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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 resistance is found to be 117Omega, given that the temperature coefficientof the material of the resistor is 1.70 xx 10^(-4) ""^@C^(-1) ? |
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Answer» Solution :Here `T_1 = 27.0^@C , R_1 = 100OMEGA , R_2 = 117 Omega " and " alpha = 1.70 xx 10^(-4) ""^@C^(-1)` `because R_2 =R_1 [1 + alpha (T_2 - T_1) ]` ` therefore 117 = 100 [1+1.7 xx 10^(-4) (T_2 -27) ]` ` RARR T_2 - 27 = (117 - 100)/(100 xx 1.7 xx 10^(-4)) = 1000^@C` `T_2 = 1000+27 = 1027^@C` |
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