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Q.
An inductor $20\, mH$, a capacitor $50\, \mu F$ and a resistor $40 \, \Omega $ are connected in series across a source of emf $V = 10 \, \sin \, 340\,t$. The power loss in A.C. circuit is :
$X _{ C }=\frac{1}{\omega C }=\frac{1}{340 \times 50 \times 10^{-6}}=58.8 \Omega$
$X _{ L }=\omega L =340 \times 20 \times 10^{-3}=6.8 \Omega$
$Z =\sqrt{ R ^{2}+\left( X _{ C }- X _{ L }\right)^{2}}$
$=\sqrt{40^{2}+(58.8-6.8)^{2}}$
$=\sqrt{4304} \Omega$
Power, $P = i _{ rms }^{2} R =\left(\frac{ V _{ rms }}{ Z }\right)^{2} R$
$=\left(\frac{10 / \sqrt{2}}{\sqrt{4304}}\right)^{2} \times 40$
$=\frac{50 \times 40}{4304} \approx 0.51 W$