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Question
Obtain 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.
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Solution
Capacitance of the capacitor, C = 100 μF = 100 × 10−6 F
Resistance of the resistor, R = 40 Ω
Supply voltage, V = 110 V
Frequency of the supply, v = 12 kHz = 12 × 103 Hz
Angular Frequency, ω = 2 πv = 2 × π × 12 × 103 Hz = 24π × 103 rad/s
Peak voltage, V0 = `"V"sqrt2 = 110 sqrt2 "V"`
Maximum current, I0 = `"V"_0/sqrt("R"^2 + 1/(ω^2"C"^2))`
= `(110 sqrt2)/sqrt((40)^2 + 1/(24π xx 10^3 xx 100 xx 10^-6)^2)`
= `(110 sqrt2)/sqrt(1600 + (10/(24π))^2)`
= 3.9 A
For an RC circuit, the voltage lags behind the current by a phase angle of Φ given as:
`tan phi = (1/(ω"C"))/"R" = 1/(ω"CR")`
= `1/(24π xx 10^3 xx 100 xx 10^-6 xx 40)`
`tan phi = 1/(96π)`
∴ `phi` ≃ 0.2°
= `(0.2π)/180 "rad"`
∴ Time lag = `phi/ω`
= `(0.2π)/(180 xx 24π xx 10^3)`
= 1.55 × 10−3 s
= 0.04 μs
Hence, Φ tends to become zero at high frequencies. At a high frequency, capacitor C acts as a conductor.
In a dc circuit, after the steady state is achieved, ω = 0. Hence, capacitor C amounts to an open circuit.
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