Physics - EMI/AC Question with Solution | TestHub

If $\frac{dI}{dt}$ is the rate of change of current through inductor $L$ of any instant, then induced emf in the inductor of the same instant is $=\left(-\frac{LdI}{dt}\right)$ To maintain the flow of current in the circuit applied voltage must be equal and opposite to the induced voltage. $\therefore V=-\left(-L\frac{dI}{dt}\right)⋐$ Given, $V=V_0\sin \omega t$
$\therefore V_0\sin \omega t=\frac{LdI}{dt}$ $\frac{dI}{dt}=\frac{V_0}{L}\sin \omega t$ or $I=\frac{-V_0}{L}\frac{\cos \omega t}{\omega }$
Instantaneous power, $\because P=VI$ $=\left(V_0\sin \omega t\right)\left(-\frac{V_0\cos \omega t}{L\omega }\right)(\mathrm{Using}$ (i) \& (ii)) $=\frac{-V_0^2\sin 2\omega t}{2L\omega }$