Electro Magnetic Induction And Alternating Currents Question 192

Question: 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

Options:

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

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

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

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

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Answer:

Correct Answer: D

Solution:

  • $ Powerfactor{{}_{( old )}} $

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

$ Powerfacto{r_{(new)}} $

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

$ \therefore ,\frac{New,power,factor}{Old,power,factor}=\frac{\frac{R}{\sqrt{2}R}}{\frac{R}{\sqrt{5}R}}=\sqrt{\frac{5}{2}} $



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