An a.c. source of angular frequency ω is connected across a resistor R and a capacitor C in series. The current is observed as I. Now the frequency of the source is changed to ω/4, (keeping the voltage unchanged) the current is found to be I/3. The ratio of resistance to reactance at frequency ω is (a) √(6/7) (b) √(3/5) (c) √(7/8) (d) √(3/4)

Answer: (c). Let X be the capacitive reactance at ω. At ω/4 it becomes 4X. Current ∝ 1/Z, and it falls to I/3, so Z₂ = 3Z₁: R² + 16X² = 9(R² + X²) 7X² = 8R² ⇒ R/X = √(7/8) So the answer is (c) √(7/8).

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RL and RC Series Circuits

An a.c. source of angular frequency $\omega$ is connected across a resistor $R$ and a capacitor $C$ in series. The current is observed as $I$ . Now the frequency of the source is changed to $\omega/4$ , (keeping the voltage unchanged) the current is found to be $I/3$ . The ratio of resistance to reactance at frequency $\omega$ is

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Answer: (c)

Let $X$ be the capacitive reactance at $\omega$ . At $\omega/4$ it becomes $4X$ .
Current $\propto 1/Z$ , and it falls to $I/3$ , so $Z_2 = 3Z_1$ : $R^2 + 16X^2 = 9(R^2 + X^2)$
$7X^2 = 8R^2 \Rightarrow \frac{R}{X} = \sqrt{\frac{7}{8}}$
So the answer is (c) $\sqrt{7/8}$ .

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