Question

A series LCR circuit with $L=4.0,C=100\mu F,R=60\Omega$ connected to a variable frequency $240V$ source as shown in figure calcualte:
(i) the angular frequency of the source which drives the circuit resonance,
(ii) the current at the resonating frequency,
(iii) the rms potential drop across the inductor at resonance.

Answer 1

Given - $L=4.0H,C=100\mu F=100\times {10}^{-6}F,R=60\Omega ,V_{rms}=240V$

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

angular frequency $\omega =1/\sqrt{LC}=1/\sqrt{4.0\times 100\times {10}^{-6}}=50rad/s$

(ii) at resonant frequency , impedence =resistance ,

$Z=R$

therefore current $I_{rms}=V_{rms}/Z=V_{rms}/R$

or $I_{rms}=240/60=4A$

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

$=4\times 15.81\times 4.0=252.96V$