An AC voltage of 10 sin omega t volt is appli | vedclass

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(A) Given,voltage $V = V_0 \sin \omega t$,where $V_0 = 10 \ V$ and inductance $L = 10 \ H$.For a pure inductor,the induced $EMF$ is $E = -L \frac{di}{

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$\frac{1}{\omega} \sin \left(\omega t-\frac{\pi}{2}\right)$

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(A) Given,voltage $V = V_0 \sin \omega t$,where $V_0 = 10 \ V$ and inductance $L = 10 \ H$.For a pure inductor,the induced $EMF$ is $E = -L \frac{di}{dt}$.According to Kirchhoff's loop rule,$V + E = 0$,so $V = L \frac{di}{dt}$.Rearranging for current $i$,we get $\frac{di}{dt} = \frac{V}{L} = \frac{V_0 \sin \omega t}{L}$.Integrating with respect to time $t$:$i = \int \frac{V_0}{L} \sin \omega t \ dt = \frac{V_0}{L} \left( -\frac{\cos \omega t}{\omega} ight) = -\frac{V_0}{\omega L} \cos \omega t$.Using the trigonometric identity $-\cos 	heta = \sin \left( 	heta - \frac{\pi}{2} ight)$:$i = \frac{V_0}{\omega L} \sin \left( \omega t - \frac{\pi}{2} ight)$.Substituting the values $V_0 = 10$ and $L = 10$:$i = \frac{10}{\omega 	imes 10} \sin \left( \omega t - \frac{\pi}{2} ight) = \frac{1}{\omega} \sin \left( \omega t - \frac{\pi}{2} ight)$.

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

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