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A coil of inductance 0.50 H and resistance 100Ω is connected to a 240 V, 50 Hz ac supply.What is the maximum current in the coil? |
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Answer» Solution :`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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