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 phase difference between the voltage across the source and the current.

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`
`tanphi=(X_(L)-X_(C))/(R)=(4)/(3)=1.3333`
`therefore phi=53.1^(@)or53^(@)6`'