InterviewSolution
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InterviewSolution
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At room temperature `(27.0^@C)` the resistance of a heating element is `100 Omega`. What is the temperature of the element if the resistane is found to be `117 Omega`, given that the temperature co-efficicent of the material of the resistor is `1.70 xx 10^(-4) .^@C^(-1)`. |
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Answer» Here, `R_(27) = 100 Omega , R_(t) = 117 Omega , t = ? , alpha = 1.70 xx 10^(-4) .^(@)C^(-1)` We know that, `alpha = R_(t)-R_(27)/(R_(27) (t - 27)) or t- 27 = (R_(t)-R_(27))/(R_(27)xx alpha)` `:. t = (R_(t)-R_(27))/( R_(27) xx alpha) + 27 = ( 117 - 100)/(100 xx 1.7 xx 10^(-4)) + 27 = 1000 + 27 = 1027^@C` |
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