In a series L.R circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capactor of value C is added in series to the L and R. Taking the value of C as `(n/(3pi))muF`, then value of n is ____________

Answer

400

Explanation

`P=V_(m) .i_(rms) . cos phi`
`400 = 250 xxi_(rms) xx0.8`
`i_(rms) = 2A`
`(i_(rms))^2.R=P`
`4xxR = 400 implies R = 100 Omega`
`cos phi =R/sqrt(R^2+X_L^2)`
`0.8=100/(sqrt(100^2+X_L^2))`
`100^2 +X_L^2=((100)/(0.8))^2`
`X_L=sqrt(-100^2+((100)/(0.8))^2)`
`X_L=75Omega`
Power factor in unity
`X_C =X_L =75`
`1/(omegaC)=75`
`implies C =1/(75 xx2H xx50)`
`implies C =1/(7500 pi)F =1/(3pi xx 2500) F`
`implies C =1/(3pi) xx 4 xx10^2mF implies C =400/(3pi) muF`
on comparing with given value of C i.e. `(n/(3pi))muF` we get n=400.