1.

One inductor (of inductance L henry) is connected to an A.C. source, then the current flowing through the inductor I = ........ A.

Answer»

`V_0/(omegaL) SIN (omegaL + pi/2)`
`V_0/(omegaL) sin (omegat-pi/2)`
`V_0omegaL sin (omegat-pi/2)`
`(omegaL)/V_0 sin (omegat+pi/2)`

SOLUTION :In `tandelta=(omegaL)/R`, R=0
`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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