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(B) In a pure capacitive circuit, the current leads the voltage by a phase angle of $\pi/2$ radians.The instantaneous voltage is given by $E = E_0 \si

Vedclass · Accepted Answer

$I = E_0 \omega C \sin(\omega t + \pi/2)$

Vedclass · Answer

(B) In a pure capacitive circuit, the current leads the voltage by a phase angle of $\pi/2$ radians.The instantaneous voltage is given by $E = E_0 \sin(\omega t)$.The capacitive reactance is $X_C = \frac{1}{\omega C}$.The instantaneous current $I$ is given by $I = \frac{E}{X_C} = \frac{E_0 \sin(\omega t)}{1/(\omega C)} = E_0 \omega C \sin(\omega t)$.Since the current leads the voltage by $\pi/2$, we add $\pi/2$ to the phase of the voltage.Therefore, the instantaneous current is $I = E_0 \omega C \sin(\omega t + \pi/2)$.

In an a.c. circuit with pure capacitance $C$ and a.c. source $E = E_0 \sin \omega t$, the equation of instantaneous current is given by

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