11.3 Solutions
519
11.73 Q = C 1 V = 5 × 10 −6 × 250 = 1.25 × 10 −3 C
For the parallel connection
C = C 1 + C 2 = 5 × 10
−6
+ 20 × 10
−6
= 25 × 10
−6 F
The resulting voltage
V
=
Q
C
=
1.25 × 10 −3
25 × 10 −6 = 50 V
11.74 The combination of 2 μF and 2 μF in parallel is equivalent to 4 μF. This in
series with 8 μF gives a combined capacitance of 8/3 μF; 12 μF and 6 μF
in series gives an equivalent capacitance of 4 μF. 4 μF with 4 μF in parallel
gives 8 μF which in series with 1 μF yields 8/9 μF.
Combination of 8/9 μF and 8/3 μF in parallel gives 32/9 μF.
Effective value of C with 32/9 μF in series gives
32
9
C
C +
32
9
= 1, by problem
∴
C = 1
9
23
μF
11.75 Combined capacitance C for the three capacitors in series:
C =
C 1 C 2 C 3
C 1 C 2 + C 2 C 3 + C 3 C 1
=
(4 × 3 × 2) × 10 −18
(4 × 3 + 3 × 2 + 2 × 4) × 10 −12
= 0.923 × 10
−6 F
(a)
q = C V = 0.923 × 10
−6
× 260 = 240 × 10
−6 C
∴
q 1 = q 2 = q 3 = 240 × 10
−6 C
(b) V 1 =
q
C 1
=
240 × 10 −6
4 × 10 −6 = 60 V
V 2 =
q
C 2
=
240 × 10 −6
3 × 10 −6 = 80 V
V 3 =
q
C 3
=
240
2
= 120 V
(c) W 1 =
1
2
C 1 V
2
1 =
1
2
× 4 × 10
−6
× (60)
2
= 0.0072 J
W 2 =
1
2
C 2 V
2
2 =
1
2
× 3 × 10
−6
× (80)
2
= 0.0096 J
W 3 =
1
2
C 3 V
2
3 =
1
2
× 2 × 10
−6
× (120)
2
= 0.0144 J