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The frequency for which a $5 \mu \mathrm{F}$ capacitor has a reactance of $\frac{1}{1000} \Omega$ is given by
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Verified Answer
The correct answer is:
$\frac{100}{\pi} \mathrm{MHz}$
Concept: reactance of capacitor is
$X_c=\frac{1}{\omega c}=\frac{1}{2 \pi f C}$
Given $X_c=10^{-3} \Omega, C=5 \times 10^{-6} \mathrm{~F}$
$\begin{aligned}
& \therefore 10^{-3} \Omega=\frac{1}{2 \pi \mathrm{f}\left(5 \times 10^{-6}\right)} \\
& \Rightarrow \mathrm{f}=\frac{1}{10 \pi \times 10^{-9}} \mathrm{~Hz}=\frac{10^8}{\pi} \mathrm{Hz}=\frac{100}{\pi} \mathrm{MHz}
\end{aligned}$
$X_c=\frac{1}{\omega c}=\frac{1}{2 \pi f C}$
Given $X_c=10^{-3} \Omega, C=5 \times 10^{-6} \mathrm{~F}$
$\begin{aligned}
& \therefore 10^{-3} \Omega=\frac{1}{2 \pi \mathrm{f}\left(5 \times 10^{-6}\right)} \\
& \Rightarrow \mathrm{f}=\frac{1}{10 \pi \times 10^{-9}} \mathrm{~Hz}=\frac{10^8}{\pi} \mathrm{Hz}=\frac{100}{\pi} \mathrm{MHz}
\end{aligned}$
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