JEE Mains · Physics · STD 12 -7. Alternating current
In a series \(LCR\) circuit, the inductive reactance \(\left( X _{ L }\right)\) is \(10\, \Omega\) and the capacitive reactance \(\left( X _{ C }\right)\) is \(4\, \Omega\). The resistance \(( R )\) in the circuit is \(6\, \Omega\). The power factor of the circuit is :
- A \(\frac{1}{2}\)
- B \(\frac{1}{2 \sqrt{2}}\)
- C \(\frac{1}{\sqrt{2}}\)
- D \(\frac{\sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
We know that power factor is \(\cos \phi\) \(\cos \phi=\frac{ R }{ Z }\) \(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \quad \ldots\) \(\Rightarrow Z =\sqrt{6^{2}+(10-4)^{2}}\) \(\Rightarrow Z =6 \sqrt{2} \mid \cos \phi=\frac{6}{6 \sqrt{2}}\) \(\cos \phi=\frac{1}{\sqrt{2}}\)
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