SATHEE: Electro Magnetic Induction And Alternating Currents Question 145

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) $\sqrt{\frac{4}{5}}$

Show Answer

Answer:

Correct Answer: A

Solution:

At angular frequency w, the current in RC circuit is given by $i_{r m s} = \frac{V_{r m s}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}}$ ……(i) Also $\frac{i_{r m s}}{2} = \frac{V_{r m s}}{\sqrt{R^{2} + {(\frac{1}{\frac{ω}{3} C})}^{2}}} = \frac{V_{r m s}}{\sqrt{R^{2} + \frac{9}{ω^{2} C^{2}}}}$ ……(ii) From equation (i) and (ii) we get $3 R^{2} = \frac{5}{ω^{2} C^{2}} \Rightarrow \frac{\frac{1}{ω C}}{R} = \sqrt{\frac{3}{5}}$

Þ $\frac{X_{C}}{R} = \sqrt{\frac{3}{5}}$

Electro Magnetic Induction And Alternating Currents Question 145

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Solution:

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Electro Magnetic Induction And Alternating Currents Question 145

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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