Question

Difficulty: Very hardElectric Current and Resistance

A resistance thermometer has a resistance of 4.0 Ω4.0\,\Omega at 0 ∘C0\,^\circ\text{C} and 6.0 Ω6.0\,\Omega at 100 ∘C100\,^\circ\text{C}. The thermometer is connected in series with a fixed 10.0 Ω10.0\,\Omega resistor across a 24.0 V24.0\,\text{V} DC power source having an internal resistance of 1.0 Ω1.0\,\Omega. If a steady current of 1.2 A1.2\,\text{A} flows through the circuit, what is the temperature of the thermometer's environment?

  1. 250 ∘C250\,^\circ\text{C}Answer
  2. B
    300 ∘C300\,^\circ\text{C}
  3. C
    450 ∘C450\,^\circ\text{C}
  4. D
    350 ∘C350\,^\circ\text{C}

Answer

The temperature of the thermometer environment is 250 ∘C250\,^\circ\text{C}.
By applying the complete circuit equation E=I(Rθ+Rfixed+r)E = I(R_\theta + R_{\text{fixed}} + r), the total circuit resistance is found to be 20.0 Ω20.0\,\Omega. Subtracting the fixed resistance of 10.0 Ω10.0\,\Omega and the cell internal resistance of 1.0 Ω1.0\,\Omega yields the thermometer resistance Rθ=9.0 ΩR_\theta = 9.0\,\Omega. Substituting this into the thermometric relation θ=Rθ−R0R100−R0×100 ∘C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100\,^\circ\text{C} gives 9.0−4.06.0−4.0×100=250 ∘C\frac{9.0 - 4.0}{6.0 - 4.0} \times 100 = 250\,^\circ\text{C}.

Step-by-Step Solution

1
Calculate total circuit resistance using Ohm's Law and internal resistance equation
Rtotal=EI=24.0 V1.2 A=20.0 ΩR_{\text{total}} = \frac{E}{I} = \frac{24.0\,\text{V}}{1.2\,\text{A}} = 20.0\,\Omega
The electromotive force (e.m.f) of the source equals total current multiplied by total resistance including internal resistance.
2
Determine the resistance of the thermometer at the unknown temperature (RθR_\theta)
Rθ=Rtotal−Rfixed−r=20.0 Ω−10.0 Ω−1.0 Ω=9.0 ΩR_\theta = R_{\text{total}} - R_{\text{fixed}} - r = 20.0\,\Omega - 10.0\,\Omega - 1.0\,\Omega = 9.0\,\Omega
The circuit components are in series, so total resistance is the sum of external resistances and internal resistance.
3
Apply the linear resistance thermometer temperature scale formula
\(\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100\,^\circ\text{C} = \frac{9.0 - 4.0}{6.0 - 4.0} \times 100 = \frac{5.0}{2.0} \times 100 = 250\,^\circ\text{C}\)
Resistance varies linearly with temperature between the ice point (0 ∘C0\,^\circ\text{C}) and steam point (100 ∘C100\,^\circ\text{C}).

Key Concept

Integration of Ohm's Law, internal resistance of a cell, and resistance thermometry
Estimated Time:2m 0s
Rate this question