Question

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:

• A. \( \frac{9}{8} \)
• B. \( \frac{27}{32} \)
• C. \( \frac{32}{27} \)
• D. \( \frac{8}{9} \)

Hint:

For resistance in series and parallel, remember that series sums add directly, while parallel resistances use the reciprocal sum formula.

Answer 1

The Correct Option is B

Consider resistance \( R \) split into three for the triangle and four for the square. The resistance between two corners of the triangle is \( \frac{R}{3} \) in series with two other \( \frac{R}{3} \) resistances in parallel. For the square, it is \( \frac{R}{4} \) in series with two \( \frac{R}{4} \) resistances in parallel. Simplifying the equivalent resistances and taking their ratio gives:

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