InterviewSolution
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InterviewSolution
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A coil of inductance 0.50 H and resistance `100Omega` is connected to a 240 V, 50Hz ac supply. IF the circuit is connected to a high frequency supply (240V,10 kHZ), what wull be the answer to (a) and (b) . From the answer explain the statement that at a very high frequency the presence of a inductor in the circuit nearly amounts to an open circuit. |
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Answer» `omega=2pif=2pitimes10^4rad.s^-1` `I_(max)=V_(max)/sqrt(R^2+omega^2L^2)=(240sqrt2)/(sqrt(100^2+4pi^2. 10^8 .5^2))=0.011A` `thereforephi=tan^-1{X_L/R}=tan^-1((2pifL)/R)` `=tan^-1 ""(2pitimes10^4times0.5)/100` `tan^-1(100pi)approxpi/2` `therefore` Time interval=`phi.T/360=phi/(360f)` `=90/(360times10times10^3)=0.25times10^-4s` `I_(max)` is very small. So it can be concluded that at high frequencies an inductance behaves as an open circuit. |
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