A heating element has a resistance of $100 Ω$ at room temperature. When it is connected to a supply of $220 V$ , a steady current of $2 A$ passes in it and temperature is $500^{\circ} C$ more than room temperature. What is the temperature coefficient of resistance of the heating element?

1 \times 10^{- 4}^{\circ} C^{- 1}

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5 \times 10^{- 4}^{\circ} C^{- 1}

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2 \times 10^{- 4}^{\circ} C^{- 1}

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0.5 \times 10^{- 4}^{\circ} C^{- 1}

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Solution

The correct option is C $2 \times 10^{- 4}^{\circ} C^{- 1}$
We know that resistance $R = \frac{V}{i} = \frac{220}{2} = R_{o} (1 + α Δ t)$
Also $R_{o}$ is given as $100 Ω$
Putting $Δ T = 500 K$
We get $α = 2 \times 10^{- 4}$

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A heating element has a resistance of 100Ω at room temperature. When it is connected to a supply of 220 V, a steady current of 2A passes in it and temperature is 500∘C more than room temperature. What is the temperature coefficient of resistance of the heating element?

The correct option is C 2×10−4 ∘C−1We know that resistance R=Vi=2202=Ro(1+αΔt)Also Ro is given as 100ΩPutting ΔT=500KWe get α=2×10−4

A heating element has a resistance of $100 Ω$ at room temperature. When it is connected to a supply of $220 V$ , a steady current of $2 A$ passes in it and temperature is $500^{\circ} C$ more than room temperature. What is the temperature coefficient of resistance of the heating element?

The correct option is C $2 \times 10^{- 4}^{\circ} C^{- 1}$
We know that resistance $R = \frac{V}{i} = \frac{220}{2} = R_{o} (1 + α Δ t)$
Also $R_{o}$ is given as $100 Ω$
Putting $Δ T = 500 K$
We get $α = 2 \times 10^{- 4}$