An inductor 20 mH20 \text{ mH}20 mH, a capacitor 100 μF100 \text{ }\mu\text{F}100 μF and a resistor 50 Ω50 \text{ }\Omega50 Ω are connected in series across a source of emf, V=10sin314 tV = 10 \sin 314 \text{ t}V=10sin314 t. The power loss in the circuit is
2.74 W
0.43 W
0.79 W
1.13 W
Pav=(VRMSZ)2RP_{av} = \left( \frac{V_{RMS}}{Z} \right)^2 RPav=(ZVRMS)2R. Z=R2+(ωL−1ωC)2=56 ΩZ = \sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2} = 56 \text{ }\OmegaZ=R2+(ωL−ωC1)2=56 Ω. Therefore, Pav=(10(2)56)2×50=0.79 WP_{av} = \left( \frac{10}{(\sqrt{2}) 56} \right)^2 \times 50 = 0.79 \text{ W}Pav=((2)5610)2×50=0.79 W.
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