q
3
μ
F
3
μ
F
=
20
.
16
μ
C
3
μ
F
≈
6
.
7
V
V
884
=
Q
884
C
884
=
20
.
16
μ
C
10
.
67
μ
F
≈
1
.
9
V
Notice that the voltage on all three of these series capacitors adds up to the total battery voltage  it must be so, based
on conservation of energy. Next, we know that
C
884
is really two capacitors in parallel  the lower
8
μ
F capacitor and
C
84
.
Parallel capacitors have the same voltage, so we know that both of these have to have
V
lower 8
μ
F
=
V
84
=
V
884
=1
.
9
V across
them.
We know the voltage and the capacitance for
C
84
and the lower
8
μ
F capacitors now, so we can find the stored
charge,
Q
=
CV
:
Q
lower 8
μ
F
= 8
μ
F
·
V
884
= 8
μ
F
·
1
.
9
V
≈
15
.
2
μ
C
Q
84
=
C
84
·
V
884
= 2
.
67
μ
F
·
1
.
9
V
≈
5
.
1
μ
C
Finally, the capacitor
C
84
is really two capacitors in series, which must both have the same charge:
Q
4
μ
F
=
Q
upper 8
μ
F
=
Q
84
.
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 Electric charge, qeq, Department of Physics and Astronomy