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
Pick an option to check your answer.
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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