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A resistor of $100 \Omega$, an inductor of $\frac{25}{\pi^2} \mathrm{mH}$ and a capacitor of $0.1 \mu \mathrm{F}$ are connected in series to an ac source. The impedance of the circuit is minimum for a frequency of
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The correct answer is:
$10 \mathrm{kHz}$
The impedance of the circuit is minimum for a frequency
$\begin{aligned}
& \omega=\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{\frac{25}{\pi^2} \times 10^{-3} \times 0.1 \times 10^{-6}}} \\
& 2 \pi \mathrm{f}=\frac{\pi}{5} \times 10^5 \Rightarrow \mathrm{f}=10 \mathrm{KHz}
\end{aligned}$
$\begin{aligned}
& \omega=\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{\frac{25}{\pi^2} \times 10^{-3} \times 0.1 \times 10^{-6}}} \\
& 2 \pi \mathrm{f}=\frac{\pi}{5} \times 10^5 \Rightarrow \mathrm{f}=10 \mathrm{KHz}
\end{aligned}$
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