14.3 Solutions
653
14.7 (a) X C =
1
ωC
=
1
2π f C
C =
1
2π f X C
=
1
2π × 250 × 4
= 1.57 × 10
−4 F
(b) X
C =
1
2π f C
=
1
2π × 100 × 1.57 × 10 −4 = 10
(c) I e =
V e
X
C
=
220
10
= 22 A
14.8
V e
I e
=
ω 2 L 2 + R 2 =
4π 2 f 2 L 2 + R 2
∴
12
0.05
=
4π 2 × (50) 2 L 2 + R 2
(1)
Also tan α =
ωL
R
=
2π fL
R
∴ tan 60
◦
=
√
3 = 2π × 50
L
R
(2)
Solving (1) and (2) we find R = 120 and L = 0.66 H.
When the capacitor of capacitance C is connected in series with the above
circuit α = 0.
tan α = tan 0
◦
=
1
R
ωL −
1
ω c
∴ C =
1
ω 2 L
=
1
(2π × 50) 2 × 0.66
= 15.37 × 10
−6 F = 15.37 μF
14.9 (a) The rms current, I e =
V e
ωL
=
V e
2π fL
=
220
2π × 50 × 0.6
= 1.17 A
(b) Peak current, I 0 =
√
2I e = 1.414 × 1.17 = 1.65 A.
14.10 (a) I e =
V e
4π 2 f 2 L 2 + R 2
=
24
4π 2 × (50) 2 × (0.01) 2 + 4 2
= 4.72 A
(b) Power, P = I
2
e R = (4.72)
2
× 4 = 89 W
(c) V R = I e R = 4.72 × 4 = 18.88 V
V L = 2π fLI e = 2π × 50 × 0.01 × 4.72 = 13.82 V
14.11 V = V m sin ωt
0.5 V m = V m sin
2π f
1
360
∴
2π f
360
=
π
6
⇒ f = 30 Hz