A sinusoidal voltage V(t)=100 (500 t) is applied across a pure...

In a pure inductive circuit current always lags behind the emf by $\frac{\pi}{2}$. If $\mathrm{v}(\mathrm{t})=\mathrm{v}_0 \sin \omega \mathrm{t}$ then $I=I_0 \sin$\left(\omega t-\frac{\pi}{2}\right)$Now, given$v(t)=100 \sin (500 t)$and$\mathrm{I}_0=\frac{\mathrm{E}_0}{\omega \mathrm{L}}=\frac{100}{500 \times 0.02}[\because \mathrm{L}=0.02 \mathrm{H}] $\mathrm{I}$_0=10 \sin \left(500 \mathrm{t}-\frac{\pi}{2}\right)$\mathrm{I}_0=-10 \cos (500 \mathrm{t})$