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At room temperature (28.0^(@) C)the resistance of a heating element is 100 Omega . What ic. the temperature of the element if the resistance is found to be 117Omega, given that the temperature coefficient of the material of the resistor is 1.70 xx 10^(-4)""^(@) C^(-1) . |
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Answer» SOLUTION :If resistances of a given wire are `R_(1) and R_(Z)` at temperatures `T_(1) and T_(2)` respectively then, we have the formula, `R_(2) = R_(1) (1 + alpha (T_(2) - T_(1))) ` `therefore 117 = 100 { 1 + (1.7 XX 10^(-4)) (T_(2) - 27)} ` `therefore 1.17 = 1 + (1.7 xx 10^(-4) ) (T_(2) - 27) ` `therefore 0.17 = 1.7 xx 10^(-4) T_(2) - 45.9 xx 10^(-4) ` `therefore 0.17459= 1.7 xx 10^(-4) T_(2)` `therefore T_(2) = (0.17459)/(1.7 xx 10^(-4)) = (1745.9)/(1.7) = 1027^(@)` C |
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