Two wires made of tinned copper having identical cross section `( = 10^(-6) m^(2))` and lengths `10 and 15 cm` are to be used as fuses. Show that the fuses will melt at the same value of current in each case.

Explanation

The temperature of the wire rises to a certain steady temperature when the heat produced per second by the current just becomes equal to the rate of loss of heat from its surface. Heat produced per second by the current is
`I^(2) R = I^(2) ( rho l )/( pi r^(2))`
Where `l` is the length , `r` is the radius , and `rho` is the specific resistance. Let `H` be the heat lost per second per unit surface area of the wire , then heat lost per second by the wire is `H xx "surface area of wire" = H xx 2 pi rl`
At steady state temperature , `H xx 2 pi r l = (I^(2) rho l )/( pi r^(2))`
or `H = (I^(2) rho ) / ( 2 pi ^(2) r^(3))` (i)
From Eq. (i) we note that the rate of loss of heat `(H)`, which , in turn , depends on the temperature of the wire . Hence , the cfuses of two wires of the same values of `r and rho` but of different lengths will melt for the same value of current in each case.