The capacitive reactance of a capacitor ' $\mathrm{C}$ ' is $\mathrm{X} \Omega$. Both, the frequency of a.c.…
The capacitive reactance of a capacitor ' $\mathrm{C}$ ' is $\mathrm{X} \Omega$. Both, the frequency of a.c. supply and capacitance of the above capacitor are doubled. The new capacitive reactance will be
$\frac{X}{4} \Omega$
$\frac{\mathrm{X}}{2} \Omega$
$2 \mathrm{X} \Omega$
$4 \mathrm{X} \Omega$
Solution
Given, $\mathrm{X}_{\mathrm{C}}=\mathrm{X} \Omega$
$\Rightarrow \frac{1}{2 \pi \mathrm{fC}}=\mathrm{X} \Omega$
New Capacitance $\mathrm{C}=2 \mathrm{C}$ and new frequency
$\mathrm{f}^{\prime}=2 \mathrm{f}$
$\begin{aligned}
\therefore \quad \text { New capacitive reactance } \mathrm{X}_{\mathrm{C}}^{\prime} & =\frac{1}{2 \pi(2 \mathrm{f})(2 \mathrm{C})} \\
& =\frac{1}{(2 \pi)(4 \mathrm{fC})} \\
& =\frac{1}{4} \mathrm{X}_{\mathrm{C}} \\
& =\frac{\mathrm{X}}{4} \Omega
\end{aligned}$
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