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.

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, 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.