Question

A series LCR circuit with `L = 0.12 H, C = 480 nF, R = 23 Omega` is connected to a 230 V variable frequency supply.
What is the source frequency for which current amplitude is maximum. Obtain this maximum vlaue.

Answer 1

Inductance `L = 0.12 H`,
Capacitance `C = 480 nF = 480xx10^(-9)F`
Resistance `R = 23 Omega`
The rms value of voltage Vrms = 230 V
Current is maimum at resonance.
At resonance impedance `Z = R = 23 Omega`
The rms value of current
`I_("rms")=(V_("rms"))/(Z)=(230)/(23)=10A`.
The maximum value of current `I_(0)=sqrt(2)`
`I_("rms")=1.414xx10=14.14A`.
At natural frequency, the current amplitude is maximum.
`omega = (1)/(sqrt(LC))(1)/(2pi sqrt(0.12xx480xx10^(-9)))=4166.6`
= 4167 rad/s.