Question

A capacitor of capacitance \( C \) has reactance \( X \) in an AC circuit. If the capacitance and the frequency of the applied voltage are doubled, the new reactance will become:

• A. 4X
• B. 2X
• C. \(\frac{X{2\)
• D. \(\frac{X{4\)

Hint:

The capacitive reactance is inversely proportional to both capacitance and frequency. Doubling both will decrease the reactance by a factor of 4.

Answer 1

The Correct Option is D

The capacitive reactance, denoted as \( X \), is calculated using the formula: \[ X = \frac{1}{2 \pi f C} \] In this formula, \( f \) represents the frequency of the applied voltage, and \( C \) signifies the capacitance of the capacitor. When both the capacitance and the frequency are doubled, the new reactance, \( X' \), is determined as follows: \[ X' = \frac{1}{2 \pi (2f) (2C)} = \frac{1}{2 \pi f C \cdot 4} \] This simplifies to: \[ X' = \frac{X}{4} \] Consequently, the new reactance is \( \frac{X}{4} \).