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Question:

The figure shows a series LCR circuit with L=54H, C=80µF, R=40Ω 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.

Solution:

Given -L=54H,C=80µF=80×10⁻⁶F,R=40Ω,Vrms=240V
(i) for an L-C-R circuit to be in resonance ,angular frequency ω=1√LC=1√54×80×10⁻⁶=5.21rad/s
(ii) at resonant frequency , impedence =resistance ,Z=R
therefore current Irms=VrmsZ=VrmsR or Irms=24040=6A
(III) the rms potential drop across inductor=IrmsωL=6×5.21×54=1688V