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

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(D) The circuit is a series $LCR$ circuit connected to an $AC$ source. The total resistance $R_{total} = 6 \Omega + 4 \Omega = 10 \Omega$.The angular

Vedclass · Accepted Answer

$5.6 ~V, 1.4 ~A$

Vedclass · Answer

(D) The circuit is a series $LCR$ circuit connected to an $AC$ source. The total resistance $R_{total} = 6 \Omega + 4 \Omega = 10 \Omega$.The angular frequency $\omega = 2000 	ext{ rad/s}$.Inductive reactance $X_L = \omega L = 2000 	imes 5 	imes 10^{-3} \Omega = 10 \Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2000 	imes 50 	imes 10^{-6}} \Omega = \frac{1}{0.1} \Omega = 10 \Omega$.Since $X_L = X_C$,the circuit is in resonance.The impedance $Z = \sqrt{R_{total}^2 + (X_L - X_C)^2} = \sqrt{10^2 + (10 - 10)^2} = 10 \Omega$.The peak voltage $V_0 = 20 	ext{ V}$,so the $RMS$ voltage $V_{rms} = \frac{20}{\sqrt{2}} \approx 14.14 	ext{ V}$.The $RMS$ current $I_{rms} = \frac{V_{rms}}{Z} = \frac{14.14}{10} = 1.414 	ext{ A} \approx 1.4 	ext{ A}$.The voltmeter is connected across the $LCR$ combination. The voltage across the $LCR$ combination is $V_{LCR} = I_{rms} 	imes Z_{LCR}$,where $Z_{LCR} = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{4^2 + (10 - 10)^2} = 4 \Omega$.Thus,$V_{voltmeter} = 1.414 	imes 4 = 5.656 	ext{ V} \approx 5.6 	ext{ V}$.Therefore,the readings are $5.6 	ext{ V}$ and $1.4 	ext{ A}$.

List$-I$	List$-II$
$(P)$ Passing $H _2 S$ in the presence of $NH _4 OH$	$(1) \ Cu ^{2+}$
$(Q) \ \left( NH _4\right)_2 CO _3$ in the presence of $NH _4 OH$	$(2) \ Al ^{3+}$
$(R) \ NH _4 OH$ in the presence of $NH _4 Cl$	$(3) \ Mn ^{2+}$
$(S)$ Passing $H _2 S$ in the presence of dilute $\text{HCl}$	$(4) \ Ba ^{2+}$
	$(5) \ Mg ^{2+}$

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

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