Question

A source of alternating emf $e = \varepsilon_0 \sin(\omega t)$ is connected to a pure capacitor. Then the instantaneous current in the circuit is:

• A. $I = I_0 \sin\left(\omega t + \frac{\pi}{2}\right)$
• B. $I = \sqrt{2}I_0 \sin\left(\omega t + \frac{\pi}{2}\right)$
• C. $I = I_0 \sin(\omega t)$
• D. $I = I_0 \sin\left(\omega t - \frac{\pi}{2}\right)$

Hint:

Remember the phrase "ICE" to keep your AC phases clear: In a Capacitor, I (current) comes before E (emf/voltage). Therefore, the phase inside the sine term must have a $+\frac{\pi}{2}$ shift!

Answer 1

The Correct Option is A

Concept: In a purely capacitive alternating current (AC) circuit, the electrical charge stored on the plates at any instant is $q = C \cdot e$. The current flowing through the circuit is the time rate of flow of this charge: \[ I = \frac{dq}{dt} \]

Step 1: Substitute emf and differentiate with respect to time.
Given $e = \varepsilon_0 \sin(\omega t)$: \[ q = C\varepsilon_0 \sin(\omega t) \] Differentiating with respect to $t$: \[ I = \frac{d}{dt}[C\varepsilon_0 \sin(\omega t)] = C\varepsilon_0 \cdot \omega \cos(\omega t) \] \[ I = \left(\frac{\varepsilon_0}{1/\omega C}\right) \cos(\omega t) = I_0 \cos(\omega t) \]

Step 2: Convert the cosine function into standard phase format.
Using the trigonometric identity $\cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right)$: \[ I = I_0 \sin\left(\omega t + \frac{\pi}{2}\right) \] This mathematically demonstrates that in a purely capacitive circuit, the alternating current leads the alternating voltage by a phase angle of $90^\circ$ ($\frac{\pi}{2}$).