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