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 1

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, ν = 12 kHz = 12 × 10³ Hz

Angular Frequency, ω = 2 πν= 2 × π × 12 × 10³03

= 24π × 10³ rad/s

Peak voltage,

Maximum current,

For an RC circuit, the voltage lags behind the current by a phase angle of Φ given as:

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.