A uniform metal wire is stretched to double its original length and then bent to form a square. If the equivalent resistance across any two adjacent corners of the square is 15 , what was the resistance of the wire before it was stretched?

Step-by-step solution

Let the original resistance of the unstretched wire be R₀ . When the wire is stretched to double its length ( n = 2 ), its volume remains constant. The new resistance becomes: R_ new = n^2 R₀ = (2)^2 R₀ = 4 R₀ . The wire is then bent into a square, which has four equal sides. The resistance of each side of the square is: R_ side = 4 R₀ 4 = R₀ . Across any two adjacent corners, one side is in parallel with the series combination of the remaining three sides. Resistance of the three sides in series = R₀ + R₀ + R₀ = 3