Question

The phase-difference between current and voltage in only capacitive alternating current circuit is

• A. 0°
• B. 90°
• C. 180°
• D. 45°

Hint:

A useful mnemonic to remember the phase relationship in AC circuits is "CIVIL": \begin{itemize} \item In a \textbf{C}apacitor (C), \textbf{I} (current) leads \textbf{V} (voltage). \item In an \textbf{I}nductor (L), \textbf{V} (voltage) leads \textbf{I} (current). \end{itemize} This helps you quickly recall that for a capacitor, the phase difference is 90° with current ahead.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In an alternating current (AC) circuit containing only a capacitor, the flow of current is related to the rate of change of voltage across the capacitor. This relationship causes a phase difference between the current and voltage waveforms.
Step 2: Detailed Explanation:
Let the alternating voltage applied to the circuit be \(V = V_m \sin(\omega t)\).
The charge on the capacitor at any instant is \(q = CV = CV_m \sin(\omega t)\).
The current in the circuit is the rate of flow of charge, so we differentiate the charge with respect to time:
\[ I = \frac{dq}{dt} = \frac{d}{dt} (CV_m \sin(\omega t)) \] \[ I = CV_m \omega \cos(\omega t) \] To compare the phase of current and voltage, we express the current in terms of a sine function:
Since \(\cos(\theta) = \sin(\theta + \frac{\pi}{2})\), we can write:
\[ I = I_m \sin(\omega t + \frac{\pi}{2}) \] Where \(I_m = V_m \omega C\) is the peak current.
Comparing the phase of voltage, \(\phi_V = \omega t\), with the phase of current, \(\phi_I = \omega t + \frac{\pi}{2}\), we find the phase difference:
\[ \Delta \phi = \phi_I - \phi_V = (\omega t + \frac{\pi}{2}) - \omega t = \frac{\pi}{2} \] A phase difference of \(\frac{\pi}{2}\) radians is equal to 90°. The positive sign indicates that the current leads the voltage by 90°.
Step 3: Final Answer:
The phase-difference between current and voltage in a purely capacitive AC circuit is 90° (or \(\pi/2\) radians), with the current leading the voltage. Therefore, option (B) is correct.