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Q7.11 (b) Figure shows a series LCR circuit connected to a variable frequency $230\: V$ source. Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. $(\text{where,}\ L=5H, C=80\mu F \ \text{and} R=40\Omega)$

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

Now, at resonance condition, impedance is minimum which means the 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, the amplitude of the current at resonance is 8.13 A.

Posted by

Pankaj Sanodiya

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