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3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70\times 10^{-4}\degree C^{-1}.

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Given,

Temperature coefficient of filament, \alpha = 1.70\times 10^{-4}\degree C^{-1} 

T_1= 27\degree C ;  R_1 = 100 \Omega

Let T_2 be the temperature of element, R_2= 117\Omega 

(Positive \alpha means that the resistance increases with temperature. Hence we can deduce that T_2 will be greater than  T_1)

We know,

R_2 = R_1[1 + \alpha \Delta T]

\implies 117 = 100[1 +(1.70\times 10^{-4})(T_2 - 27) ]

\\ \Rightarrow T_2-27=\frac{117-100}{1.7\times10^{-4}}\\\Rightarrow T_2-27=1000\\\Rightarrow T_2=1027^\circ C

Hence, the temperature of the element is 1027 °C.

Posted by

HARSH KANKARIA

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