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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

  1. A \(\sqrt {\frac{2}{3}} \)
  2. B \(\sqrt {\frac{2}{5}} \)
  3. C \(\sqrt {\frac{3}{2}} \)
  4. D \(\sqrt {\frac{5}{2}} \)
Verified Solution

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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