Question

In a series $LR$ circuit $X_L=R$ and power factor of the circuit is $P_1$. When capacitor with capacitance C such that $X_L=X_C$ is put in series, the power factor becomes $P_2$. The ratio $\frac {P_1}{P_2}$ is

• A. $\frac 12$
• B. $\frac {1}{\sqrt 2}$
• C. $\frac {\sqrt 3}{\sqrt 2}$
• D. $2:1$

Answer 1

Answer: (B)

$\frac {1}{\sqrt 2}$

Answer 2

$P_1=cos⁡\ Φ= \frac {1}{\sqrt 2}$ $(X_L=R)$
$P_2=cos\ ⁡Φ‘=1$ (will become resonance circuit)
Then,
$\frac {P_1}{P_2}=\frac {1}{\sqrt 2}$

So, the correct option is (B): $\frac {1}{\sqrt 2}$