In an AC circuit containing only a capacitor, | vedclass

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(A) In a purely capacitive $AC$ circuit,the voltage across the capacitor is given by $V = V_m \sin(\omega t)$.The current in the circuit is given by $

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Leads the potential

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(A) In a purely capacitive $AC$ circuit,the voltage across the capacitor is given by $V = V_m \sin(\omega t)$.The current in the circuit is given by $I = C \frac{dV}{dt} = C \frac{d}{dt}(V_m \sin(\omega t)) = \omega C V_m \cos(\omega t)$.This can be written as $I = I_m \sin(\omega t + \frac{\pi}{2})$,where $I_m = \omega C V_m$.Comparing the phases,the current leads the potential difference by a phase angle of $\frac{\pi}{2}$ radians (or $90^{\circ}$).Therefore,the current is forward relative to the potential.

In an $AC$ circuit containing only a capacitor,how does the current relate to the potential difference?

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