An a.c. source of angular… - JEE Main 2026 Physics PYQ

JEE Main 2026 — Physics Electromagnetism

2026mcqmedium

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

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$\sqrt{\dfrac{6}{7}}$
B
$\sqrt{\dfrac{3}{5}}$
C
$\sqrt{\dfrac{7}{8}}$
D
$\sqrt{\dfrac{3}{4}}$

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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: $\sqrt{\dfrac{7}{8}}$

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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: $\sqrt{\dfrac{7}{8}}$

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Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism