Question

The figure shows a series LCR circuit with $L=54H,C=80\mu F,R=40\Omega$ connected to a variable frequency $240V$ source, calculate
(i) the angular frequency of the source which drives the circuit at resonance,
(ii) the current at the resonating frequency,
(iii) the rms potential drop across the inductor at resonance.

Answer 1

Given - $L=54H,C=80\mu F=80\times {10}^{-6}F,R=40\Omega ,V_{rms}=240V$

(i) for an L-C-R circuit to be in resonance ,

angular frequency $\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{54\times 80\times {10}^{-6}}}=5.21rad/s$

(ii) at resonant frequency , impedence =resistance ,

$Z=R$

therefore current $I_{rms}=\frac{V_{rms}}{Z}=\frac{V_{rms}}{R}$

or $I_{rms}=\frac{240}{40}=6A$

(III) the rms potential drop across inductor $=I_{rms}\omega L$

$=6\times 5.21\times 54=1688V$