An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf V = 10sin 340 t. The power loss in A.C. circuit is :

A
0.51 W
B
0.67 W
C
0.76 W
D
0.89 W

Solution:

$X_{C} = \frac{1}{ωC} = \frac{1}{340 x 50 x 10^{- 6}} = 58.8 Ω X_{L} = ωL = 340 x 20 x 10^{- 3} = 6.8 Ω Z = \sqrt{R^{2} + {(X_{C} - X_{L})}^{2}} = \sqrt{40^{2} + {(58.8 - 6.8)}^{2}} = \sqrt{4304} Ω P = {i^{2}}_{rms} R = {(\frac{V_{rms}}{Z})}^{2} R = {(\frac{10 / \sqrt{2}}{\sqrt{4304}})}^{2} x 40 = \frac{50 x 40}{4304} = 0.47 W$

So best answer( nearest answer) will be (1).

An inductor 20 mH, a capacitor 50 μF and a resistor 40 $Ω$ are connected in series across a source of emf V = 10sin 340 t. The power loss in A.C. circuit is :

Solution:

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