Question

A fuse wire with a radius of $0.1mm$ blows up at $5A$. The radius of another fuse wire made of the same material which will blow up at $40A$ is _____.

• A. $0.8mm$
• B. $0.6mm$
• C. $0.45mm$
• D. $0.4mm$

Answer 1

Answer: (D)

$0.4mm$

When the circuit is switched on, the temperature of the fuse will increase up to a certain value (below the melting point) and then remains constant At this equilibrium stage, the heat lost by the radiation per second balances the heat liberated per second. That is
$\left(2\pi rI\right)H=I^{2}R$
where $H$ is the rate of loss of heat per unit area, $I$ is the current and $R$ the resistance of the fuse wire.
But $R=\frac{\rho l}{\pi r^2}$
$\therefore H\left(2\pi rl\right)=\frac{I^2\rho l}{\pi r^2}$ or $r^3=\frac{\rho I_2}{2{\pi }^{2}H}$
For a given material of the fuse wire, $\rho$ and $H$ are constants.
$r^3\propto I^2$
If $r_1$ and $r_2$ are the radii of two fuse wires of the same material which blow $u$ at currents $I_1$ and $I_2$,
respectively, then
${\left(\frac{r_2}{r_1}\right)}^3=\left(\frac{I_2}{I_1}\right),$ i.e., $r_2=r_1{\left(\frac{I_2}{I_1}\right)}^{\frac{2}{3}}$
Given: $I_1=5A,r_1=0.1mm,I_2=40A$,
$r_2=0.1{\left(\frac{40}{5}\right)}^{\frac{2}{3}}=0.1\left(4\right)=0.4mm$