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JEE Main 2025 (7 Apr, Shift 1) · Combination of Resistors

Combination of Resistors question from JEE Main 2025 (7 Apr, Shift 1)

Q10JEE Main 2025 · 7 APR · SHIFT 1Medium

Combination of Resistors

A wire of resistance RR is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points AA and BB is R/nR/n. The value of nn is:

Figure

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Answer: (c)

  1. The pyramid (tetrahedron) has 6 equal edges, each of resistance r=R6r = \dfrac{R}{6}.
  2. By symmetry the other two vertices are at the same potential, so the edge joining them carries no current.
  3. Paths: direct rr, plus two paths of 2r2r each, all in parallel: 1RAB=1r+12r+12r=2r\dfrac{1}{R_{AB}} = \dfrac{1}{r} + \dfrac{1}{2r} + \dfrac{1}{2r} = \dfrac{2}{r}
  4. RAB=r2=R12R_{AB} = \dfrac{r}{2} = \dfrac{R}{12}, so n=12n = 12.
  5. So the answer is (c) 12.
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