An inductor `20 mH`, a capacitor `100muF` and a resistor `50Omega` are connected in series across a source of emf `V=10sin314t`. The power loss in the circuit is

Options

A. 2.74 W
B. 0.43 W
C. 0.79 W
D. 1.13 W

Answer

C

Explanation

Here, inductance, L=20mH=`20 xx 10^(-3)H`
Capacitance, C=`100muF=100 xx 10^(-6)`F
Resistance, `R= 50Omega`
emf, `V=10 sin 314t`…………(i)
`therefore` The general equation of the emf is given as
`V-V_(0)sinomegat` ……………(ii)
`therefore` comparing Eqs. (i) and (ii), wet get
`V_(0) = 10V, omega=314rads^(-1)`
The power loss associated with the given AC circuit is given as
`P=V_(rms)I_(rms)cosphi`
`=V_(rms)(V_(rms))/Z(R/Z)`
`=(V_(rms)/Z)^(2)R= (V_(0)/(sqrt(2).Z))^(2)R`.........(i)
`therefore` Impedance, `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))`
`=sqrt(R^(2)+(omegaL-1/(omegaC))^(2)`
`therefore` Substituting the given values in the above equation, we get
`=sqrt((50)^(2)+[314 xx 20 xx 10^(-3)-1/(314 xx 10^(-4))]^(2))`
`=sqrt(2500 +[6280 xx 10^(-3)-0.00318 xx 10^(4)]^(2))`
`sqrt(2500 + (25.56)^(2))=56.15 Omega ~~ 56 Omega`
Now, substituting this values in Eq.(i), we get
`P=10/(sqrt(2)xx 56)^(2) xx 50 = 100/(2 xx 3136) xx 50 = 0.79` W