Two uniform wires of same material , each weighing `1 g` but one having double the length of the other, are connected in series , carrying a current of `10 A`. The length of the longer wire is `20 cm`. Calculate the rate of consumption of energy in each of the two wires. which wire gets hotter ? The density of the material of the wire is `20 xx 10^(-5) Omega cm`.

Explanation

Let `a_(1) and a_(2)` be the areas of cross section of the short and the long wieres , respectively. Since ` mass = volume xx density = alrho`,
` 1 = a_(1) xx 10 xx 11 = a_(2) xx 20 xx 11`
or `a_(1) = (1) / ( 10 xx 11) cm^(2) and a_(2) = (1) /( 20 xx 11 ) cm^(2)`
`:. R_(1) = 20 xx 10^(-5) xx ( 10)/( 1//( 10 xx 11 ))`
`= 20 xx 10^(-5) xx 10 xx 10 xx 11 = 22 xx 10 ^(-2) Omega`
`R_(2) = 20 xx 10 ^(-5) xx ( 20 ) /( 1//( 20 xx 11)) = 88 xx 10 ^(-2) Omega`
And rate of heat produced is
`H_(1) = I^(2) R _(1) = (10)^(2) xx 22 xx 10 ^(-2) = 22 W`
`H_(2) = I^(2)R_(2) = (10)^(2) xx 88 xxx 10^(-2) = 88 W`
Thus , the wire of longer length gets hotter.