Physics - EMI/AC Question with Solution | TestHub

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 $v(t)=100\sin (500t)$ and ${\mathrm{I}}_0=\frac{{\mathrm{E}}_0}{\omega \mathrm{L}}=\frac{100}{500\times 0.02}[\because \mathrm{L}=0.02\mathrm{H}]$ ${\mathrm{I}}_0=10\sin \left(500\mathrm{t}-\frac{\pi }{2}\right)$ ${\mathrm{I}}_0=-10\cos (500\mathrm{t})$