Electrical Energy and Capacitance
51
(b) From Figure 3:
( ) ( )( )
=
2.00 F
12.0 V
24.0 C
ac
ac
ac
QC V
µ
∆=
=
=
24.0 C
ab
bc
ac
QQ Q
From Figure 2:
=
=
3.00 F
Thus, the charge on the
capacitor is
3
=24
.0
C
Q
Continuing to use Figure 2,
()
24.0 C
=4
.
6.00 F
ab
ab
ab
Q
V
C
0
0
V
∆==
and
3
24.0 C
=8
.
0
0
V
3.00 F
bc
bc
bc
Q
VV
C
∆
=
=
()()
From Figure 1,
42
.00
V
ab
VVV
∆
()( )
( )
and
44
=
QC
4
4.00 F
4.00 V
16.0 C
V
=
(
)
22
=
2
V
8.00 C
=
16.32
(1)
12
1
9.00 pF
9.00 pF
parallel
CC
C
C
=+=
⇒ =
−
2
C
11
1
2.00 pF
series
series
C
C
C
C
=+ ⇒
=
=
+
Thus, using equation (1),
( )
9.00 pF
9.00 pF
series
C
−
=
−+
(
=
which reduces to

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