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NCERT Class XII Chapter
Alternating Current
Questions With Solutions

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Question : 16
Total: 26
Obtain the answer to (a) and (b) in Q.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
Solution:  
Given, Erms = 110 V, u = 12 kHz = 12 × 103 Hz
(a) XC =
1
2×3.14×(12×103)×10−4
 = 0.1326 Ω
As, R = 40 Ω, Xc ⋘ R, Z ≈ R = 40 Ω
Irms =
110
40
 = 2.75 A , I0 = √2Irms = 1.414 (2.75) = 3.9 A
This value of current is same as that without capacitor in the circuit. So, at high frequency, a capacitor offer negligible resistance (0.1326 W in this case), it behave like a conductor.
(b) As tan ϕ =
XC
R
 =
0.1326Ω
40Ω
 = 0.0033
ϕ = 0.189° = 0°
Time lag =
0.189°
360°
 ×
1
12×103
 = 43.8 × 10−9 s
In dc circuit, after steady state, v = 0 and accordingly, XC = ∞, i.e., a capacitor amounts to an open circuit, i.e., it is a perfect insulator of current.
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