JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

Two known resistance of R $Ω$ and 2R $Ω$ and and one unknown resistance X $Ω$ are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is X $Ω$ , then the value of X is ________ $Ω$ .

A
$(3 - 1) R$
B
$R$
C
$2 (3 - 1) R$
D
$(3 + 1) R$

Solution

Upper branch resistance: $2 R + X$ . Lower branch: $R$ . They are in parallel between A and B, and the equivalent equals $X$ :

$\frac{( 2 R + X ) \cdot R}{( 2 R + X ) + R} = X$ , i.e. $\frac{R ( 2 R + X )}{3 R + X} = X$ .

Cross-multiplying: $R (2 R + X) = X (3 R + X)$ , so $2 R^{2} + R X = 3 R X + X^{2}$ , giving $X^{2} + 2 R X - 2 R^{2} = 0$ .

Quadratic in $X$ : $X = \frac{- 2 R + 4 R ^{2} + 8 R ^{2}}{2} = \frac{- 2 R + 2 R 3}{2} = (3 - 1) R$ .

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