An inductor `20 mH`, a capacitor `50 muF` and a resistor `40 Omega`are connected in series across of emf `V=10 sin 340 t`. The power loss in `A.C.` circuit isA. 0.67 WB. 0.76 WC. 0.89 WD. 0.51 W

An inductor `20 mH`, a capacitor `50 muF` and a resistor `40 Omega`are

1.	An inductor `20 mH`, a capacitor `50 muF` and a resistor `40 Omega`are connected in series across of emf `V=10 sin 340 t`. The power loss in `A.C.` circuit isA. 0.67 WB. 0.76 WC. 0.89 WD. 0.51 W
Answer» Correct Answer - D Given, inductance, L=20mH Capacitance, C=`50muF` Resistance, R=`40Omega` emf, V`=10sin340t` `therefore` Power loss in AC circuit will be given as `P_(av)=I_(V)^(2)R= [E_(V)^(2)/Z]^(2).R` `(10/sqrt(2))^(2). 40[1/(40^(2) + (340xx20xx10^(-3) - 1/(340 xx 50 xx 10^(-6))))]` =`100/sqrt(2) x 40 xx 1/(1600 + (6.8-58.8)^(2))` `2000/(1600+2704) ~~ 0.46W ~~ 0.51 W`

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An inductor `20 mH`, a capacitor `50 muF` and a resistor `40 Omega`are connected in series across of emf `V=10 sin 340 t`. The power loss in `A.C.` circuit isA. 0.67 WB. 0.76 WC. 0.89 WD. 0.51 W

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