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(A) In a purely capacitive circuit,the current $I_{C}$ leads the voltage $V_{C}$ across the capacitor by a phase angle of $\frac{\pi}{2}$ radians $(90

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(A) In a purely capacitive circuit,the current $I_{C}$ leads the voltage $V_{C}$ across the capacitor by a phase angle of $\frac{\pi}{2}$ radians $(90^{\circ})$. Mathematically,if $V_{C} = V_{0} \sin \omega t$,then $I_{C} = I_{0} \sin(\omega t + \frac{\pi}{2})$. This means that the phasor representing current $I_{C}$ is rotated $90^{\circ}$ counter-clockwise relative to the phasor representing voltage $V_{C}$. Looking at the provided options,the correct phasor diagram is the one where $I_{C}$ is $90^{\circ}$ ahead of $V_{C}$ in the counter-clockwise direction.

In a circuit consisting of a capacitance and a generator with alternating emf $E_{g}=E_{g0} \sin \omega t$,$V_{C}$ and $I_{C}$ are the voltage and current. The correct phasor diagram for such a circuit is:

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