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

`omega = (E_0)/(sqrt(R^2 + (2pi v L)^2) ) = (240sqrt2)/(sqrt(100^2 + (2pi xx 10^4 xx 0.5)^2)) = 0.11A`
` tan phi = (2pi vL)/( R) = (2pi xx 10^4 xx 0.5)/(100) = 100 pi`
`phi` is close to `pi/2` , then time lag ` = (2pi)/(omega) .pi/2. (qv)/(360) = 0.25 xx 10^(-4) s `
`I_0` being very small we can conclude that at high frequencies , an inductor behaves as an open circuit .
In dc circuit , ` v = 0` . so inductor acts as a conductor .