Physics · Current Electricity

JEE Main 2026 — 21 January, Evening Shift — Question 39

Two known resistance of $\mathrm{R} \Omega$ and $2 \mathrm{R} \Omega$ and and one unknown resistance $\mathrm{X} \Omega$ are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is $\mathrm{X} \Omega$ , then the value of X is $\_\_\_\_$ $\Omega$ .

Option A:
$(\sqrt{3}-1) \mathrm{R}$
Correct
Option B:
R
Option C:
$2(\sqrt{3}-1) \mathrm{R}$
Option D:
$(\sqrt{3}+1) \mathrm{R}$

Answer: A

Step-by-step solution

$\frac{(2 \mathrm{R}+\mathrm{x}) \cdot(\mathrm{R})}{3 \mathrm{R}+\mathrm{x}}=\mathrm{x}$ $x^{2}+2 R x-2 R^{2}=0$ $\mathrm{x}=(\sqrt{3}-1) \mathrm{R}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge

A battery with EMF E and internal resistance r is connected across a resistance R . The power consumption in R will be maximum when :

Answer: A

Step-by-step solution

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