There is a 100,Omega resistance in an L-C-R A | vedclass

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(C) Given resistance $R = 100\,\Omega$,voltage $V = 200\,V$.When the capacitor is removed,the circuit becomes an $L-R$ circuit. The phase angle $\phi

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(C) Given resistance $R = 100\,\Omega$,voltage $V = 200\,V$.When the capacitor is removed,the circuit becomes an $L-R$ circuit. The phase angle $\phi = 60^o$ is given by $	an 60^o = \frac{X_L}{R}$.$\Rightarrow X_L = R 	an 60^o = 100 	imes \sqrt{3} = 100\sqrt{3}\,\Omega$.When the inductor is removed,the circuit becomes a $C-R$ circuit. The phase angle $\phi = 60^o$ is given by $	an 60^o = \frac{X_C}{R}$.$\Rightarrow X_C = R 	an 60^o = 100 	imes \sqrt{3} = 100\sqrt{3}\,\Omega$.Since $X_L = X_C$,the circuit is in resonance.In resonance,the impedance $Z = R = 100\,\Omega$.The current $I = \frac{V}{Z} = \frac{200}{100} = 2\,A$.

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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