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An inductor `20 mH`, a capacitor `100 muF` 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 isA. 2.74 WB. 0.43 WC. 0.79 WD. 1.13 W |
Answer» Correct Answer - C `P_(av)=((V_(RMS))/(Z))^(2)R` `Z=sqrt(R^(2)+(omegaL-(1)/(omegaC))^(2))=56 Omega` `therefore P_(av)=((10)/((sqrt(2))56))^(2)xx50=0.79 W` |
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