When an A.C. source is connected across a pur | vedclass

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(B) For a pure capacitor connected to an $A$.$C$. source,the voltage is given by $e_c = V_0 \sin(\omega t)$.The charge on the capacitor is $q = C e_c

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$(B)$

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(B) For a pure capacitor connected to an $A$.$C$. source,the voltage is given by $e_c = V_0 \sin(\omega t)$.The charge on the capacitor is $q = C e_c = C V_0 \sin(\omega t)$.The current in the circuit is $i_c = \frac{dq}{dt} = \frac{d}{dt} [C V_0 \sin(\omega t)] = \omega C V_0 \cos(\omega t) = \omega C V_0 \sin(\omega t + \frac{\pi}{2})$.This shows that the current $i_c$ leads the voltage $e_c$ by a phase angle of $\frac{\pi}{2}$ radians.In the given phasor diagrams,the vector representing $i_c$ should be $90^{\circ}$ ahead of the vector representing $e_c$ in the counter-clockwise direction.Looking at the options,in diagram $(B)$,the current vector $i_c$ is $90^{\circ}$ ahead of the voltage vector $e_c$.Therefore,the correct option is $(B)$.

When an $A$.$C$. source is connected across a pure capacitor,the correct phase relation between current $(i_c)$ and voltage $(e_c)$ is shown in the figure.

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