Q7.20 (a)  A series LCR circuit with L=0.12H, C=480nFR=23\Omega  is connected to a 230V variable frequency supply.

What is the source frequency for which current amplitude is maximum. Obtain this maximum value.

 

Answers (1)

The inductance of the inductor L=0.12H

The capacitance of the capacitor C=480n F

The resistance of the resistor R=23\Omega

Voltage supply V = 230V

Frequency of voltage supply f=50Hz

As we know,

the current amplitude is maximum at the natural frequency of oscillation, which is 

w_{natural}=\sqrt\frac{1}{LC}=\frac{1}{\sqrt{0.12*480*10^{-9}})}=4166.67rad/sec

Also, at this frequency,

Z=R=23

SO,

The maximum current in the circuit :

I_{max}=\frac{V_{max}}{Z}=\frac{V_{max}}{R}=\frac{\sqrt{2*}230}{23}=14.14A

Hence maximum current is 14.14A.

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