A series LCR circuit is connected to a 220 V variable frequency (a.c) supply. If \( L = 20\,\mathrm{mH},\ C = \left(\displaystyle\frac{800}{\pi^2}\right)\,\mu\mathrm{F} \) and \( R = 110\,\Omega \). The frequency of the source, for which the average power absorbed by the circuit is maximum will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 125 Hz
The average power in a series LCR circuit is greatest at resonance, where \( X_L = X_C \), the impedance is the least (\( Z = R \)) and the current is the largest. The resonant frequency depends only on \( L \) and \( C \); the voltage and \( R \) do not affect it.
\( \displaystyle \nu_0 = \frac{1}{2\pi\sqrt{LC}} \)
\( L = 20\,\mathrm{mH} = 20\times10^{-3}\,\mathrm{H} \)
\( \displaystyle C = \frac{800}{\pi^{2}}\,\mu\mathrm{F} = \frac{800\times10^{-6}}{\pi^{2}}\,\mathrm{F} \)
\( \displaystyle LC = \frac{20\times10^{-3} \times 800\times10^{-6}}{\pi^{2}} = \frac{16\times10^{-6}}{\pi^{2}} \;\Rightarrow\; \sqrt{LC} = \frac{4\times10^{-3}}{\pi} \)
\( \displaystyle \nu_0 = \frac{1}{2\pi \times \displaystyle\frac{4\times10^{-3}}{\pi}} = \frac{1}{8\times10^{-3}} = 125\,\mathrm{Hz} \)