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`.

In a series LR circuit, `X_L=R` and the power factor of the circuit is

1.	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» 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`

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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`.

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