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

An inductor 20 mH , a capacitor `50 muF` and a resistor `40 Omega` are connected in series across a source of emf v = 10 sin 340 t. The power loss in AC circuit is ………. .

A

`0.76W`

B

`0.89W`

C

`0.46W`

D

`0.67W`

Text Solution

Verified by Experts

The correct Answer is:
C
`L=20times10^(-3)H`
`C=50times10^(-6)F`
`R=40Omega`
emf `V=10 sin 340 t`
`P_(av)=V_(m)I_(m)(cosphi)/(2)`
`ImpdanceZ=sqrt(R^(2)+(X_(L)-X_(C))^(2))`
`V=10 sin 340 t` From `V_(m)=10V,omega=340 rad//s`
`X_(L)=L_(omega)=20times10^(-3)times340`
`=6800times10^(-3)`
`=6.8Omega`
`X_(C)=1/(Comega)`
=`1/(50times10^(-6)times340`
`=10^(6)/17000`
`10^(3)/17=1000/17`
`=58.823`
`Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))`
`=sqrt((40)^(2)+(6-8-58.8)^(2)`
`=sqrt((40)^(2)+(-52)^(2)`
`=sqrt(1600+2704)=sqrt9304`
`96.4thereforeI_(m)=10/96.4`
Average Power` = V_(m)I_(m)"times power factors"`
`P_(av)=V_(m)I_(m)cosphi`
`=0.46W`
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