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

Options

A. `V_0/(omegaL) sin (omegaL + pi/2)`
B. `V_0/(omegaL) sin (omegat-pi/2)`
C. `V_0omegaL sin (omegat-pi/2)`
D. `(omegaL)/V_0 sin (omegat+pi/2)`

Answer

B

Explanation

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.