Circuit identification

The circuit contains only a capacitor connected to an AC source: $E_g = E_{g0}\sin \omega t$

Let the capacitor voltage be $V_C$ and current be $I_C$.

Relation between current and voltage in a capacitor

For a capacitor, $i = C\frac{dv}{dt}$

If $v = V_0\sin \omega t$ then $i = C\frac{d}{dt}(V_0\sin \omega t) = \omega C V_0 \cos \omega t$

Using $\cos \omega t = \sin\left(\omega t + \frac{\pi}{2}\right)$ we get $i = I_0\sin\left(\omega t + \frac{\pi}{2}\right)$

So, current leads voltage by $90^\circ$ in a purely capacitive circuit.

Phasor relation

Since current leads voltage by $\frac{\pi}{2}$,

• $I_C$ is ahead of $V_C$ by $90^\circ$.
• Equivalently, $V_C$ lags behind $I_C$ by $90^\circ$.

Also, in this simple capacitor-generator circuit, the generator emf phasor is in phase with the capacitor voltage phasor: $E_g \equiv V_C$

Correct phasor diagram

Therefore the correct diagram must show:

• $I_C$ leading,
• $V_C$ and $E_g$ together,
• angle between current and voltage equal to $90^\circ$.

This corresponds to Option C.

Comparison with stored answer

Derived answer: C
Stored correct answer: C

Hence, they agree.