Question:medium

A wire of resistance R 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:

Show Hint

For resistance in series and parallel, remember that series sums add directly, while parallel resistances use the reciprocal sum formula.
Updated On: Mar 25, 2026
  • \( \frac{9}{8} \)
  • \( \frac{27}{32} \)
  • \( \frac{32}{27} \)
  • \( \frac{8}{9} \)
Show Solution

The Correct Option is B

Solution and Explanation

The total resistance \( R \) is divided into three parts for a triangle configuration and four parts for a square configuration. For the triangle, the resistance between two corners is \( \frac{R}{3} \) in series with two parallel \( \frac{R}{3} \) resistances. For the square, the resistance between two corners is \( \frac{R}{4} \) in series with two parallel \( \frac{R}{4} \) resistances. The ratio of the simplified equivalent resistances is as follows:

\[ \text{Ratio} = \frac{\frac{R/3}{2} + R/3}{\frac{R/4}{2} + R/4} = \frac{27}{32} \]

Was this answer helpful?
0


Questions Asked in JEE Main exam