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(C) The heat produced in a fuse wire is given by $H = I^2 R t$. The fuse wire blows off when the heat generated reaches a critical value, which is pro

Vedclass · Accepted Answer

$5 \sqrt{\frac{27}{8}} \,A$

Vedclass · Answer

(C) The heat produced in a fuse wire is given by $H = I^2 R t$. The fuse wire blows off when the heat generated reaches a critical value, which is proportional to the surface area of the wire for heat dissipation. Thus, $I^2 R \propto r^2$. Since resistance $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$, we have $I^2 \left(\frac{1}{r^2}\right) \propto r^2$, which implies $I^2 \propto r^3$ or $I \propto r^{3/2}$.Given $r_1 = 0.2 \,mm$, $I_1 = 5 \,A$, and $r_2 = 0.3 \,mm$.Using the ratio $\frac{I_2}{I_1} = \left(\frac{r_2}{r_1}\right)^{3/2}$:$\frac{I_2}{5} = \left(\frac{0.3}{0.2}\right)^{3/2} = \left(\frac{3}{2}\right)^{3/2} = \sqrt{\frac{27}{8}}$.Therefore, $I_2 = 5 \sqrt{\frac{27}{8}} \,A$.

$A$ fuse wire of radius $0.2 \,mm$ blows off with a current of $5 \,A$. The fuse wire of the same material but of radius $0.3 \,mm$ will blow off with a current of:

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