Question

In a series LR circuit, `X_L=R` and the power factor of the circuit is `P_1`.When a capacitor with capacitance C such that `X_C=X_L` is put in series , the power factor becomes `P_2`.Find out `P_1//P_2`.

Answer 1

In a series LR circuit, power factor `(P_1)=R/Z`
Here, Z =impedance = `sqrt(R^2+X_L^2)`
Here, `Z=sqrt(R^2+R^2)=sqrt2R`
`thereforeP_1=1/sqrt2`
In a series LCR circuit power factor `(P_2)=R/Z`
where,`Z=sqrt(R^2+(X_L-X_C)^2)=R`
`thereforeP_2=1`
Hence,`P_1/P_2=1/sqrt2`