An inductor 20 , mH,a capacitor 100 , mu F an | vedclass

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(A) The given values are $L = 20 \, mH = 20 	imes 10^{-3} \, H$,$C = 100 \, \mu F = 100 	imes 10^{-6} \, F$,$R = 50 \, \Omega$,and $V = 10 \sin(314

Vedclass · Accepted Answer

$0.79$

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(A) The given values are $L = 20 \, mH = 20 	imes 10^{-3} \, H$,$C = 100 \, \mu F = 100 	imes 10^{-6} \, F$,$R = 50 \, \Omega$,and $V = 10 \sin(314 \, t)$.Comparing with $V = V_0 \sin(\omega t)$,we get $V_0 = 10 \, V$ and $\omega = 314 \, rad/s$.The inductive reactance is $X_L = \omega L = 314 	imes 20 	imes 10^{-3} = 6.28 \, \Omega$.The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{314 	imes 100 	imes 10^{-6}} = \frac{1}{0.0314} \approx 31.85 \, \Omega$.The impedance $Z$ is given by $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{50^2 + (31.85 - 6.28)^2} = \sqrt{2500 + (25.57)^2} = \sqrt{2500 + 653.8} = \sqrt{3153.8} \approx 56.16 \, \Omega$.The $RMS$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, V$.The power loss $P_{av}$ is given by $P_{av} = I_{rms}^2 R = \left(\frac{V_{rms}}{Z}ight)^2 R = \left(\frac{10}{\sqrt{2} 	imes 56.16}ight)^2 	imes 50$.$P_{av} = \left(\frac{10}{79.42}ight)^2 	imes 50 = (0.1259)^2 	imes 50 = 0.01585 	imes 50 \approx 0.79 \, W$.

An inductor $20 \, mH$,a capacitor $100 \, \mu F$ and a resistor $50 \, \Omega$ are connected in series across a source of emf,$V = 10 \sin(314 \, t)$. The power loss in the circuit is ...... $W$.

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