In a series $LR$ circuit with $X_L = R$. Power factor is $P_1$. If a capacitor of capacitance $C$ with $X_c = X_L$ is added to the circuit the power factor becomes $P_2$. The ratio of $P_1$ to $P_2$ will be :

  • A
    $1 : 3$
  • B
    $1 : \sqrt{2}$
  • C
    $1 : 1$
  • D
    $1 : 2$

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