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    				| 1. | A coil of wire has a resistance of `25.00Omega at 20^@ C` and a resistance of `25.17Omega at 150^@C`. What is its temperauture cofficient of resistance? | 
| Answer» `R = R_0[1+alpha(T - T_0)] or alpha = Delta R |(R_0 DeltaT),` with `Delta R =R -R_0 = 0.17Omega and Delta T =T - T_0 = 15^@C.` then `alpha = (0.17)/(25.00xxx15) = 4.5xx10^(-4)C^(-1)` | |