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: Think in terms of the phasor diagram.
In a pure capacitor we draw the current phasor and the voltage phasor and ask how they sit relative to each other.
Step 2: Use the capacitor reactance idea.
A capacitor opposes change in voltage, so it must first draw current before voltage builds up across it.
Step 3: Read the impedance angle.
The capacitive impedance is $Z_C = \dfrac{1}{j\omega C} = -\dfrac{j}{\omega C}$. The factor $-j$ corresponds to an angle of $-90^\circ$ for voltage relative to current.
Step 4: Translate the sign.
Voltage lagging by $90^\circ$ is the same statement as current leading by $90^\circ$.
Step 5: Confirm frequency is irrelevant here.
The given $50\ \text{Hz}$ only affects the magnitude of reactance, not the phase angle, which stays a clean $90^\circ$ for any frequency.
Step 6: Conclude.
In a pure capacitor the current leads the voltage by a quarter cycle. \[ \boxed{90^\circ\ \text{(current leads)}} \]