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(C) In a purely inductive circuit,the voltage $V$ leads the current $i$ by a phase angle of $90^o$ (or $\pi/2$ radians).If the voltage is represented

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(C) In a purely inductive circuit,the voltage $V$ leads the current $i$ by a phase angle of $90^o$ (or $\pi/2$ radians).If the voltage is represented as $V = V_m \sin(\omega t + \phi)$,then the current is represented as $i = I_m \sin(\omega t + \phi - 90^o) = -I_m \cos(\omega t + \phi)$.Looking at the provided voltage graph,the voltage is at its maximum at $t = 0$. This corresponds to a cosine function $V = V_m \cos(\omega t)$.Therefore,the current must be $i = I_m \sin(\omega t - 90^o) = -I_m \cos(\omega t)$.At $t = 0$,$i = -I_m$,which means the current graph should start from the negative peak. Among the given options,the graph that represents a negative cosine wave is option $C$.

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

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