Question

Obtain if the circuit is connected to a high-frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?

Answer 1

Inductance of the inductor, L = 0.5 Hz

Resistance of the resistor, R = 100 Ω

Potential of the supply voltages, V = 240 V

Frequency of the supply, v = 10 kHz = 10⁴ Hz

Angular frequency, ω = 2πv = 2π × 10⁴ rad/s

(a) Peak voltage, V₀ = `sqrt2 xx "V" = 240 sqrt2 "V"`

Maximum current, I₀ = `"V"_0/sqrt("R"^2 + ω^2"L"^2)`

= `(240 sqrt2)/sqrt((100)^2 + (2π xx 10^4)^2 xx (0.50)^2)`

= 1.1 × 10⁻² A

(b) For phase difference Φ, we have the relation:

`tan phi = (ω"L")/"R"`

= `(2π xx 10^4 xx 0.5)/100`

= 100π

`phi` = 89.82° = `(89.82π)/180 "rad"`

ωt = `(89.82π)/180`

t = `(89.82π)/(180 xx 2π xx 10^4)`

= 25 μs

It can be observed that I₀ is very small in this case. Hence, at high frequencies, the inductor amounts to an open circuit.

In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure conducting object.