University Physics II · topic UPII-07.02
Kirchhoff's loop rule
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.
Kirchhoff's Loop Rule: predict a finite change
Kirchhoff's loop rule states that the algebraic sum of the potential changes around any closed path in a circuit is zero, provided no changing magnetic flux passes through that path. Use the supplied topic model Y_out = (e1-e2)/30, where Y_out is LOOP CURRENT I (A). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. e1 is Driving emf ε₁, measured in V. Keep e2 (Opposing emf ε₂) = 6 V; r1 (Resistance R₁ (R₂ = 30 Ω − R₁)) = 10 Ω fixed. For this exercise use only 2.55 V ≤ e1 ≤ 21.45 V 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 e1 = 6.708 V and 9.354 V. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δe1 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LOOP CURRENT I against e1
Predictions from the supplied Kirchhoff's Loop Rule 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): LOOP CURRENT I / A
Horizontal axis (x): e1: Driving emf ε₁ / V
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| e1: Driving emf ε₁ / V | LOOP CURRENT I / A |
|---|---|
| 2.55 | -0.115 |
| 7.2749999999999995 | 0.04249999999999998 |
| 12 | 0.2 |
| 16.724999999999998 | 0.35749999999999993 |
| 21.45 | 0.515 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of e1 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At e1 = 6.708 V, substitute into Y_out = (e1-e2)/30: Y₀ = 0.0236 A.
- At e1 = 9.354 V, the same fixed-parameter model gives Y₁ = 0.1118 A.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0882 A; the input change is Δe1 = 2.646 V.
- Average sensitivity is ΔY_out/Δe1 = 0.0333333 A per V. 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 = 0.0236 A, Y₁ = 0.1118 A; ΔY_out = 0.0882 A; average sensitivity = 0.0333333 A per V.
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.
Kirchhoff's Loop Rule: infer an input with uncertainty
Kirchhoff's loop rule states that the algebraic sum of the potential changes around any closed path in a circuit is zero, provided no changing magnetic flux passes through that path. Use the supplied topic model Y_out = (e1-e2)/30, where Y_out is LOOP CURRENT I (A). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. e1 is Driving emf ε₁, measured in V. Keep e2 (Opposing emf ε₂) = 6 V; r1 (Resistance R₁ (R₂ = 30 Ω − R₁)) = 10 Ω fixed. For this exercise use only 2.55 V ≤ e1 ≤ 21.45 V 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 = 0.2693 A, with a bounded reading uncertainty ±0.036855 A. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown e1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00189 V, 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.
LOOP CURRENT I against e1
Predictions from the supplied Kirchhoff's Loop Rule 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): LOOP CURRENT I / A
Horizontal axis (x): e1: Driving emf ε₁ / V
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| e1: Driving emf ε₁ / V | LOOP CURRENT I / A |
|---|---|
| 2.55 | -0.115 |
| 7.2749999999999995 | 0.04249999999999998 |
| 12 | 0.2 |
| 16.724999999999998 | 0.35749999999999993 |
| 21.45 | 0.515 |
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(e1) = 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(2.55) = -0.115 A and Y_out(21.45) = 0.515 A. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives e1_line = 2.55 + (0.2693 − -0.115)(21.45 − 2.55)/(0.515 − -0.115) = 14.079 V. 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 e1 = 14.079 V; substituting back gives Y_out = 0.2693 A.
- The interpolation error in the input is |e1_line − e1| = 0 V. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.232445 A and 0.306155 A. Sort the roots even if the response decreases: 12.9733 V ≤ e1 ≤ 15.1846 V.
- The lower and upper input deviations are 1.10565 V and 1.10565 V. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer e1 ≈ 14.079 V; compatible interval [12.9733, 15.1846] V. Endpoint interpolation gives 14.079 V.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.