Question:
The maximum current that can flow in the fuse wire before it blows out, varies with the radius \(r\) as
The maximum current that can flow in the fuse wire before it blows out, varies with the radius \(r\) as
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Fuse wire current rating increases with radius. Thicker wire can carry more current.
Updated On: Apr 23, 2026
- \(r^{3/2}\)
- \(r\)
- \(r^{2/3}\)
- \(r^{1/2}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Heat generated \(H = I^2 R\). Heat dissipated \(\propto\) surface area \(\propto r\). Also \(R \propto \frac{1}{r^2}\).
Step 2: Detailed 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: Final Answer:
Thus, \(I_{\text{max}} \propto r^{3/2}\).
Heat generated \(H = I^2 R\). Heat dissipated \(\propto\) surface area \(\propto r\). Also \(R \propto \frac{1}{r^2}\).
Step 2: Detailed 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: Final Answer:
Thus, \(I_{\text{max}} \propto r^{3/2}\).
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