University Physics II · topic UPII-06.05
Resistivity, geometry, and temperature
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Resistivity, geometry, and temperature: predict a finite change
Resistivity ρ is a material's own opposition to current: the electric field needed inside it per unit of current density, E = ρJ, measured in ohm metres and independent of how much of the material there is or what shape it takes. Use the supplied topic model Y_out = 0.0213901*len/(dia*dia)*(1+0.0039*(temp-20)), where Y_out is RESISTANCE R (Ω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dia is Wire diameter d, measured in mm. Keep len (Wire length L) = 12 m; temp (Temperature T) = 20 °C fixed. For this exercise use only 1.10625 mm ≤ dia ≤ 1.89375 mm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at dia = 1.2795 mm and 1.38975 mm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdia across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RESISTANCE R against dia
Predictions from the supplied Resistivity, geometry, and temperature model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RESISTANCE R / Ω
Horizontal axis (x): dia: Wire diameter d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dia: Wire diameter d / mm | RESISTANCE R / Ω |
|---|---|
| 1.10625 | 0.2097430087139711 |
| 1.3031249999999999 | 0.15115478770940086 |
| 1.5 | 0.11408053333333333 |
| 1.696875 | 0.08914445997781915 |
| 1.8937499999999998 | 0.0715729255301768 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of dia separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At dia = 1.2795 mm, substitute into Y_out = 0.0213901*len/(dia*dia)*(1+0.0039*(temp-20)): Y₀ = 0.156788 Ω.
- At dia = 1.38975 mm, the same fixed-parameter model gives Y₁ = 0.132899 Ω.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0238895 Ω; the input change is Δdia = 0.11025 mm.
- Average sensitivity is ΔY_out/Δdia = -0.216685 Ω per mm. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.156788 Ω, Y₁ = 0.132899 Ω; ΔY_out = -0.0238895 Ω; average sensitivity = -0.216685 Ω per mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Resistivity, geometry, and temperature: infer an input with uncertainty
Resistivity ρ is a material's own opposition to current: the electric field needed inside it per unit of current density, E = ρJ, measured in ohm metres and independent of how much of the material there is or what shape it takes. Use the supplied topic model Y_out = 0.0213901*len/(dia*dia)*(1+0.0039*(temp-20)), where Y_out is RESISTANCE R (Ω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dia is Wire diameter d, measured in mm. Keep len (Wire length L) = 12 m; temp (Temperature T) = 20 °C fixed. For this exercise use only 1.10625 mm ≤ dia ≤ 1.89375 mm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.10196367 Ω, with a bounded reading uncertainty ±0.0045586 Ω. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dia by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875 mm, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
RESISTANCE R against dia
Predictions from the supplied Resistivity, geometry, and temperature model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RESISTANCE R / Ω
Horizontal axis (x): dia: Wire diameter d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dia: Wire diameter d / mm | RESISTANCE R / Ω |
|---|---|
| 1.10625 | 0.2097430087139711 |
| 1.3031249999999999 | 0.15115478770940086 |
| 1.5 | 0.11408053333333333 |
| 1.696875 | 0.08914445997781915 |
| 1.8937499999999998 | 0.0715729255301768 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(dia) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.10625) = 0.209743 Ω and Y_out(1.89375) = 0.0715729 Ω. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dia_line = 1.10625 + (0.101964 − 0.209743)(1.89375 − 1.10625)/(0.0715729 − 0.209743) = 1.72054 mm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives dia = 1.58663 mm; substituting back gives Y_out = 0.10196367 Ω.
- The interpolation error in the input is |dia_line − dia| = 0.133913 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.09740507 Ω and 0.10652227 Ω. Sort the roots even if the response decreases: 1.5523 mm ≤ dia ≤ 1.62333 mm.
- The lower and upper input deviations are 0.0343209 mm and 0.0367029 mm. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer dia ≈ 1.58663 mm; compatible interval [1.5523, 1.62333] mm. Endpoint interpolation gives 1.72054 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.