Obtain the answers to (a) and (b) Q.13, if the circuit is connected to a high frequency supply (240 V , 10 kHz). Hence explain statement that at very high frequency. Inductor in circuit nearly amount to open circuit. How does an indcutor behave in a d.c. 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.