Question

Obtain the answers (a) to (b) in Exercise 7.13 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, ν = 10 kHz = 10⁴ Hz 

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

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.