Question

An AC source of frequency \( 50 \, \text{Hz} \) is connected to a pure capacitor. The phase difference between voltage and current is:

• A. \( 0^\circ \)
• B. \( 45^\circ \)
• C. \( 90^\circ \) (current leads)
• D. \( 90^\circ \) (voltage leads)

Hint:

Remember the classic mnemonic "ELI the ICE man":
- In an Inductor (L), Voltage (E) leads Current (I).
- In a Capacitor (C), Current (I) leads Voltage (E).
For pure components, the lead/lag angle is always \( 90^\circ \).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the phase difference between current and voltage in an AC circuit consisting of an AC source and a pure capacitor.

Step 2: Detailed Explanation:
In a purely capacitive AC circuit, the relationship between charge \( q \) and voltage \( v \) is given by:
\[ q = C v \] If the alternating voltage is:
\[ v = V_0 \sin(\omega t) \] Then the current \( i \) in the circuit is:
\[ i = \frac{dq}{dt} = C \frac{dv}{dt} = C \frac{d}{dt} [V_0 \sin(\omega t)] = C V_0 \omega \cos(\omega t) \] Using trigonometric identities:
\[ i = I_0 \sin\left(\omega t + \frac{\pi}{2}\right) \] Comparing the equations for \( v \) and \( i \), we see that the phase of current is ahead of the phase of voltage by \( \frac{\pi}{2} \) radians or \( 90^\circ \).
Therefore, the current leads the voltage by \( 90^\circ \).

Step 3: Final Answer:
(C) \( 90^\circ \) (current leads)