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Karnataka Board PUCPUC Science 2nd PUC Class 12

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

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Question

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 × 10−4 °C−1.

Numerical
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Solution

Room temperature, T = 27°C

Resistance of the heating element at T, R = 100 Ω

Let T1 is the increased temperature of the filament.

Resistance of the heating element at T1, R1 = 117 Ω

Temperature co-efficient of the material of the filament,

α = 1.70 × 10−4 °C−1

α is given by the relation 

α = `("R"_1 - "R")/("R"("T"_1 - "T"))`

`"T"_1 - "T" = ("R"_1 - "R")/("R"α)`

`"T"_1 - 27 = (117 - 100)/(100(1.7 xx 10^-4))`

`"T"_1 - 27 = 1000`

T1 = 1027°C

Therefore, at 1027°C, the resistance of the element is 117 Ω.

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Chapter 3: Current Electricity - EXERCISES [Page 105]

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NCERT Physics Part I and II [English] Class 12
Chapter 3 Current Electricity
EXERCISES | Q 3.3 | Page 105
NCERT Physics Part I and II [English] Class 12
Chapter 3 Current Electricity
Exercise | Q 3.5 | Page 127

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