In the circuit diagram shown,X_C = 100 , Omeg | vedclass

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(D) The circuit consists of a resistor $R$ in parallel with a series combination of an inductor $L$ and a capacitor $C$.The peak voltage is $V_0 = 200

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$2 \, 	ext{A}$

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(D) The circuit consists of a resistor $R$ in parallel with a series combination of an inductor $L$ and a capacitor $C$.The peak voltage is $V_0 = 200 \, 	ext{V}$. The $RMS$ voltage is $V_{	ext{rms}} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \, 	ext{V}$.The current through the resistor is $I_R = \frac{V_{	ext{rms}}}{R} = \frac{100\sqrt{2}}{100} = \sqrt{2} \, 	ext{A}$.The current through the $LC$ branch is $I_{LC} = \frac{V_{	ext{rms}}}{|X_L - X_C|} = \frac{100\sqrt{2}}{|200 - 100|} = \frac{100\sqrt{2}}{100} = \sqrt{2} \, 	ext{A}$.Since the current through the resistor is in phase with the voltage and the current through the $LC$ branch is $90^\circ$ out of phase with the voltage,the total $RMS$ current is given by $I_{	ext{rms}} = \sqrt{I_R^2 + I_{LC}^2}$.$I_{	ext{rms}} = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \, 	ext{A}$.

In the circuit diagram shown,$X_C = 100 \, \Omega$,$X_L = 200 \, \Omega$,and $R = 100 \, \Omega$. The source voltage is $V = 200 \sin(\omega t) \, \text{V}$. The effective $(RMS)$ current through the source is:

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