When an a.c. source is connected across a pur | vedclass

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(B) When an alternating current $(a.c.)$ source is connected across a pure capacitor, the voltage $(e_c)$ across the capacitor lags behind the current

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(B) When an alternating current $(a.c.)$ source is connected across a pure capacitor, the voltage $(e_c)$ across the capacitor lags behind the current $(I_c)$ by a phase angle of $90^{\circ}$ (or $\pi/2$ radians).Alternatively, we can say that the current leads the voltage by $90^{\circ}$.Mathematically, if the voltage is $e_c = E_0 \sin(\omega t)$, then the current is $I_c = I_0 \sin(\omega t + \pi/2)$.In a phasor diagram, if the voltage vector is along the positive $x$-axis, the current vector points along the positive $y$-axis, indicating a $90^{\circ}$ lead.Therefore, option $B$ correctly represents this phase relationship.

When an $a.c.$ source is connected across a pure capacitor, the correct phase relation between current and $emf$ is:

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