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57. Initially the capacitors
C
1
,
C
2
, and
C
3
form a series combination equivalent to a single
capacitor which we denote
C
123
. Solving the equation
12
23
13
123
1
2
3
1
2
3
11
1
1
CC
C
C
C
C
C
+
+
=++=
,
we obtain
C
123
= 2.40
µ
F.
With
V =
12.0 V, we then obtain
q
=
C
123
V
= 28.8
µ
C.
In the
final situation,
C
2
and
C
4
are in parallel and are thus effectively equivalent to
24
12.0 F
C
µ
=
.
Similar to the previous computation, we use
4
2
43
1234
1
24
3
1
24
3
1
1
C C
C
C
C
C
C
+
+
=+
+=
and find
C
1234
= 3.00
µ
F.
Therefore, the final charge is
q
=
C
1234
V
= 36.0
µ
C.
(a) This represents a change (relative to the initial charge) of
∆
q
= 7.20
µ
C.
(b) The capacitor
C
24
which we imagined to replace the parallel pair
C
2
and
C
4
is in series
with
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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