A sinusoidal voltage `V (t) = 100 sin (500 t)` is applied across a pure inductance of `L = 0.02 H`. The current through the coil is :

Options

A. 10 cos (500 t)
B. `-10 cos (500 t)`
C. `10 sin (500 t)`
D. `-10 sin (500 t)`

Answer

B

Explanation

In a pure inductive circuit current always lags behind the emf by `(pi)/(2)`
If v(t) = `v_(0) sin omega t`
then `I = I_(0) sin (omega t- (pi)/(2))`
Now, given `v(t) = 100 sin (500 t)`
and `I_(0) = (E_(0))/(omegaL) = (100)/(500 xx 0.02) [ because L = 0.02 H]`
`I_(0) = 10 sin (500 t - (pi)/(2)) Rightarrow I_(0) = - 10 cos(500 t)`