University Physics II · topic UPII-07.04
Potential dividers and circuit measuring instruments
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.
Potential Dividers & Measuring Instruments: predict a finite change
A potential divider is two resistors in series across a supply with the output taken across one of them, so the output is the supply voltage multiplied by that resistor's share of the total resistance. Use the supplied topic model Y_out = 12*f*rl/max(rl+f*(1-f)*rt,0.01), where Y_out is LOADED OUTPUT (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. rl is Load resistance R_L, measured in kΩ. Keep rt (Track resistance R_T) = 10 kΩ; f (Wiper fraction f) = 0.5 fixed. For this exercise use only 6.20625 kΩ ≤ rl ≤ 44.7937 kΩ 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 rl = 14.6955 kΩ and 20.0977 kΩ. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrl across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LOADED OUTPUT against rl
Predictions from the supplied Potential Dividers & Measuring Instruments 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): LOADED OUTPUT / V
Horizontal axis (x): rl: Load resistance R_L / kΩ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rl: Load resistance R_L / kΩ | LOADED OUTPUT / V |
|---|---|
| 6.20625 | 4.277099784637472 |
| 15.853124999999999 | 5.182700493785118 |
| 25.5 | 5.464285714285714 |
| 35.146874999999994 | 5.6015605544948945 |
| 44.793749999999996 | 5.682833355358795 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rl separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rl = 14.6955 kΩ, substitute into Y_out = 12*f*rl/max(rl+f*(1-f)*rt,0.01): Y₀ = 5.12768 V.
- At rl = 20.0977 kΩ, the same fixed-parameter model gives Y₁ = 5.33622 V.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.208538 V; the input change is Δrl = 5.40225 kΩ.
- Average sensitivity is ΔY_out/Δrl = 0.0386021 V per kΩ. Its positive sign means that this response increases 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 = 5.12768 V, Y₁ = 5.33622 V; ΔY_out = 0.208538 V; average sensitivity = 0.0386021 V per kΩ.
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.
Potential Dividers & Measuring Instruments: infer an input with uncertainty
A potential divider is two resistors in series across a supply with the output taken across one of them, so the output is the supply voltage multiplied by that resistor's share of the total resistance. Use the supplied topic model Y_out = 12*f*rl/max(rl+f*(1-f)*rt,0.01), where Y_out is LOADED OUTPUT (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. rl is Load resistance R_L, measured in kΩ. Keep rt (Track resistance R_T) = 10 kΩ; f (Wiper fraction f) = 0.5 fixed. For this exercise use only 6.20625 kΩ ≤ rl ≤ 44.7937 kΩ 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 = 5.5348062 V, with a bounded reading uncertainty ±0.022204 V. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00385875 kΩ, 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.
LOADED OUTPUT against rl
Predictions from the supplied Potential Dividers & Measuring Instruments 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): LOADED OUTPUT / V
Horizontal axis (x): rl: Load resistance R_L / kΩ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rl: Load resistance R_L / kΩ | LOADED OUTPUT / V |
|---|---|
| 6.20625 | 4.277099784637472 |
| 15.853124999999999 | 5.182700493785118 |
| 25.5 | 5.464285714285714 |
| 35.146874999999994 | 5.6015605544948945 |
| 44.793749999999996 | 5.682833355358795 |
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(rl) = 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(6.20625) = 4.2771 V and Y_out(44.7937) = 5.68283 V. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives rl_line = 6.20625 + (5.53481 − 4.2771)(44.7937 − 6.20625)/(5.68283 − 4.2771) = 40.7304 kΩ. 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 rl = 29.7446 kΩ; substituting back gives Y_out = 5.5348062 V.
- The interpolation error in the input is |rl_line − rl| = 10.9858 kΩ. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 5.5126022 V and 5.5570102 V. Sort the roots even if the response decreases: 28.2757 kΩ ≤ rl ≤ 31.3608 kΩ.
- The lower and upper input deviations are 1.46894 kΩ and 1.6162 kΩ. 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 rl ≈ 29.7446 kΩ; compatible interval [28.2757, 31.3608] kΩ. Endpoint interpolation gives 40.7304 kΩ.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.