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A capacitor of capacity C has reactance X. If capacitance and frequency become double then reactance will be : -
A
4X
B
$\dfrac{X}{2}$
C
$\dfrac{X}{4}$
D
2X
Detailed Solution
Capacitive reactance: $X_C = \dfrac{1}{2\pi fC}$, so $X_C \propto \dfrac{1}{fC}$.
When both f and C are doubled: $X' = \dfrac{1}{2\pi(2f)(2C)} = \dfrac{1}{4} \times \dfrac{1}{2\pi fC}$
$X' = \dfrac{X}{4}$
When both f and C are doubled: $X' = \dfrac{1}{2\pi(2f)(2C)} = \dfrac{1}{4} \times \dfrac{1}{2\pi fC}$
$X' = \dfrac{X}{4}$