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JEE Mains · Physics · STD 12 -7. Alternating current
A series \(LR\) circuit is connected to an ac source of frequency \(\omega \) and the inductive reactance is equal to \(2R\). A capacitance of capacitive reactance equal to \(R\) is added in series with \(L\) and \(R\). The ratio of the new power factor to the old one is
- A \(\sqrt {\frac{2}{3}} \)
- B \(\sqrt {\frac{2}{5}} \)
- C \(\sqrt {\frac{3}{2}} \)
- D \(\sqrt {\frac{5}{2}} \)
Answer & Solution
Correct Answer
(D) \(\sqrt {\frac{5}{2}} \)
Step-by-step Solution
Detailed explanation
Power factor \(_{(old)}\) \(=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+(2 \mathrm{R})^{2}}}=\frac{\mathrm{R}}{\sqrt{5 \mathrm{R}}}\) Power factor \(_{(\text {new })}\)…
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