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One inductor (of inductance L henry) is connected to an A.C. source, then the current flowing through the inductor I = ........ A. |
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Answer» `V_0/(omegaL) SIN (omegaL + pi/2)` `THEREFORE tan DELTA` = infinite `therefore delta = pi/2` `therefore` From `I=V/"|Z|" sin (omegat-delta)` `therefore I=V_0/(omegaL) sin (omegat-pi/2)` Amp. |
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