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.

Capacitance of the capacitor, C = 100 μF = 100 x 10^-6 F =10⁻⁴F
Resistance of the resistor, R = 40Ω
Supply voltage, V = 110 V
Frequency of the supply, ν = 12 kHz = 12 x 10³ Hz
Angular Frequency, ω = 2π ν = 2 x π x 12 x 10³ Hz
= 24π x 10³ rad/s 
Peak voltage, V₀ = V√2 = 110√2 V
Maximum Current ,I₀=  V₀/√(R² +1/ω²c²)

=110 √2 /√[(40)² +1/(24π x 10³)²(10⁻⁴)²]

=110 √2/√[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 φ = (1/ωC)/R = 1/ωCR = 1/(24π x 10³  x 10⁻⁴ x 40)

tan φ = 1/96π

Therefore , φ = 0.2º or 0.2 π/180 rad

Therefore ,Time lag = φ/ω  =  0.2 π/(180 x 24π x 10³)

=4.69 x 10⁻³s

= 0.04μs

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.