JEE Physics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: The resistance network is connected between A and B.

Find: The equivalent resistance between A and B.

The wheat stone bridge is in balanced condition.

So the central branch does not affect the equivalent resistance, and the circuit reduces to two series branches in parallel.

Upper branch resistance:

$$R + 3R = 4R$$

Lower branch resistance:

$$2R + 6R = 8R$$

Hence,

$$\frac{1}{R_{eq}} = \frac{1}{4R} + \frac{1}{8R}$$

Therefore,

$$R_{eq} = \frac{8R}{3}$$

The working gives equivalent resistance $\frac{8R}{3}$. Since this value corresponds to option D in the listed options, the correct option by the extracted working is D. The solution label showing C is inconsistent with the computed result.

Common mistakes

• Treating the bridge as unbalanced and including the $9R$ branch in current flow is incorrect because the solution states the wheat stone bridge is balanced. First check the balance condition, then remove the central branch from consideration.

• Adding all resistors directly is wrong because the network is not a single series path. After reducing the balanced bridge, identify the two series branches and then combine those branches in parallel.

• Using the parallel formula incorrectly for $4R$ and $8R$ leads to an incorrect equivalent resistance. Use

  $$\frac{1}{R_{eq}} = \frac{1}{4R} + \frac{1}{8R}$$

  and then invert carefully.