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
- A$I=E_0 \omega C \sin (\omega t)$
- B$I=E_0 \omega C \sin \left(\omega t+\frac{\pi}{2}\right)$
- C$I=\frac{E_0}{\omega C} \sin (\omega t)$
- D$I=\frac{E_0}{\omega C} \sin \left(\omega t+\frac{\pi}{2}\right)$
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