Electro Magnetic Induction And Alternating Currents Question 197
Question: An ac source of angular frequency co 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 $ \omega /3 $ (but maintaining the same voltage), the current in the circuit is found to be halved. Then the ratio of reactance to resistance at the original frequency $ \omega $ is
Options:
A) $ \sqrt{3/5} $
B) $ \sqrt{5/3} $
C) $ \sqrt{2/3} $
D) $ \sqrt{3/2} $
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Answer:
Correct Answer: A
Solution:
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According to given problem,
$ I,=\frac{V}{Z},=\frac{V}{{{[R^{2}+{{(1/C\omega )}^{2}}]}^{1/2}}} $ …(i) and
$ \frac{I}{2},=\frac{V}{{{[R^{2}+{{(3/C\omega )}^{2}}]}^{1/2}}} $ …(ii) Substituting the value of I from eq. (i) in (11),
$ 4,( R^{2}+\frac{1}{C^{2}{{\omega }^{2}}} )=R^{2}+\frac{9}{C^{2},{{\omega }^{2}}} $ i.e.,
$ \frac{1}{C^{2},{{\omega }^{2}}},=\frac{3}{5},R^{2} $ so that $ \frac{X}{R} $
BETA
