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An inductor 20 mH, a capacitor 50 muF an...

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 is

A

0.67 W

B

0.76 W

C

0.89 W

D

0.51 W

Text Solution

Verified by Experts

The correct Answer is:
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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