Current Electricity

JEE Main level, single correct, medium

258 of 380

Question 258 of 380

A heating element has a resistance of 100Ω100\Omega at room temperature. When it is connected to a supply of 220V220V, a steady current of 2A2A passes in it and temperature is 500oC500{}^{o}C more than the room temperature. The temperature coefficient of resistance of the heating element is

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Answer: (B) 2×10−4oC−12\times {10}^{-4}{}^{o}{C}^{-1}

Resistance of heating element after temperature increases by 500oC=2202=110Ω500{}^{o}C=\dfrac{220}{2}=110\Omega (from Ohm's law). The new resistance is related to the temperature coefficient of the heating element as 110=100(1+α500)110=100(1+\alpha 500)

α=10100×500\alpha =\dfrac{10}{100\times 500}\Rightarrow \alpha =2\times {10}^{-4}^{\circ}{C}^{-1}

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