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

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(D) The circuit consists of a resistor $R_1 = 6\,\Omega$ in series with an $LCR$ branch containing $L = 5\,mH$,$R_2 = 4\,\Omega$,and $C = 50\,\mu F$.G

Vedclass · Accepted Answer

$5.6\,V, 1.4\,A$

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(D) The circuit consists of a resistor $R_1 = 6\,\Omega$ in series with an $LCR$ branch containing $L = 5\,mH$,$R_2 = 4\,\Omega$,and $C = 50\,\mu F$.Given $\omega = 2000\,rad/s$.Inductive reactance $X_L = \omega L = 2000 	imes 5 	imes 10^{-3} = 10\,\Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{2000 	imes 50 	imes 10^{-6}} = 10\,\Omega$.Total resistance $R_{total} = R_1 + R_2 = 6 + 4 = 10\,\Omega$.Since $X_L = X_C$,the circuit is in resonance.Total impedance $Z = \sqrt{R_{total}^2 + (X_L - X_C)^2} = \sqrt{10^2 + 0^2} = 10\,\Omega$.Peak current $I_0 = \frac{V_0}{Z} = \frac{20}{10} = 2\,A$.$RMS$ current $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2}{1.414} \approx 1.41\,A$.The voltmeter is connected across the $LCR$ branch. The impedance of this branch is $Z_{LCR} = \sqrt{R_2^2 + (X_L - X_C)^2} = \sqrt{4^2 + 0^2} = 4\,\Omega$.The voltage across this branch is $V_{LCR} = I_{rms} 	imes Z_{LCR} = 1.41 	imes 4 = 5.64\,V \approx 5.6\,V$.

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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