Question

A series L-C-R circuit is connected to a 220 V variable frequency AC supply. If L = 20 mH, C = $\left(800/{\pi }^2\right)\mu F$ and R = 110 $\Omega$.
(i) Find the frequency of the source for which average power absorbed by the circuit is maximum.
(ii) Calculate the value of maximum current amplitude.

Answer 1

Maximum power is absorbed at resonant frequency.

Resonant frequency $f_r=\frac{1}{2\pi \sqrt{LC}}$

Or $f_r=\frac{1}{2\pi \sqrt{20\times {10}^{-3}\times 800\times {10}^{-6}/{\pi }^2}}=125Hz$

At resonance, impedance of the circuit $Z=R=110\Omega$

Rms value of voltage $V_{rms}=220V$

Peak value of voltage $V_o=\frac{V_{rms}}{\sqrt{2}}=110\sqrt{2}V$

So, maximum current amplitude $I_{max}=\frac{V_o}{Z}=\frac{110\sqrt{2}}{110}=\sqrt{2}A$