P26_069.PDF
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Chapter 26 - 26.69
69. (a) The voltage across
C
1
is 12 V, so the charge is
q
1
=
C
1
V
1
=24
µ
C
.
(b) We reduce the circuit, starting with
C
4
and
C
3
(in parallel) which are equivalent to 4
µ
F. This is
then in series with
C
2
, resulting in an equivalence equal to
3
µ
F which would have 12 V across it.
The charge on this
3
µ
F capacitor (and therefore on
C
2
)is(
3
µ
F)(12V) = 16
µ
C. Consequently,
the voltage across
C
2
is
C
2
=
16
µ
C
2
µ
F
=8V
.
This leaves 12
−
8=4Vacross
C
4
(similarly for
C
3
).
V
2
=
q
2
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P26_005.PDF
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P26_002.PDF
(52 KB)
P26_003.PDF
(57 KB)
P26_001.PDF
(52 KB)
P26_019.PDF
(64 KB)
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