In the circuit shown below, the inductance $L$ is connected to an ac source. The current flowing in

Question

In the circuit shown below, the inductance $L$ is connected to an ac source. The current flowing in the circuit is $I=I_0 \sin \omega t$. The velocity drop ( $V_L$ ) across $L$ is

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d)Given, $I=I_0 \sin \omega t$ The voltage drop ( $V_L$ ) across inductor is given by, \ $ \begin{aligned} V_L & =L \frac{d I}{d t}=L \frac{d}{d t}\left(I_0 \sin \omega tight) \ V_L & =\omega L I_0 \cos \omega t \end{aligned} $