In a series LCR circuit,R = 200 , Omega and t | vedclass

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Question

(A) When capacitance is removed,the circuit becomes an $LR$ circuit.Given the phase angle $\phi = 30^\circ$,we have $	an \phi = \frac{\omega L}{R}$.$

Vedclass · Accepted Answer

$242$

Vedclass · Answer

(A) When capacitance is removed,the circuit becomes an $LR$ circuit.Given the phase angle $\phi = 30^\circ$,we have $	an \phi = \frac{\omega L}{R}$.$\omega L = R 	an 30^\circ = 200 	imes \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}} \, \Omega$.When the inductor is removed,the circuit becomes a $CR$ circuit.Given the phase angle $\phi = 30^\circ$,we have $	an \phi = \frac{1}{\omega C R}$.$\frac{1}{\omega C} = R 	an 30^\circ = 200 	imes \frac{1}{\sqrt{3}} = \frac{200}{\sqrt{3}} \, \Omega$.In the original $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.Substituting the values: $Z = \sqrt{200^2 + (\frac{200}{\sqrt{3}} - \frac{200}{\sqrt{3}})^2} = \sqrt{200^2 + 0} = 200 \, \Omega$.The power dissipated in an $AC$ circuit is $P = V_{rms} I_{rms} \cos \phi$,where $\cos \phi = \frac{R}{Z}$.Since $\omega L = \frac{1}{\omega C}$,the circuit is in resonance,so $\phi = 0^\circ$ and $\cos \phi = 1$.$P = \frac{V_{rms}^2}{Z^2} 	imes R = \frac{220^2}{200^2} 	imes 200 = \frac{220 	imes 220}{200} = 242 \, W$.

List-$I$	List-$II$
$A$. $AC$ generator	$I$. Detects the presence of current in the circuit
$B$. Galvanometer	$II$. Converts mechanical energy into electrical energy
$C$. Transformer	$III$. Works on the principle of resonance in $AC$ circuit
$D$. Metal detector	$IV$. Changes an alternating voltage for smaller or greater value

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