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?

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 x 10⁴ rad/s

Ans (a).

Peak voltage, V₀ = √2 V= 240 √2 V

Maximum Current, I₀ = V₀ /√ (R² + ω² L²)

=240 √2/√[(100)² + (2π x 10⁴ )² x (0.50)² ]

=1.1 x 10⁻²A

Ans (b).

For phase difference φ,we have the relation :

tanφ = Lω/R =(2π x 10⁴  x 0.5)/100 = 100π

=> φ = 89.82º =89.82π /180 rad

ωt = 89.82π /180

=> t = 89.82π /(180 x 2π x 10⁴) = 25μs
It can be observed that Io 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.