Question 172: AC fundamentals

Question

In a series $$L R$$ 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:

Accepted Answer

$${P_1} = \cos \phi  = {1 \over {\sqrt 2 }}({X_L} = R)$$
$${P_2} = \cos \phi ' = 1$$ (will become resonance circuit)
So, $${{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}$$

Answer

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

Answer

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

Answer

$$\frac{\sqrt{3}}{\sqrt{2}}$$

Answer

2 : 1

AC fundamentals

In a series $L R circuit X_{L}=R and power factor of the circuit is P_{1}. When capacitor with capacitance C such that X_{L}=X_{C} is put in series, the power factor becomes P_{2}. The ratio \frac{P_{1}}{P_{2}}$ is:

Choose the correct answer: