JEE Mains · Physics · STD 12 -7. Alternating current
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
- A \(\frac{1}{2}\)
- B \(\frac{1}{\sqrt{2}}\)
- C \(\frac{\sqrt{3}}{\sqrt{2}}\)
- D \(2: 1\)
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 \(L-C-R\)…
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