An ac voltage of $10 \sin \omega t$ volt is applied to a pure inducfor of inductance $10 \mathrm{H}$. The current through the inductor in ampere is

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Apply the kirchhoff's loop rule $ \begin{aligned} & V+E=0 \\ & V-L \frac{d i}{d t}=0 \\ & V=L \frac{d i}{d t} \end{aligned} $ $ \begin{aligned} & \frac{\mathrm{di}}{\mathrm{dt}}=\frac{\mathrm{V}}{\mathrm{L}}=\mathrm{V}_0 \frac{\sin \omega \mathrm{t}}{\mathrm{L}} \\ & \mathrm{di}=\frac{\mathrm{V}_0}{\mathrm{~L}} \sin \omega \mathrm{t} \end{aligned} $ Integration with respect to time $t$ $ \begin{aligned} & i=\frac{V_0}{L}\left(-\frac{\cos \omega t}{\omega}\right)=\frac{-V_0}{\omega L} \cos \omega t \\ & i=\frac{V_0}{\omega L} \sin \left(\omega t-\frac{\pi}{2}\right)=\frac{10}{\omega \times 10} \sin \left(\omega t-\frac{\pi}{2}\right) \\ & i=\frac{1}{\omega} \sin \left(\omega t-\frac{\pi}{2}\right) \end{aligned} $

An ac voltage of $10 \sin \omega t$ volt is applied to a pure inducfor of inductance $10 \mathrm{H}$. The…

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