University Physics II · topic UPII-06.07
Series and parallel resistor networks
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.
Series & Parallel Resistor Networks: predict a finite change
Two resistors are in series when the same current must pass through both in turn, and in parallel when both are tied to the same pair of nodes, so the same potential difference sits across both; the equivalent resistance is the single resistor that would draw the same current from the same terminal voltage. Use the supplied topic model Y_out = 12/(r1+r2*r3/(r2+r3)), where Y_out is CURRENT FROM 12 V (A). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r2 is Parallel branch R₂, measured in Ω. Keep r1 (Series resistor R₁) = 20 Ω; r3 (Parallel branch R₃) = 40 Ω fixed. For this exercise use only 15.0938 Ω ≤ r2 ≤ 89.9063 Ω 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 r2 = 31.5525 Ω and 42.0262 Ω. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr2 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CURRENT FROM 12 V against r2
Predictions from the supplied Series & Parallel Resistor Networks 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): CURRENT FROM 12 V / A
Horizontal axis (x): r2: Parallel branch R₂ / Ω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r2: Parallel branch R₂ / Ω | CURRENT FROM 12 V / A |
|---|---|
| 15.09375 | 0.38761451080982046 |
| 33.796875 | 0.31316167532324013 |
| 52.5 | 0.2810126582278481 |
| 71.203125 | 0.26308915039122666 |
| 89.90625 | 0.2516597719705378 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r2 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r2 = 31.5525 Ω, substitute into Y_out = 12/(r1+r2*r3/(r2+r3)): Y₀ = 0.31882 A.
- At r2 = 42.0262 Ω, the same fixed-parameter model gives Y₁ = 0.29634 A.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0224801 A; the input change is Δr2 = 10.4737 Ω.
- Average sensitivity is ΔY_out/Δr2 = -0.00214633 A per Ω. 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 = 0.31882 A, Y₁ = 0.29634 A; ΔY_out = -0.0224801 A; average sensitivity = -0.00214633 A per Ω.
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.
Series & Parallel Resistor Networks: infer an input with uncertainty
Two resistors are in series when the same current must pass through both in turn, and in parallel when both are tied to the same pair of nodes, so the same potential difference sits across both; the equivalent resistance is the single resistor that would draw the same current from the same terminal voltage. Use the supplied topic model Y_out = 12/(r1+r2*r3/(r2+r3)), where Y_out is CURRENT FROM 12 V (A). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r2 is Parallel branch R₂, measured in Ω. Keep r1 (Series resistor R₁) = 20 Ω; r3 (Parallel branch R₃) = 40 Ω fixed. For this exercise use only 15.0938 Ω ≤ r2 ≤ 89.9063 Ω 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 = 0.27201105 A, with a bounded reading uncertainty ±0.0030527 A. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00748125 Ω, 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.
CURRENT FROM 12 V against r2
Predictions from the supplied Series & Parallel Resistor Networks 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): CURRENT FROM 12 V / A
Horizontal axis (x): r2: Parallel branch R₂ / Ω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r2: Parallel branch R₂ / Ω | CURRENT FROM 12 V / A |
|---|---|
| 15.09375 | 0.38761451080982046 |
| 33.796875 | 0.31316167532324013 |
| 52.5 | 0.2810126582278481 |
| 71.203125 | 0.26308915039122666 |
| 89.90625 | 0.2516597719705378 |
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(r2) = 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(15.0938) = 0.387615 A and Y_out(89.9063) = 0.25166 A. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r2_line = 15.0938 + (0.272011 − 0.387615)(89.9063 − 15.0938)/(0.25166 − 0.387615) = 78.7074 Ω. 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 r2 = 60.7294 Ω; substituting back gives Y_out = 0.27201105 A.
- The interpolation error in the input is |r2_line − r2| = 17.9781 Ω. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.26895835 A and 0.27506375 A. Sort the roots even if the response decreases: 57.7174 Ω ≤ r2 ≤ 64.008 Ω.
- The lower and upper input deviations are 3.01199 Ω and 3.27866 Ω. 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 r2 ≈ 60.7294 Ω; compatible interval [57.7174, 64.008] Ω. Endpoint interpolation gives 78.7074 Ω.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.