If the electrical conductivity of a fuse wire is doubled (all dimensions unchanged), by what factor does the safe (maximum) current it can carry change?

2 times
sqrt(2) times
1/2 times
1/sqrt(2) times

Correct answer: sqrt(2) times

Solution

At the fusing limit, I² R = constant (rate of heat dissipation). Since R = 1/(sigma) * (L/A), R is inversely proportional to conductivity sigma, so I_safe proportional to sqrt(sigma); doubling sigma multiplies I_safe by sqrt(2).

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