A series AC circuit containing an inductor (2 | vedclass

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(C) Given: $L = 20\,mH = 20 	imes 10^{-3}\,H,$ $C = 120\,\mu F = 120 	imes 10^{-6}\,F,$ $R = 60\,\Omega,$ $V_{rms} = 24\,V,$ $f = 50\,Hz,$ $t = 60\,

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$5.17	imes 10^2\,J$

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(C) Given: $L = 20\,mH = 20 	imes 10^{-3}\,H,$ $C = 120\,\mu F = 120 	imes 10^{-6}\,F,$ $R = 60\,\Omega,$ $V_{rms} = 24\,V,$ $f = 50\,Hz,$ $t = 60\,s.$Inductive reactance: $X_L = 2\pi fL = 2\pi 	imes 50 	imes 20 	imes 10^{-3} = 2\pi \approx 6.28\,\Omega.$Capacitive reactance: $X_C = \frac{1}{2\pi fC} = \frac{1}{2\pi 	imes 50 	imes 120 	imes 10^{-6}} = \frac{1}{0.012\pi} \approx 26.53\,\Omega.$Impedance: $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{60^2 + (26.53 - 6.28)^2} = \sqrt{3600 + (20.25)^2} = \sqrt{3600 + 410.06} = \sqrt{4010.06} \approx 63.33\,\Omega.$Power dissipated: $P = I_{rms}^2 R = \left(\frac{V_{rms}}{Z}ight)^2 R = \left(\frac{24}{63.33}ight)^2 	imes 60 \approx (0.379)^2 	imes 60 \approx 0.1436 	imes 60 \approx 8.616\,W.$Energy dissipated: $E = P 	imes t = 8.616 	imes 60 \approx 516.96\,J \approx 5.17 	imes 10^2\,J.$

$A$ series $AC$ circuit containing an inductor $(20\,mH),$ a capacitor $(120\,\mu F)$ and a resistor $(60\,\Omega)$ is driven by an $AC$ source of $24\,V$ and $50\,Hz.$ The energy dissipated in the circuit in $60\,s$ is

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