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JEE Main 2013 (25 Apr, Shift 1) · Power, Power Factor & Wattless Current

Power, Power Factor & Wattless Current question from JEE Main 2013 (25 Apr, Shift 1)

Q71JEE Main 2013 · 25 APR · SHIFT 1Difficult

Power, Power Factor & Wattless Current

A series LR circuit is connected to an ac source of frequency ω\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:

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Answer: (d)

  1. Old: Z=R2+4R2=5RZ = \sqrt{R^2+4R^2}=\sqrt{5}R, cos⁡ϕ1=1/5\cos\phi_1 = 1/\sqrt{5}.
  2. New: X=2R−R=RX = 2R - R = R, Z=2RZ=\sqrt{2}R, cos⁡ϕ2=1/2\cos\phi_2 = 1/\sqrt{2}.
  3. Ratio =52=52= \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}}, so the answer is (d).
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