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Difficulty: MediumElectric Current and Resistance

A tungsten filament lamp operating at a room temperature of 20 ∘C20\,^\circ\text{C} draws a current of 0.50 A0.50\,\text{A} when connected to a 120 V120\,\text{V} power source. When the lamp reaches its steady operating temperature, the current drops to 0.10 A0.10\,\text{A} under the same voltage. If the temperature coefficient of resistance of tungsten is 4.0×10−3 ∘C−14.0 \times 10^{-3}\,^\circ\text{C}^{-1}, what is the operating temperature of the filament?

  1. 1020 ∘C1020\,^\circ\text{C}Answer
  2. B
    1000 ∘C1000\,^\circ\text{C}
  3. C
    270 ∘C270\,^\circ\text{C}
  4. D
    200 ∘C200\,^\circ\text{C}

Answer

The operating temperature of the tungsten filament is 1020 ∘C1020\,^\circ\text{C}.
Using Ohm's law (R=V/IR = V/I), the initial resistance at 20 ∘C20\,^\circ\text{C} is 240 Ω240\,\Omega and the operating resistance is 1200 Ω1200\,\Omega. Substituting these into R2=R1[1+α(T2−T1)]R_2 = R_1[1 + \alpha(T_2 - T_1)] yields 1200=240[1+4.0×10−3(T2−20)]1200 = 240[1 + 4.0\times 10^{-3}(T_2 - 20)]. Solving gives T2−20=1000 ∘CT_2 - 20 = 1000\,^\circ\text{C}, so the operating temperature is 1020 ∘C1020\,^\circ\text{C}.

Step-by-Step Solution

1
Calculate the resistance of the filament at room temperature (20 ∘C20\,^\circ\text{C}) and at the operating temperature using Ohm's Law (R=V/IR = V/I).
Cold resistance R1=120 V0.50 A=240 ΩR_1 = \frac{120\,\text{V}}{0.50\,\text{A}} = 240\,\Omega; Hot resistance R2=120 V0.10 A=1200 ΩR_2 = \frac{120\,\text{V}}{0.10\,\text{A}} = 1200\,\Omega.
Ohm's law relates voltage, current, and resistance for a conductor.
2
Apply the temperature dependence equation of electrical resistance: R2=R1[1+α(T2−T1)]R_2 = R_1[1 + \alpha(T_2 - T_1)].
1200240=1+(4.0×10−3)(T2−20)⇒5=1+(4.0×10−3)(T2−20)\frac{1200}{240} = 1 + (4.0 \times 10^{-3})(T_2 - 20) \Rightarrow 5 = 1 + (4.0 \times 10^{-3})(T_2 - 20).
Resistance increases linearly with temperature according to the material's temperature coefficient.
3
Solve for the temperature difference ΔT=T2−20\Delta T = T_2 - 20 and find the final temperature T2T_2.
4.0×10−3(T2−20)=4⇒T2−20=44.0×10−3=1000 ∘C⇒T2=1020 ∘C4.0 \times 10^{-3}(T_2 - 20) = 4 \Rightarrow T_2 - 20 = \frac{4}{4.0 \times 10^{-3}} = 1000\,^\circ\text{C} \Rightarrow T_2 = 1020\,^\circ\text{C}.
Adding the initial temperature to the temperature change gives the final absolute operating temperature.

Key Concept

Temperature Dependence of Electrical Resistance and Ohm's Law
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Topics

Electric Current and Resistance

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