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  4. An inductor 20 mH, a capacitor 100 μ F and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is

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Q. 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

NEETNEET 2018Electromagnetic Induction

Solution:

$P_{av} = \left(\frac{V_{RMS}}{Z}\right)^{2} R$
$ Z = \sqrt{R^{2} + \left(\omega L - \frac{1}{\omega C}\right)^{2}} = 56\, \Omega $
$\therefore P_{av} = \left(\frac{10}{\left(\sqrt{2}\right)56}\right)^{2} \times50 = 0.79 \,W $