In the circuit shown in the figure,the AC sou | vedclass

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(C) The given voltage is $V = 20 \cos(2000t)$,so the peak voltage $V_0 = 20 \, V$. The $RMS$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{20}{\s

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$5.6 \, V, 1.4 \, A$

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(C) The given voltage is $V = 20 \cos(2000t)$,so the peak voltage $V_0 = 20 \, V$. The $RMS$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, V \approx 14.14 \, V$.The angular frequency is $\omega = 2000 \, rad/s$. The inductive reactance is $X_L = \omega L = 2000 	imes 5 	imes 10^{-3} = 10 \, \Omega$. The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{2000 	imes 50 	imes 10^{-6}} = 10 \, \Omega$.Since $X_L = X_C$,the series $LC$ branch is in resonance. The impedance of the $LC$ branch is $Z_{LC} = \sqrt{R_{LC}^2 + (X_L - X_C)^2} = \sqrt{4^2 + 0^2} = 4 \, \Omega$.The total impedance of the circuit is $Z = R_{total} + Z_{LC} = 6 \, \Omega + 4 \, \Omega = 10 \, \Omega$.The $RMS$ current in the circuit is $I_{rms} = \frac{V_{rms}}{Z} = \frac{10\sqrt{2}}{10} = \sqrt{2} \approx 1.414 \, A$. Thus,the ammeter reading is $1.4 \, A$.The voltmeter is connected across the $LC$ branch. The voltage across this branch is $V_{LC} = I_{rms} 	imes Z_{LC} = 1.414 	imes 4 = 5.656 \, V \approx 5.6 \, V$.

In the circuit shown in the figure,the $AC$ source provides a voltage $V = 20 \cos(2000t)$. Neglecting source resistance,the voltmeter and ammeter readings will be:

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