A series LCR circuit with L=4.0H, C=100µF, R=60Ω connected to a variable frequency 240V source. Calculate: (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.<p></p>

Given -L=4.0H,C=100µF=100×10⁻⁶F,R=60Ω,Vrms=240V
(i) for an L-C-R circuit to be in resonance ,angular frequency ω=1/√LC=1/√4.0×100×10⁻⁶=50rad/s
(ii) at resonant frequency , impedence =resistance ,Z=R
therefore current Irms=Vrms/Z=Vrms/R
or Irms=240/60=4A
(III) the rms potential drop across inductor=IrmsωL=4×50×4.0=800V

Eduprobe

Question:

A series LCR circuit with L=4.0H, C=100µF, R=60Ω connected to a variable frequency 240V source. Calculate: (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.

Solution: