A sinusoidal voltage of peak value 283V and frequency 50Hz is applied to a seres LCR circuit in which R`=3Omega,L=25.48mH`. And C=`796muF`. Find (a) the impedance of the circuit: (b) the phase difference between the voltage across the source and the current: (c) the power dissipated in the circuit: and (d) the power factor.

Explanation

a) To find the impedance of the circuit, we first calculate `X_(L) and X_(C)`
`X_(L)=2pivL=2xx3.14xx50xx25.48xx10^(-3)Omega=8Omega`
`X_(L)=(1)/(2pivC)=(1)/(2xx3.14xx50xx796xx10^(-6))=4Omega`
Therefore, `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))=sqrt(3^(2)+(8-4)^(2))=5Omega`
b) Phase difference,
`phi=tan^(-1)(K_(C)-X_(L))/(R)=tan^(-1)((4-8)/(3))=-53.1^(@)`
Since `phi` is negative, the current in the circuit lags the voltage across the source.
c) The power dissipated in the circuit is `P=I^(2)R`
Now, `I=(i_(m))/(sqrt2)=(1)/(sqrt2)((283)/(5))=40A`
Therefore, `P=(40A)^(2)xx3Omega=4800W`
d) Power factor `=cos phi=cos53.1^(0)=0.6`