An L-C-R series circuit is connected to an AC | vedclass

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(D) For an $L-C-R$ series circuit,the phase difference $\phi$ is given by $	an \phi = \frac{|X_C - X_L|}{R}$.Given $\phi = 45^{\circ}$ and $R = 100 \

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$1.2$

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(D) For an $L-C-R$ series circuit,the phase difference $\phi$ is given by $	an \phi = \frac{|X_C - X_L|}{R}$.Given $\phi = 45^{\circ}$ and $R = 100 \ \Omega$,we have $	an 45^{\circ} = 1 = \frac{|X_C - X_L|}{R}$,which implies $|X_C - X_L| = R = 100 \ \Omega$.The impedance $Z$ of the circuit is $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{R^2 + R^2} = R\sqrt{2} = 100\sqrt{2} \ \Omega$.The peak voltage $V_m = 240 \ V$. The $rms$ voltage is $V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{240}{\sqrt{2}} \ V$.The $rms$ current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{240 / \sqrt{2}}{100\sqrt{2}} = \frac{240}{100 	imes 2} = \frac{240}{200} = 1.2 \ A$.

An $L-C-R$ series circuit is connected to an $AC$ source of peak voltage $240 \ V$. The phase difference between voltage and current of this circuit is $45^{\circ}$ and the resistance is $100 \ \Omega$. The $rms$ value of current through the circuit is . . . . . . . (in $A$)

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