JEE Physics 2026 Question with Solution

Given: Two known resistances are $R$ and $2R$, and the unknown resistance is $X$. The equivalent resistance between $A$ and $B$ is given to be $X$.

Find: The value of $X$.

From the circuit, the resistor $R$ is in parallel with the series combination $2R + X$. Therefore,

$$R_{eq} = \frac{(2R + X)R}{(2R + X) + R}$$

Since $R_{eq} = X$,

$$X = \frac{(2R + X)R}{3R + X}$$

Cross-multiplying,

$$X(3R + X) = 2R^2 + RX$$ $$3RX + X^2 = 2R^2 + RX$$ $$X^2 + 2RX - 2R^2 = 0$$

Using the quadratic formula,

$$X = \frac{-2R \pm \sqrt{(2R)^2 - 4(1)(-2R^2)}}{2}$$ $$X = \frac{-2R \pm \sqrt{4R^2 + 8R^2}}{2}$$ $$X = \frac{-2R \pm \sqrt{12R^2}}{2}$$ $$X = \frac{-2R \pm 2R\sqrt{3}}{2}$$ $$X = -R \pm R\sqrt{3}$$

Since resistance cannot be negative, we take the positive root.

$$X = (\sqrt{3} - 1)R$$

Therefore, the value of $X$ is $(\sqrt{3} - 1)R$, so the correct option is B.

Given: The equivalent resistance of the network is itself equal to the unknown resistance $X$.

Find: The value of $X$ by forming and solving the circuit equation.

The key idea is to first write the equivalent resistance of the visible series-parallel combination. The upper branch contains $2R$ and $X$ in series, so that branch has resistance

$$2R + X$$

This branch is in parallel with the lower branch of resistance $R$. Hence,

$$R_{eq} = \frac{R(2R + X)}{R + (2R + X)}$$ $$R_{eq} = \frac{R(2R + X)}{3R + X}$$

Now use the given condition $R_{eq} = X$:

$$X = \frac{R(2R + X)}{3R + X}$$

Multiply both sides by $3R + X$:

$$X(3R + X) = R(2R + X)$$

Expand both sides:

$$3RX + X^2 = 2R^2 + RX$$

Bring all terms to one side:

$$X^2 + 2RX - 2R^2 = 0$$

This is a quadratic in $X$. Solving,

$$X = \frac{-2R \pm \sqrt{4R^2 + 8R^2}}{2}$$ $$X = \frac{-2R \pm \sqrt{12R^2}}{2}$$ $$X = \frac{-2R \pm 2R\sqrt{3}}{2}$$ $$X = -R \pm R\sqrt{3}$$

The physically acceptable value is the positive one:

$$X = (\sqrt{3} - 1)R$$

Therefore, the correct option is B.