Electro Magnetic Induction And Alternating Currents Question 145
Question: An ac source of angular frequency w is fed across a resistor r and a capacitor C in series. The current registered is I. If now the frequency of source is changed to w/3 (but maintaining the same voltage), the current in then circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency w [Roorkee 1996]
Options:
A) $ \sqrt{\frac{3}{5}} $
B) $ \sqrt{\frac{2}{5}} $
C) $ \sqrt{\frac{1}{5}} $
D) $ \sqrt{\frac{4}{5}} $
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Answer:
Correct Answer: A
Solution:
At angular frequency w, the current in RC circuit is given by $ i_{rms}=\frac{V_{rms}}{\sqrt{R^{2}+{{( \frac{1}{\omega C} )}^{2}}}} $ ……(i) Also $ \frac{i_{rms}}{2}=\frac{V_{rms}}{\sqrt{R^{2}+{{( \frac{1}{\frac{\omega }{3}C} )}^{2}}}}=\frac{V_{rms}}{\sqrt{R^{2}+\frac{9}{{{\omega }^{2}}C^{2}}}} $ ……(ii) From equation (i) and (ii) we get $ 3R^{2}=\frac{5}{{{\omega }^{2}}C^{2}}\Rightarrow \frac{\frac{1}{\omega C}}{R}=\sqrt{\frac{3}{5}} $
Þ $ \frac{X_{C}}{R}=\sqrt{\frac{3}{5}} $
BETA
