Physics - Current Electricity Question with Solution | TestHub

The heat produced by a current $I$ flowing through the fuse wire in time $t$ is given by Joule's law of heating:

$H=I^{2}Rt=I^2\left(\rho \frac{l}{\pi r^2}\right)t$

where $\rho$ is the resistivity, $l$ is the length, and $r$ is the radius of the wire.

The heat required to raise the temperature of the fuse wire to its melting point is:

$Q=ms\Delta \theta =(\pi r^{2}ld)s\Delta \theta$

where $m$ is the mass, $d$ is the density, $s$ is the specific heat, and $\Delta \theta$ is the required change in temperature.

Assuming negligible heat loss to the surroundings (adiabatic heating for a fast-blowing fuse), we equate the heat produced to the heat required:

$I^2\left(\rho \frac{l}{\pi r^2}\right)t=\pi r^{2}lds\Delta \theta$

Solving for $t$, we get:

$t=\frac{{\pi }^2{r}^{4}ds\Delta \theta }{I^2\rho }$

For a given current $I$, the time of heating $t$ is proportional to $r^4$ and is independent of the length $l$.

Therefore, $t\propto r^4{l}^0$.

Answer: $t\propto r^4{l}^0$