Chapter 23 Problem 31 † Given C1 = 0.02 µF C2 = 0.01 µF C3

```Chapter 23
Problem 31
†
C1
C2
C3
Given
C1 = 0.02 µF
C2 = 0.01 µF
C3 = 0.02 µF
V = 100 V
Solution
a) Find the equivalent capacitance of the combination.
C1
C'
C2 and C3 are in parallel, their equivalent capacitance is
C 0 = C2 + C3 = 0.01 µF + 0.02 µF = 0.03 µF
This leaves us with a combination of C1 and C 0 in series. This gives us an equivalent capacitance of
1
1
1
=
+ 0
Ce
C1 C
Ce =
1
C1
1
+
1
C0
=
1
0.02 µF
1
+
1
0.03 µF
= 0.012 µF
Note: Although the textbook asks to solve for the charge on each capacitor, I find it easier to solve for
the voltages first. Therefore, I am doing part c) first followed by part b).
c) Find the voltage across each capacitor.
From the second diagram the voltage across C1 and C 0 must total 100 V .
V1 + V 0 = 100 V
†
Problem from Essential University Physics, Wolfson
From the definition of capacitance
C=
Q
V
then
V =
Q
C
Therefore,
Q1 Q0
+
= 100 V
C1 C 0
When the voltage is first applied, the charge drawn off of C1 will have to accumulate on C 0 , therefore, the
charge on each of these capacitors will be the same.
Q
Q
+ 0 = 100 V
C1 C
Solving for Q gives
Q=
100 V
=
+ C10
1
C1
100 V
= 1.2 µC
+ 0.031 µF
1
0.02 µF
The voltage across C1 is then
V1 =
1.2 µC
= 60 V
0.02 µF
Therefore, the voltage across C 0 is
V 0 = 100 V − V1 = 100 V − 60 V = 40 V
Since C2 and C3 are in parallel, the voltage across each of them must be the same. Therefore,
V2 = V3 = 40 V
b) Find the charge on each capacitor.
From part b the charge on C1 is 1.2 µC. The charge on C2 is
C2 =
Q2
V2
Solving for Q2 gives
Q2 = C2 V2 = (0.01 µF )(40 V ) = 0.4 µC
By the same process the charge on C3 is
Q3 = C3 V3 = (0.02 µF )(40 V ) = 0.8 µC
Notice that the charge flowing off of C1 is equal to the total charge on C2 and C3 .
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