The reactance of capacitor at $50 \mathrm{~Hz}$ is $5 \Omega$. If the frequency is increased to $100…
The reactance of capacitor at $50 \mathrm{~Hz}$ is $5 \Omega$. If the frequency is increased to $100 \mathrm{~Hz}$, the new reactance is
- $5 \Omega$
- $2.5 \Omega$
- $10 \Omega$
- $125 \Omega$
Solution
$\begin{aligned}
\mathrm{X}_{\mathrm{C}} & =\frac{1}{2 \pi \mathrm{fc}} \\
\therefore \quad \mathrm{C} & =\frac{1}{2 \pi \mathrm{f}(5)} \quad \ldots .\left(\because \mathrm{X}_{\mathrm{C}}=5 \Omega\right)
\end{aligned}$
New reactance,
$\mathrm{X}_{\mathrm{C}}^{\prime}=\frac{1}{2 \pi \mathrm{f}^{\prime} \mathrm{C}}=\frac{1}{2 \pi(2 \mathrm{f}) \mathrm{C}}=\frac{\mathrm{X}_{\mathrm{C}}}{2}=\frac{1}{2} \times 5=2.5 \Omega$
^
Asked in: MHT CET 2023 (09 May Shift 2)
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