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

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(B) Given: $V = 20 \cos(2000t) \, V$. Peak voltage $V_m = 20 \, V$. Angular frequency $\omega = 2000 \, rad/s$.Inductive reactance: $X_L = \omega L =

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$0 \, V, 1.4 \, A$

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(B) Given: $V = 20 \cos(2000t) \, V$. Peak voltage $V_m = 20 \, V$. Angular frequency $\omega = 2000 \, rad/s$.Inductive reactance: $X_L = \omega L = 2000 	imes 5 	imes 10^{-3} = 10 \, \Omega$.Capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{2000 	imes 50 	imes 10^{-6}} = \frac{1}{0.1} = 10 \, \Omega$.Since $X_L = X_C$,the circuit is in resonance. The total impedance $Z$ of the circuit is the sum of the resistances: $Z = R_{total} = 6 \, \Omega + 4 \, \Omega = 10 \, \Omega$.The $RMS$ voltage is $V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, V$.Ammeter reading $(I_{rms})$: $I_{rms} = \frac{V_{rms}}{Z} = \frac{10\sqrt{2}}{10} = \sqrt{2} \approx 1.414 \, A \approx 1.4 \, A$.Voltmeter reading: The voltmeter is connected across the resistor $(4 \, \Omega)$ and the capacitor $(X_C = 10 \, \Omega)$. The voltage across the series combination of $R=4 \, \Omega$ and $C=10 \, \Omega$ is $V_V = I_{rms} 	imes \sqrt{R^2 + X_C^2} = 1.414 	imes \sqrt{4^2 + 10^2} = 1.414 	imes \sqrt{116} \approx 1.414 	imes 10.77 \approx 15.2 \, V$. Wait,re-evaluating the circuit diagram: The voltmeter is across the inductor $(L)$ and capacitor $(C)$. Since $X_L = X_C$,the net reactance is zero. The voltage across the $LC$ combination is $V_{LC} = I_{rms} |X_L - X_C| = 0 \, V$. Thus,the voltmeter reading is $0 \, V$.

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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