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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? |
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Answer» Solution :(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 caseshowing 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. |
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