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 :
Pick an option to check your answer.
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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