InterviewSolution
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InterviewSolution
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If the circuit is connected to a high frequencysupply (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 circuitafter the steady state? |
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Answer» Solution :`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 . |
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