Q7.11 (b) Figure  shows a series LCR circuit connected to a variable frequency 230\: V source.

                L=5.0HC=80\mu F, R=40\Omega.

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

            

Answers (1)

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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