Question

A capacitor of capacitance $C$ and reactance $X$ is connected to an $\text{AC}$ source of frequency $f$. If the capacitance and frequency becomes double, then the reactance becomes $\frac{X}{n}$. The value of $n$ is :

Answer 1

The reactance is given by:

$X=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

When frequency and capacitance is doubled:

$X^{\prime }=\frac{1}{2\pi \left(2f\right)\left(2C\right)}=\frac{1}{4\left(2\pi fC\right)}$

$\therefore X^{\prime }=\frac{X}{4}$

Hence, $n=4$.