A wire has a resistance of `2.5 Omega ` at `100^@` C. Temperature coefficient of resistance of the material `alpha = 3.6 xx 10^(-3) K^(-1) ` at `25^@` C. Find its resistance at `25^@ C`.

Explanation

`R = 2.5 Omega , t_2 = 100^@ C`
`alpha = 3.6 xx 10^(-3) K^(-1) , R_1 = ?`
`t_2 = 100+ 273 = 373 K . T_1 = 25^@ C = 273 + 25 = 298 K`
`alpha =(R_2 -t_1)/(R_1 (t_2 -t_1))`
`3.6xx10^(-3) = (2.5 - R_1)/(R_1 (373-298)) = (2.5 - R_1)/(R_1 xx 75)`
`3.6xx10^(-3) xx 75 xx R_1 = 2.5 - R_1`
`0.270 xx R_1 = 2.5 - R_1`
`0.270 xx R_1 + R_1 = 2.5`
`1.27 R_1 = 2.5 implies R_1 = 1.96 Omega`
Resistance at `25^@ C , R_1 = 1.97 Omega`