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?

Explanation

(a) `I_(0)=1.1xx10^(-2)A`
(b) `tan phi=100pi, phi` is close to `pi//2`.
`I_(0)` is much smaller than the low frequency case showing thereby that at high frequencies, L nearly amounts to an open circuit. In a dc circuit (after steady state) `omega=0`, so here L acts like a pure conductor.