An inductor 20 mH, a capacitor 100 and a resistor 50 are connected in series across a source of emf, The power loss in the circuit is [2018]

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Solution

Q.
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 [2018]

1 0.79 W

2 0.43 W

3 2.74 W

4 1.13 W

Ans.
(1)

Impedance $Z$ in an ac circuit is

$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}};$ where $X_{C} =$ capacitive reactance and $X_{L} =$ inductive reactance.

Also $X_{C} = \frac{1}{ω C} and X_{L} = ω L$

$∴$ $Z = \sqrt{{(50)}^{2} + {(\frac{1}{314 \times 100 \times 10^{- 6}} - 314 \times 20 \times 10^{- 3})}^{2}}$

or $Z = 56$ $Ω$

The power loss in the circuit is $P_{a v} = {(\frac{V_{rms}}{Z})}^{2} R$

$∴$ $P_{a v} = {(\frac{10}{(\sqrt{2}) 56})}^{2} \times 50 = 0.79$ $W$

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