Question:
Obtain the answers to (a) and (b) in Exercise 7.15 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.
Answer:
Capacitor C = 100μF = 100 x 10-6 F
Resistor R = 40 Ω
Voltage V = 110 V
Frequency v = 12 KHz = 12 x 103 Hz
Angular frequency ω = 2πv = 2 x π x 12 x 103 = 24 π x 103 rad/s
Peak voltage V0 = V√2 = 110√2 V
Maximum current I0 = V0/√ R2 + ω2L2
= 110√2/ √(40)2 + 1/ (24 π x 103 x 100 x 10-6 )2 =3.9 A
For an RC circuit, the voltage lags behind the current by a phase angle
tan ɸ = 1/ωCR = 1/ 24
π x 103 x 100 x 10-6 x 40 = 1/96 π
ɸ = 0.2 ° = 0.2 π /180 rad
Time lag = ɸ/ω = 0.2 π /180 x 24 π x 103 = 1.55 x 10-3 s = 0.04 μs
Hence, ɸ tends to become 0 at high frequencies. Capacitor C amount to an open circuit.
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