A sinusoidal voltage of peak vale 283V and frequency 50Hz is applied to a series LCR circuit in which `R=3Omega,L=25.48mHandC=786muF`.
The power factor.

Explanation

Similar to problem 9 on
Given `R=3Omega,L=25.48xx10^(-3)H,C=786xx10^(-6)F`
`f=50Hz,V_("peak")=283V`
We know that `X_(L)=2pifL=2xx3.142xx50xx25.48xx10^(-3)Omega`
i.e., `X_(L)=8.000Omega=8Omega`
and `X_(C)=(1)/(2pifC)=(1)/(2xx3.142xx50xx786xx10^(-6))Omega`
i.e. `X_(C)=(1)/(3.142xx7.86xx10^(-2))=(1)/(0.24696)`
i.e. `X_(C)=4.049Omega~~4Omega`
Hence `X_(L)-X_(C)=8-4=4Omega`
Power factor `cosphi=(R)/(Z)=(3)/(5)=0.6`
Hence reactance of the circuit = `4Omega`, impedance of the circuit = `5Omega`, phase angle between V and I is `53^(@)6'` and power factor of the circuit = 0.6.