660
14 Electromagnetism II
f 0 =
1
2π
√
LC
=
1
2π
√
0.2 × 10 × 10 −6
= 112.6 Hz
14.22 X C =
1
2π f C
=
1
(2π)(300)(50 × 10 −6 )
= 10.6
Z =
R 2 + X 2
C =
(300) 2 + (10.6) 2 = 300.19
I =
V
Z
=
5
300.19
= 0.0166 A
14.23 Impedance of a capacitor
i. Let an AC emf be applied across a capacitor. The potential difference
across the capacitor will be
V C = V 0 sin ωt
(1)
where V 0 is the amplitude of the AC voltage of angular frequency ωt =
2π f , across the capacitor.
q C = C V C = C V 0 sin ωt
(2)
The current i c =
dq C
dt
= ωC V 0 cos ωt = ωC V 0 sin(ωt + 90
◦
)
(3)
or i c =
V 0
X c
sin (ωt + 90
◦
)
(4)
where X C = 1/ωC
(5)
Comparison of (4) with (1) shows that i c leads V C by 90 ◦ or quarter of a
cycle. Further, the current amplitude
I 0 =
V 0
X C
(6)
By Ohm’s law X C is to be regarded as impedance offered by the capacitor. In complex plane
Z c =
−j
ωC
(7)
where j is imaginary.
Impedance of an inductance
On applying an AC across an inductance the potential difference will be
V L = V 0 sin ωt
(8)
where V 0 is the amplitude of V L . By Faraday’s law of induction (ξ =
−L di/dt) we can write