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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2026 · 21 January · Evening Shift · Physics · Current Electricity

Two known resistances of R and 2 R and one unknown resistance X are...

Two known resistances of RΩR \Omega and 2RΩ2 R \Omega and one unknown resistance XΩX \Omega are connected in a circuit as shown in the figure. If the equivalent resistance between points AA and BB in the circuit is XΩX \Omega, then the value of XX is ____\_\_\_\_ Ω\Omega.
Two known resistances of R and 2 R and one unknown resistance X are connected in a circuit as shown in the figure. If

Options

  1. A(3−1)R(\sqrt{3}-1) R
  2. B2(3−1)R2(\sqrt{3}-1) R
  3. C(3+1)R(\sqrt{3}+1) R
  4. DRR

Correct answer

A. (3−1)R(\sqrt{3}-1) R

Step-by-step solution

The circuit has 2R and X in series (top branch) parallel to R (bottom branch).
The equivalent resistance is Req=(2R+X)⋅R3R+X=XR_{eq} = \frac{(2R + X) \cdot R}{3R + X} = X.
Setting up the equation: X(3R+X)=R(2R+X)X(3R + X) = R(2R + X), which simplifies to X2+2RX−2R2=0X^2 + 2RX - 2R^2 = 0.
Using the quadratic formula: X=−2R±2R32=R(−1±3)X = \frac{-2R \pm 2R\sqrt{3}}{2} = R(-1 \pm \sqrt{3}).
Since resistance must be positive, X=(3−1)RX = (\sqrt{3} - 1)R Ω.
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