SATHEE: Electro Magnetic Induction And Alternating Currents Question 192

Electro Magnetic Induction And Alternating Currents Question 192

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

Options:

A) $\sqrt{\frac{2}{3}}$

B) $\sqrt{\frac{2}{5}}$

C) $\sqrt{\frac{3}{2}}$

D) $\sqrt{\frac{5}{2}}$

Show Answer

Answer:

Correct Answer: D

Solution:

$P o w e r f a c t o r_{(o l d)}$

$= \frac{R}{\sqrt{R^{2} + X_{L}^{2}}} = \frac{R}{\sqrt{R^{2} + {(2 R)}^{2}}} = \frac{R}{\sqrt{5 R}}$

$P o w e r f a c t o r_{(n e w)}$

$= \frac{R}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} = \frac{R}{\sqrt{R^{2} + {(2 R - R)}^{2}}} = \frac{R}{\sqrt{2} R}$

$∴, \frac{N e w, p o w e r, f a c t o r}{O l d, p o w e r, f a c t o r} = \frac{\frac{R}{\sqrt{2} R}}{\frac{R}{\sqrt{5} R}} = \sqrt{\frac{5}{2}}$

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