# Q7.11 (b) Figure  shows a series LCR circuit connected to a variable frequency $230\: V$ source.                $L=5.0H$,  $C=80\mu F$, $R=40\Omega$.(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

P Pankaj Sanodiya

Given,

Variable frequency supply voltage $V$ = 230V

Inductance $L=5.0H$

Capacitance $C=80\mu F=80*10^{-6}F$

Resistance $R=40\Omega$

Now,

The impedance of the circuit is

$Z=\sqrt{(wL-\frac{1}{wC})^2+R^2}$

at Resonance Condition

:$wL=\frac{1}{wC}$

$Z=R=40\Omega$

Hence, Impedance at resonance is 40$\Omega$.

Now, at resonance condition, impedance is minimum which means current is maximum which will happen when we have a peak voltage, so

Current in the Resonance circuit is Given by

$I_{resonance}=\frac{V_{peak}}{Z}=\frac{\sqrt{2}* 230}{40}=8.13A$

Hence amplitude of the current at resonance is 8.13A.

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