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MCQEasyJEE 2025Combination of Resistors

JEE Physics 2025 Question with Solution

A wire of resistance RR is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is:

  • A

    98\frac{9}{8}

  • B

    2732\frac{27}{32}

  • C

    3227\frac{32}{27}

  • D

    89\frac{8}{9}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Each wire has total resistance RR. One wire is bent into an equilateral triangle and the other into a square. Resistance is required between the two end points of one edge in each case.

Find: The ratio RtriangleRsquare\dfrac{R_{\text{triangle}}}{R_{\text{square}}}.

For the equilateral triangle, the wire is divided into 33 equal sides, so resistance of each side is

R3\frac{R}{3}

Between the end points of one side, there are two parallel paths:

  • direct side: R3\frac{R}{3}
  • remaining two sides in series: 2R3\frac{2R}{3}

Therefore, the equivalent resistance of the triangle is

Rtriangle=(R3)(2R3)R3+2R3=2R29R=2R9R_{\text{triangle}} = \frac{\left(\frac{R}{3}\right)\left(\frac{2R}{3}\right)}{\frac{R}{3} + \frac{2R}{3}} = \frac{\frac{2R^2}{9}}{R} = \frac{2R}{9}

For the square, the wire is divided into 44 equal sides, so resistance of each side is

R4\frac{R}{4}

Between the end points of one side, there are again two parallel paths:

  • direct side: R4\frac{R}{4}
  • remaining three sides in series: 3R4\frac{3R}{4}

Therefore, the equivalent resistance of the square is

Rsquare=(R4)(3R4)R4+3R4=3R216R=3R16R_{\text{square}} = \frac{\left(\frac{R}{4}\right)\left(\frac{3R}{4}\right)}{\frac{R}{4} + \frac{3R}{4}} = \frac{\frac{3R^2}{16}}{R} = \frac{3R}{16}

Now the required ratio is

RtriangleRsquare=2R93R16=2R9×163R=3227\frac{R_{\text{triangle}}}{R_{\text{square}}} = \frac{\frac{2R}{9}}{\frac{3R}{16}} = \frac{2R}{9} \times \frac{16}{3R} = \frac{32}{27}

Therefore, the correct option is C, that is 3227\frac{32}{27}.

The solution marks option B, but the worked calculation shown in the solution gives 3227\frac{32}{27}, which matches option C.

Reciprocal Parallel Form

Given: Total resistance of each wire is RR.

Find: Effective resistance ratio for adjacent vertices of the triangle and the square.

For the triangle,

1Rtriangle=1R/3+12R/3=3R+32R=92R\frac{1}{R_{\text{triangle}}} = \frac{1}{R/3} + \frac{1}{2R/3} = \frac{3}{R} + \frac{3}{2R} = \frac{9}{2R}

Hence,

Rtriangle=2R9R_{\text{triangle}} = \frac{2R}{9}

For the square,

1Rsquare=1R/4+13R/4=4R+43R=163R\frac{1}{R_{\text{square}}} = \frac{1}{R/4} + \frac{1}{3R/4} = \frac{4}{R} + \frac{4}{3R} = \frac{16}{3R}

Hence,

Rsquare=3R16R_{\text{square}} = \frac{3R}{16}

Now,

RtriangleRsquare=2R93R16=3227\frac{R_{\text{triangle}}}{R_{\text{square}}} = \frac{\frac{2R}{9}}{\frac{3R}{16}} = \frac{32}{27}

Therefore, the required ratio is 3227\frac{32}{27} and the correct option is C.

Common mistakes

  • Treating the two paths between the chosen vertices as series instead of parallel. This is wrong because current can split into the direct edge and the alternate path simultaneously. First identify the two distinct paths, then combine them using the parallel formula.

  • Assuming each side has resistance RR. This is wrong because RR is the resistance of the entire wire, which is divided equally among 33 sides for the triangle and 44 sides for the square. Use side resistances R/3R/3 and R/4R/4 respectively.

  • Using the incorrect alternate-path resistance for the square or triangle. The non-direct path must include the remaining sides in series: 2R/32R/3 for the triangle and 3R/43R/4 for the square. Count the sides carefully before applying parallel combination.

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