Q69 / 75 · JEE Main 2026 (05 April Shift 1) · Physics
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
- A$\sqrt{\dfrac{6}{7}}$
- B$\sqrt{\dfrac{3}{5}}$
- Correct answerC$\sqrt{\dfrac{7}{8}}$
- D$\sqrt{\dfrac{3}{4}}$
Solution
The current in the RC circuit at frequency $\omega$ is given by
$I = \dfrac{V}{\sqrt{R^2 + X_C^2}}$
where $X_C = \dfrac{1}{\omega C}$ is the capacitive reactance.
When the frequency is changed to $\omega/4$, the new reactance becomes $X_C' = \dfrac{1}{(\omega/4)C} = 4X_C$.
The new current is given as $I/3$, so
$\dfrac{I}{3} = \dfrac{V}{\sqrt{R^2 + (4X_C)^2}}$
Substituting the expression for $I$, we get
$\dfrac{V}{3\sqrt{R^2 + X_C^2}} = \dfrac{V}{\sqrt{R^2 + 16X_C^2}}$
Squaring both sides, we obtain
$\dfrac{1}{9(R^2 + X_C^2)} = \dfrac{1}{R^2 + 16X_C^2}$
$9R^2 + 9X_C^2 = R^2 + 16X_C^2$
$8R^2 = 7X_C^2$
$\dfrac{R^2}{X_C^2} = \dfrac{7}{8}$
$\dfrac{R}{X_C} = \sqrt{\dfrac{7}{8}}$
Answer: $\sqrt{\dfrac{7}{8}}$
$I = \dfrac{V}{\sqrt{R^2 + X_C^2}}$
where $X_C = \dfrac{1}{\omega C}$ is the capacitive reactance.
When the frequency is changed to $\omega/4$, the new reactance becomes $X_C' = \dfrac{1}{(\omega/4)C} = 4X_C$.
The new current is given as $I/3$, so
$\dfrac{I}{3} = \dfrac{V}{\sqrt{R^2 + (4X_C)^2}}$
Substituting the expression for $I$, we get
$\dfrac{V}{3\sqrt{R^2 + X_C^2}} = \dfrac{V}{\sqrt{R^2 + 16X_C^2}}$
Squaring both sides, we obtain
$\dfrac{1}{9(R^2 + X_C^2)} = \dfrac{1}{R^2 + 16X_C^2}$
$9R^2 + 9X_C^2 = R^2 + 16X_C^2$
$8R^2 = 7X_C^2$
$\dfrac{R^2}{X_C^2} = \dfrac{7}{8}$
$\dfrac{R}{X_C} = \sqrt{\dfrac{7}{8}}$
Answer: $\sqrt{\dfrac{7}{8}}$
Hence, the correct answer is option: 'C'.