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. \( \frac{32}{27} \)
• C. \( \frac{27}{32} \)
• 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

Let the resistance of the wire be $ R $.

Equilateral Triangle:
Each side has a resistance of $ \frac{R}{3} $.
If we consider two endpoints of one side of the triangle:

• Resistance along that side is $ \frac{R}{3} $.
• Resistance along the other two sides is $ \frac{2R}{3} $.

Since these two paths are in parallel:

$$ R_{\text{triangle}} = \frac{\left(\frac{R}{3}\right)\left(\frac{2R}{3}\right)}{\frac{R}{3} + \frac{2R}{3}} = \frac{\frac{2R^2}{9}}{R} = \frac{2R}{9} $$

Square:
Each side has a resistance of $ \frac{R}{4} $.
If we consider two endpoints of one side of the square:

• Resistance along that side is $ \frac{R}{4} $.
• Resistance along the other three sides is $ \frac{3R}{4} $.

Since these two paths are in parallel:

$$ R_{\text{square}} = \frac{\left(\frac{R}{4}\right)\left(\frac{3R}{4}\right)}{\frac{R}{4} + \frac{3R}{4}} = \frac{\frac{3R^2}{16}}{R} = \frac{3R}{16} $$

The ratio of the resistance of the triangle to that of the square is:

$$ \frac{R_{\text{triangle}}}{R_{\text{square}}} = \frac{\frac{2R}{9}}{\frac{3R}{16}} = \frac{2R}{9} \times \frac{16}{3R} = \frac{32}{27} $$

Final Answer:
The final answer is $ \frac{32}{27} $.