JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 5)
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:
$$\frac{1}{2}$$
$$\frac{1}{\sqrt{2}}$$
$$\frac{\sqrt{3}}{\sqrt{2}}$$
2 : 1
Explanation
$${P_1} = \cos \phi = {1 \over {\sqrt 2 }}({X_L} = R)$$
$${P_2} = \cos \phi ' = 1$$ (will become resonance circuit)
So, $${{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}$$
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