In the given LCR AC circuit,the effective cur | vedclass

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(A) From the given circuit,the voltage is $V = 200 \sin(100t) \, V$. Comparing this with $V = V_m \sin(\omega t)$,we get $V_m = 200 \, V$ and $\omega

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$2 \, A$

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(A) From the given circuit,the voltage is $V = 200 \sin(100t) \, V$. Comparing this with $V = V_m \sin(\omega t)$,we get $V_m = 200 \, V$ and $\omega = 100 \, rad/s$.The $RMS$ voltage is $V_{	ext{rms}} = \frac{V_m}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \, V$.The circuit components are $R = 50 \, \Omega$,$L = 0.5 \, H$,and $C = 100 \, \mu F = 100 	imes 10^{-6} \, F$.Inductive reactance $X_L = \omega L = 100 	imes 0.5 = 50 \, \Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 	imes 100 	imes 10^{-6}} = \frac{1}{0.01} = 100 \, \Omega$.The impedance $Z$ of the $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.$Z = \sqrt{50^2 + (50 - 100)^2} = \sqrt{50^2 + (-50)^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \, \Omega$.The effective $(RMS)$ current is $I_{	ext{rms}} = \frac{V_{	ext{rms}}}{Z} = \frac{100\sqrt{2}}{50\sqrt{2}} = 2 \, A$.

In the given $LCR$ $AC$ circuit,the effective current flowing through the circuit will be:

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