The voltage across a pure inductor is represe | vedclass

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(C) For a pure inductor,the current $i$ lags behind the voltage $V$ by a phase angle of $\pi / 2$. If the voltage is represented as $V = V_m \sin(\ome

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(C) For a pure inductor,the current $i$ lags behind the voltage $V$ by a phase angle of $\pi / 2$. If the voltage is represented as $V = V_m \sin(\omega t)$,then the current is $i = I_m \sin(\omega t - \pi / 2) = -I_m \cos(\omega t)$. In the given diagram,the voltage $V$ is shown as a sine wave starting from $0$ and increasing. Therefore,the current $i$ must be a negative cosine wave,which starts at a negative maximum value at $t = 0$ and increases to $0$ at $t = \pi / (2\omega)$. This corresponds to the graph shown in option $C$.

The voltage across a pure inductor is represented by the following diagram. Which of the following diagrams will represent the current?

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