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:
Solution
<p>${P_1} = \cos \phi = {1 \over {\sqrt 2 }}({X_L} = R)$</p>
<p>${P_2} = \cos \phi ' = 1$ (will become resonance circuit)</p>
<p>So, ${{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}$</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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