MHT CET · Physics · Alternating Current
When alternating current is passed through L-R series circuit, the power factor is \(\frac{\sqrt{3}}{2}\) and \(R=50 \Omega\), then the value of \(L\) is
\(\left[\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6}=\frac{1}{2}, \quad \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]\)
- A \(\frac{1}{2} \pi\)
- B \(\frac{\sqrt{3}}{2} \pi\)
- C \(\frac{1}{2 \sqrt{3} \pi}\)
- D \(\frac{1}{\sqrt{3} \pi}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2 \sqrt{3} \pi}\)
Step-by-step Solution
Detailed explanation
\(\cos \phi = \frac{\sqrt{3}}{2} \implies \phi = \frac{\pi}{6}\) \(\tan \phi = \frac{X_L}{R}\)
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