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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.

Q7.20 (a)  A series LCR circuit with $L=0.12H$, $C=480nF$$R=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.

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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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