Question

If an 18 $\Omega$ resistance wire is bent into the shape of an equilateral triangle, then the equivalent resistance across the ends of any side = ?

• A. $ 2 \, \Omega $
• B. $ 4 \, \Omega $
• C. $ 6 \, \Omega $
• D. $ 12 \, \Omega $

Hint:

Key points to remember:
- Total resistance divides equally among triangle sides
- Any two vertices have one direct path and one indirect path
- Parallel resistance formula is crucial for such problems
- For equilateral triangle, symmetry simplifies calculations

Answer 1

The Correct Option is B

Step 1: Understand the configuration
When the wire is bent into an equilateral triangle: - Total resistance = 18 $\Omega$ - Each side of triangle = 6 $\Omega$ (since 18 $\Omega$/3 sides = 6 $\Omega$ per side)
Step 2: Visualize the circuit
When measuring resistance across any two vertices (let's say A and B): - One direct path: side AB = 6 $\Omega$ - Two parallel paths: side AC + side CB = 6 $\Omega$ + 6 $\Omega$ = 12 $\Omega$
Step 3: Calculate equivalent resistance
The circuit becomes: \[ R_{eq} = R_{AB} \parallel (R_{AC} + R_{CB}) \] \[ R_{eq} = 6 \, \Omega \parallel 12 \, \Omega \] \[ R_{eq} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \, \Omega \]
Step 4: Verify the calculation
Parallel resistance formula confirms: \[ \frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \] \[ R_{eq} = 4 \, \Omega \]