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A circuit of negligible resistance has an inductor of $0.16 \mathrm{H}$ and capacitor of $25 \mu \mathrm{F}$ connected in series with an alternating voltage source. The resonating frequency of the circuit is
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The correct answer is:
$\frac{250}{\pi}$
At resonance inductive and capacitive reactance's are equal:
$\begin{aligned} & X_C=\omega C=X_L=\frac{1}{\omega L} \\ & \Rightarrow \omega=\frac{1}{\sqrt{L C}}=2 \pi f\end{aligned}$
Given, $L=0.16 \mathrm{H} \& C=25 \times 10^{-6} \mathrm{~F}$
$\begin{aligned} & \Rightarrow \omega=\frac{1}{\sqrt{4 \times 10^{-6}}} \mathrm{~s}^{-1}=\frac{1000}{2} \mathrm{~s}^{-1}=2 \pi f \\ & \Rightarrow f=\left(\frac{250}{\pi}\right) \mathrm{sec}^{-1}\end{aligned}$
$\begin{aligned} & X_C=\omega C=X_L=\frac{1}{\omega L} \\ & \Rightarrow \omega=\frac{1}{\sqrt{L C}}=2 \pi f\end{aligned}$
Given, $L=0.16 \mathrm{H} \& C=25 \times 10^{-6} \mathrm{~F}$
$\begin{aligned} & \Rightarrow \omega=\frac{1}{\sqrt{4 \times 10^{-6}}} \mathrm{~s}^{-1}=\frac{1000}{2} \mathrm{~s}^{-1}=2 \pi f \\ & \Rightarrow f=\left(\frac{250}{\pi}\right) \mathrm{sec}^{-1}\end{aligned}$
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