In a series LR circuit with $\mathrm{X_L=R}$, power factor P1. If a capacitor of capacitance C with $\mathrm{X_C=X_L}$ is added to the circuit the power factor becomes P2. The ratio of P1 to P2 will be :
Solution
<p>${X_L} = R$</p>
<p>$\Rightarrow {P_1} = {R \over {\sqrt {X_L^2 + {R^2}} }} = {1 \over {\sqrt 2 }}$</p>
<p>Now, ${X_L} = {X_C} = R$</p>
<p>$\Rightarrow {P_2} = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }} = 1$</p>
<p>$\Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}$</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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