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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2023
  • Medium
  • MCQ

In a series \(L R\) circuit with \(X_L=R\). power factor is \(P_1\). If a capacitor of capacitance \(C\) with \(X _{ C }= X _{ L }\) is added to the circuit the power factor becomes \(P _2\). The ratio of \(P_1\) to \(P_2\) will be :

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Answer & Solution

Correct Answer

(B) \(1: \sqrt{2}\)

Step-by-step Solution

Detailed explanation

\(P=\frac{R}{Z} \Rightarrow P_1=\frac{R}{\sqrt{R^2+X_L^2}}=\frac{R}{R \sqrt{2}}\left(\text { as } X_L=R\right)\)

\(P_1=\frac{1}{\sqrt{2}}\)

\(P_2=\frac{R}{\sqrt{R^2+\left(X_L-X_L\right)^2}}=P_2=1\)

\(\frac{P_1}{P_2}=\frac{1}{\sqrt{2}}\)

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