JEE Physics 2026 Question with Solution

Given: The equivalent resistance between points $A$ and $B$ is $\frac{x}{5}\,\Omega$.

Find: The value of $x$.

Concept: Use symmetry in the resistor network. In a symmetric bridge, the resistor joining two equipotential points carries no current.

From the solution, the circuit is symmetric about the vertical line through the middle. Hence, the potentials at the midpoints of the top and bottom branches are equal, so no current flows through the central $3\,\Omega$ resistor.

After removing that branch, the network reduces to two parallel branches between $A$ and $B$:

• Top branch: $6\,\Omega + 3\,\Omega = 9\,\Omega$
• Bottom branch: $3\,\Omega + 6\,\Omega = 9\,\Omega$

Now the equivalent resistance is

$$R_{\text{eq}}=\frac{9\times 9}{9+9}=\frac{81}{18}=4.5\,\Omega$$

Given that

$$\frac{x}{5}=4.5$$

So,

$$x=22.5$$

Therefore, the value of $x$ is $22.5$.

Given: A symmetric resistor network is connected between $A$ and $B$.

Find: The value of $x$ if the equivalent resistance is $\frac{x}{5}\,\Omega$.

Because the circuit is symmetric, the two central junctions are at the same potential. Therefore, the bridge resistor of $3\,\Omega$ has zero current and can be ignored.

That leaves two equal branches of $9\,\Omega$ each in parallel, so identical resistors in parallel give half their value:

$$R_{\text{eq}}=\frac{9}{2}=4.5\,\Omega$$

Now compare with the given form:

$$\frac{x}{5}=4.5 \Rightarrow x=22.5$$

Therefore, the correct numerical value is $22.5$.