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:
Show Hint
- 4X
- 2X
- \(\frac{X{2\)
- \(\frac{X{4\)
The Correct Option is D
Solution and Explanation
- \( f \) is the frequency of the applied voltage,
- \( C \) is the capacitance of the capacitor. Now, if the capacitance and the frequency are doubled, the new reactance \( X' \) will be: \[ X' = \frac{1}{2 \pi (2f) (2C)} = \frac{1}{2 \pi f C \cdot 4} \] \[ X' = \frac{X}{4} \] Therefore, the new reactance will be \( \frac{X}{4} \).
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