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} $.

Answer 2

Step 1: Understand the setup.
We are given two identical wires, each having resistance \( R \).
- The first wire is bent into an equilateral triangle.
- The second wire is bent into a square.
We need to find the ratio of the effective resistance between two adjacent vertices (end points of one side) of the triangle to that of the square.

Step 2: For the triangular wire.
When the wire is bent into an equilateral triangle, the total length of the wire is divided into three equal parts. Hence, the resistance of each side is: \[ R_t = \frac{R}{3}. \] Between two adjacent vertices, say \( A \) and \( B \):
- One direct path is the side \( AB \) having resistance \( \frac{R}{3} \).
- The other path from \( A \) to \( B \) goes through the remaining two sides (via the third vertex), which are in series, giving total resistance \( \frac{R}{3} + \frac{R}{3} = \frac{2R}{3}. \)

These two paths are in parallel, so the effective resistance between \( A \) and \( B \) is: \[ \frac{1}{R_{AB}} = \frac{1}{R/3} + \frac{1}{2R/3} = \frac{3}{R} + \frac{3}{2R} = \frac{9}{2R}. \] Hence, \[ R_{AB} = \frac{2R}{9}. \]

Step 3: For the square wire.
When the same wire is bent into a square, each side has resistance: \[ R_s = \frac{R}{4}. \] Between two adjacent vertices, say \( A \) and \( B \):
- The direct path has resistance \( \frac{R}{4} \).
- The other path (around the remaining three sides) has total resistance \( 3 \times \frac{R}{4} = \frac{3R}{4}. \)

These two paths are in parallel, so: \[ \frac{1}{R'_{AB}} = \frac{1}{R/4} + \frac{1}{3R/4} = \frac{4}{R} + \frac{4}{3R} = \frac{16}{3R}. \] Hence, \[ R'_{AB} = \frac{3R}{16}. \]

Step 4: Calculate the ratio.
The required ratio of resistances is: \[ \frac{R_{triangle}}{R_{square}} = \frac{\frac{2R}{9}}{\frac{3R}{16}} = \frac{2}{9} \times \frac{16}{3} = \frac{32}{27}. \]

Final Answer:
\[ \boxed{\frac{32}{27}} \]