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An inductor 20 mH, a capacitor 100 $\mu F$ and a resistor 50 $\Omega$ are connected in series across a source of emf, $V = 10\sin 314t$. The power loss in the circuit is
Explanation
Compute Z from $X_L$ and $X_C$, then $P = (V_{rms}/Z)^2R$.
Detailed Solution
$P_{av} = \left(\frac{V_{rms}}{Z}\right)^2 R$
$X_L = \omega L = 314\times20\times10^{-3} = 6.28\ \Omega$, $X_C = \frac{1}{\omega C} = \frac{1}{314\times100\times10^{-6}} = 31.85\ \Omega$
$Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \approx 56\ \Omega$
$P_{av} = \left(\frac{10}{\sqrt{2}\times56}\right)^2\times50 = 0.79$ W