Physics · Alternating Current

JEE Main 2026 — 5 April, Morning Shift — Question 17

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

Option A:
$\frac{6}{7}$
Option B:
$\frac{3}{5}$
Option C:
$\frac{7}{8}$
Correct
Option D:
$\frac{3}{4}$

Answer: C

Step-by-step solution

At ω: $I = V/\sqrt{R^2 + (1/(ωC))^2}$ . At ω/4: $I/3 = V/\sqrt{R^2 + (4/(ωC))^2}$ . Dividing gives $3 = \sqrt{R^2 + (4/(ωC))^2} / \sqrt{R^2 + (1/(ωC))^2}$ . Squaring and solving yields $R/(1/(ωC)) = 7/8$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Physics
Chapter: Alternating Current
Topic: Series L-R, R-C, L-C Circuits with AC Source