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}} $
Show Answer
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}} $
BETA
