First slide

Power in a.c. circuit

Question

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

Easy

A

0.79 W
B

0.43 W
C

2.74 W
D

1.13 W

Solution

Impedance Z in an ac circuit is
$Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}}; where X_{C} = capacitive reactance and X_{L} = inductive reactance.$
$Also X_{C} = \frac{1}{ω C} and X_{L} = ω L$
$∴ Z = \sqrt{(50)^{2} + {(\frac{1}{314 \times 100 \times 10^{- 6}} - 314 \times 20 \times 10^{- 3})}^{2}} or Z = 56 Ω$
$The power loss in the circuit is P_{a v} = {(\frac{V_{r m s}}{Z})}^{2} R$
$∴ P_{a v} = {(\frac{10}{(\sqrt{2}) 56})}^{2} \times 50 = 0.79 W$

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