Question

The maximum current that can flow in the fuse wire before it blows out, varies with the radius \(r\) as

• A. \(r^{3/2}\)
• B. \(r\)
• C. \(r^{2/3}\)
• D. \(r^{1/2}\)

Hint:

Fuse wire current rating increases with radius. Thicker wire can carry more current.

Answer 1

The Correct Option is A

Step 1: Concept:
Heat generated \(H = I^2 R\). Heat dissipated \(\propto\) surface area \(\propto r\). Also \(R \propto \frac{1}{r^2}\).
Step 2: Explanation:
At melting point: \(I^2 R \propto \text{surface area} \propto r\).
\(R = \rho \frac{l}{A} \propto \frac{1}{r^2}\). So \(I^2 \cdot \frac{1}{r^2} \propto r \Rightarrow I^2 \propto r^3 \Rightarrow I \propto r^{3/2}\).
Step 3: Conclusion:
Thus, \(I_{\text{max}} \propto r^{3/2}\).