MHT CET · Physics · Alternating Current
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
- A \(\frac{150}{\pi}\)
- B \(\frac{200}{\pi}\)
- C \(\frac{250}{\pi}\)
- D \(\frac{400}{\pi}\)
Answer & Solution
Correct Answer
(C) \(\frac{250}{\pi}\)
Step-by-step Solution
Detailed explanation
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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