Obtain the answers to (a) and (b) in Exercise 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.

Given: Capacitance C = 100 μF

In Farad, Capacitance C = 100 × 10^-6F

Resistance = 40Ω

Voltage V = 110 V

Frequency = 12 kHz

In Hz, frequency v = 12 × 10³Hz

Angular frequency ω = 2πv = 2 × π × 12 × 10³(Hz)

ω = 24 × 10³ rad/s

Peak voltage is calculated as follows:

V₀ = V/ √2

V₀ = 110√2V

The peak current is calculated as follows:

I₀ =

I₀ =

On calculating, we get

I₀ = 3.9 A

Since for RC circuit, the voltage lags behind the current by a phase angle of Ф

tanФ =

tanФ = 1/ωCR

substituting the values, we get

tanФ = 1/(24π × 10³(Hz) × 100 × 10^-6(F) × 40 (Ω))

On calculating, we get

tanФ = 1/96π

or Ф = 0.2⁰

in radians, Ф = (0.2 π) /180

At high frequency, Ф approaches to zero and capacitor acts as a conductor.

In d.c. circuit, ω is zero, and the steady state is attained and therefore, capacitor C tends to be an open circuit.