Make the fuse wire twice as thick.

A fuse wire blows at 5 A. A wire of the same material but twice the diameter is fitted. Assuming heat is lost only through the surface, at a fixed rate per unit area at the melting point, it blows at about:

Answer: 14 A.

At the melting point, the heat produced per second equals the heat lost per second through the surface. The heat produced falls as 1/r² through the resistance, and the heat lost grows as r through the surface area.

I^{2} \propto r^{3}

\Rightarrow \frac{I_{2}}{I_{1}} = {(\frac{2 r}{r})}^{3 / 2} = 2 \sqrt{2} \approx 2.83

So the thicker wire blows at about 2.83 × 5 ≈ 14 A, not 10 A. Doubling the diameter nearly triples the safe current, which is why a fuse's rating is set by the thickness of its wire.

Original practice question · JEE Physics Decoded

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