MCQMediumJEE 2024Combination of Resistors

JEE Physics 2024 Question with Solution

The effective resistance between A and B, if the resistance of each resistor is RR, will be:

  • A

    2R3\frac{2R}{3}

  • B

    8R3\frac{8R}{3}

  • C

    5R3\frac{5R}{3}

  • D

    4R3\frac{4R}{3}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The resistance of each resistor is RR.

Find: The effective resistance between A and B.

To find the effective resistance between points A and B, where each resistor has a resistance RR, we analyze the given circuit diagram step-by-step:

Circuit between A and B with outer resistors and a symmetric diamond bridge containing multiple resistors of value R, including horizontal and vertical middle branches.
  1. Identify the configuration: The circuit forms a bridge.
  2. Analyze symmetry: From symmetry, two middle resistances can be removed.
  3. The simplified circuit becomes:
Stepwise simplification of the resistor network showing the symmetric bridge reduced to three parallel branches of 2R, then to 2R by 3, then total 8R by 3 between A and B.

After symmetry-based reduction, the network between the two inner junctions consists of three branches, each of resistance

2R2R

in parallel.

Hence their equivalent resistance is

1Req=12R+12R+12R=32R\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{2R} + \frac{1}{2R} = \frac{3}{2R}

so,

Req=2R3R_{eq} = \frac{2R}{3}

Now this equivalent resistance is in series with the left and right resistors, each of resistance RR. Therefore,

RAB=R+2R3+RR_{AB} = R + \frac{2R}{3} + R RAB=8R3R_{AB} = \frac{8R}{3}

Therefore, the effective resistance between A and B is 8R3\frac{8R}{3}. The correct option is B.

Common mistakes

  • Assuming the entire network can be reduced directly without using symmetry is wrong because the middle branches must first be identified as ineffective or removable. Use symmetry before applying series-parallel combinations.

  • Treating the three inner branches as series instead of parallel is incorrect because they connect across the same two junctions. Combine them using the parallel formula.

  • Forgetting to add the two outer resistors in series after simplifying the bridge gives only the inner equivalent resistance. After finding 2R3\frac{2R}{3}, add one RR on each side.

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