When a wire is stretched, its volume remains constant. Resistance R ∝ L/A. Since V=AL is constant, A ∝ 1/L. So R ∝ L². Also, A ∝ d², so L ∝ 1/d². This means R ∝ (1/d²)² = 1/d⁴. If the diameter is halved, the resistance increases by a factor of (1/(1/2))⁴ = 2⁴ = 16. [cite: 186, 187, 191, 192, 193]