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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2022
  • Medium
  • MCQ

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

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Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

In case of \(L-R\) circuit

\(Z=\sqrt{ X _{ L }^{2}+ R ^{2}} \&\) power factor

\(P _{1}=\cos \phi=\frac{ R }{ Z }\)

\(\text { As } X _{ L }= R\)

\(\Rightarrow Z =\sqrt{2} R\)

\(\Rightarrow P _{1}=\frac{ R }{\sqrt{2} R } \Rightarrow P _{1}=\frac{1}{\sqrt{2}}\)

In case of \(L-C-R\) circuit

\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

As \(X_{L}=X_{C}\)

\(\Rightarrow Z=R\)

\(\Rightarrow P _{2}=\cos \phi=\frac{ R }{ R }=1\)

\(\Rightarrow \frac{ P _{1}}{ P _{2}}=\frac{1}{\sqrt{2}}\)

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