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.

A series LCR circuit with `L = 0.12 H, C = 480 nF, R = 23 Omega` is co

1.	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» 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.

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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.

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