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

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(D) The circuit is a series $LCR$ circuit with $R = 6 \, \Omega + 4 \, \Omega = 10 \, \Omega$,$L = 5 \, 	ext{mH} = 5 	imes 10^{-3} \, 	ext{H}$,and

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$5.6 \, V, 1.4 \, A$

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(D) The circuit is a series $LCR$ circuit with $R = 6 \, \Omega + 4 \, \Omega = 10 \, \Omega$,$L = 5 \, 	ext{mH} = 5 	imes 10^{-3} \, 	ext{H}$,and $C = 50 \, \mu	ext{F} = 50 	imes 10^{-6} \, 	ext{F}$.Given $\omega = 2000 \, 	ext{rad/s}$.Inductive reactance $X_L = \omega L = 2000 	imes 5 	imes 10^{-3} = 10 \, \Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2000 	imes 50 	imes 10^{-6}} = \frac{1}{0.1} = 10 \, \Omega$.Total impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{10^2 + (10 - 10)^2} = 10 \, \Omega$.Peak current $I_0 = \frac{V_0}{Z} = \frac{20}{10} = 2 \, 	ext{A}$.$RMS$ current $I_{	ext{rms}} = \frac{I_0}{\sqrt{2}} = \frac{2}{1.414} \approx 1.414 \, 	ext{A} \approx 1.4 \, 	ext{A}$.The voltmeter is connected across the $LCR$ combination (inductor,resistor,and capacitor). The voltage across this combination is $V_{LCR} = I_{	ext{rms}} 	imes Z_{LCR} = I_{	ext{rms}} 	imes \sqrt{R_{LCR}^2 + (X_L - X_C)^2}$.Since $X_L = X_C$,$Z_{LCR} = R_{LCR} = 4 \, \Omega$.$V_{	ext{rms}} = 1.414 	imes 4 = 5.656 \, 	ext{V} \approx 5.6 \, 	ext{V}$.

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

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