Q.48 A metallic wire of electrical resistance 40 Ω is bent in the form of a square loop. The resistance
between any two diagonally opposite corners is ________ Ω.
A metallic wire of electrical resistance 40 Ω bent into a square loop presents a classic physics problem on equivalent resistance between diagonally opposite corners. This calculation is crucial for competitive exams like JEE Main, NEET, and CSIR NET, testing series-parallel combinations in symmetric circuits.
Problem Setup
A uniform metallic wire with total resistance 40 Ω bends into a square loop ABCD. Each side has resistance R_side = 40/4 = 10 Ω since resistance distributes proportionally along the length. Measure resistance between corners A and C (diagonally opposite).
Circuit Analysis
Label corners A, B, C, D clockwise. Between A and C:
- Path 1: A → B → C (
10 Ω + 10 Ω = 20 Ω) - Path 2: A → D → C (
10 Ω + 10 Ω = 20 Ω)
These paths connect in parallel. Equivalent resistance:
R_eq = (20 × 20) / (20 + 20) = 400/40 = 10 Ω.
Common Options Explained
- 20 Ω: Resistance of one path (ignores parallel combination).
- 5 Ω: Incorrect parallel of sides (
10 ∥ 10, misses series segments). - 40 Ω: Total wire resistance (ignores reconfiguration).
- 10 Ω: Correct, as two 20 Ω paths in parallel.
Verification
This matches examples like 12 Ω wire (diagonal 3 Ω) scaled proportionally. Symmetry confirms no other paths contribute significantly.
Step-by-Step Solution
- Divide Total Resistance
The uniform wire forms four equal sides:R_side = 40/4 = 10 Ω. - Identify Paths
For diagonal points A-C:
Path ABC:10 + 10 = 20 Ω
Path ADC:10 + 10 = 20 Ω. - Parallel Combination
R_eq = (20 × 20) / (20 + 20) = 10 Ω.
Why 10 Ω?
Two identical 20 Ω paths halve the effective resistance via parallel rule 1/R = 1/R1 + 1/R2. Adjacent corners yield 20 Ω (series), but diagonal symmetry creates parallelism.
Exam Tips
- Always split uniform wire resistance by segments.
- Visualize current paths for diagonals.
- Practice variants (e.g., 12 Ω → 3 Ω).
