Question

A fuse with a circular cross-sectional radius of $0.15\,mm$ blows at $15\,A$. What is the radius of a fuse, made of the same material which will blow at $30\,A$ ?
A. $0.45$
B. $0.23$
C. $0.35$
D. $0.24$

Answer 1

Here we have to use the relation that the radius of the wire is directly proportional to the current. From this relation we can get the radius of the fuse.

Complete step by step answer:
Given,
Radius, ${r_1} = 0.15\,mm$
Current, ${I_1} = 15\,A$
Current, ${I_2} = 30\,A$
Radius, ${r_2} = ?$
Here the heat loss of the wire is given by:
$h = \dfrac{{{I^2}\rho }}{{2{\pi ^2}{r^3}}}$
Therefore, from the above equation we can see that:
{r^3} \propto {I^3} \\\ \Rightarrow \dfrac{{{r_2}^3}}{{{r_1}^3}} = \dfrac{{{I_2}^2}}{{{I_1}^2}} \\\
\implies {r_2}^3 = \dfrac{{{I_2}^2}}{{{I_1}^2}} \times {r_1}^3 \\\
$= {\left( {\dfrac{{30}}{{15}}} \right)^2} \times {\left( {0.15} \right)^3} \\\$
$\therefore {r_2} = {\left( 4 \right)^{\dfrac{1}{3}}} \times \left( {0.15} \right)\,mm = 0.24\,mm$

So, the correct answer is “Option D”.

Additional Information:
Heat loss: Heat loss is a measure of the complete flow of heat from inside to outside, whether from conduction, convection, radiation or some other processes.
There is a relationship between surface area and heat loss. The higher the surface area, the greater is the heat loss.
Heat conduction happens inside a body or between two bodies in contact, also called as diffusion. That is the direct exchange of particles with kinetic energy across the boundary of two systems. Heat transfer by the conduction process to reach thermal equilibrium when an object is at a temperature different from its atmosphere or some other body.
Fuse: A fuse is an electrical safety system that acts to provide an electrical circuit with over current protection. A metal wire or strip that melts when too much current passes through it, thereby blocking or interrupting the current is a fuse.
An electrical conductor’s electrical resistance is the opposition to the passing through the conductor of an electric current.

Note:
Here we have to be careful while calculating the radius as there are two currents and two radii. So, we have to be alert while putting the values.
The thickness of a wire is inversely proportional to the cross-sectional area of the wire. Although the wire cross-section area is proportional to the radius square, the wire resistance is inversely proportional to the radius square of the wire.