University Physics II · topic UPII-07.07
Energy and power in RC transients
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.
Energy and Power in RC Transients: predict a finite change
Energy accounting for an RC transient: multiply each term of the loop equation by the current to get instantaneous powers, then integrate them to compare the source's work, the resistor's heat, and the capacitor's stored energy. Use the supplied topic model Y_out = 14.4*(1-exp(-10*t/r)), where Y_out is BATTERY WORK εq (mJ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. t is Cursor time t, measured in s. Keep r (Series resistance R) = 2 kΩ fixed. For this exercise use only 0.1275 s ≤ t ≤ 1.0725 s 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 t = 0.3354 s and 0.4677 s. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
BATTERY WORK εq against t
Predictions from the supplied Energy and Power in RC Transients 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): BATTERY WORK εq / mJ
Horizontal axis (x): t: Cursor time t / s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Cursor time t / s | BATTERY WORK εq / mJ |
|---|---|
| 0.1275 | 6.787982818569638 |
| 0.36374999999999996 | 12.063910899485876 |
| 0.5999999999999999 | 13.683066215502759 |
| 0.8362499999999999 | 14.179976692138833 |
| 1.0724999999999998 | 14.332475973306074 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 0.3354 s, substitute into Y_out = 14.4*(1-exp(-10*t/r)): Y₀ = 11.7082 mJ.
- At t = 0.4677 s, the same fixed-parameter model gives Y₁ = 13.0108 mJ.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.30265 mJ; the input change is Δt = 0.1323 s.
- Average sensitivity is ΔY_out/Δt = 9.84617 mJ per s. 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 = 11.7082 mJ, Y₁ = 13.0108 mJ; ΔY_out = 1.30265 mJ; average sensitivity = 9.84617 mJ per s.
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.
Energy and Power in RC Transients: infer an input with uncertainty
Energy accounting for an RC transient: multiply each term of the loop equation by the current to get instantaneous powers, then integrate them to compare the source's work, the resistor's heat, and the capacitor's stored energy. Use the supplied topic model Y_out = 14.4*(1-exp(-10*t/r)), where Y_out is BATTERY WORK εq (mJ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. t is Cursor time t, measured in s. Keep r (Series resistance R) = 2 kΩ fixed. For this exercise use only 0.1275 s ≤ t ≤ 1.0725 s 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 = 13.973662 mJ, with a bounded reading uncertainty ±0.053822 mJ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000945 s, 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.
BATTERY WORK εq against t
Predictions from the supplied Energy and Power in RC Transients 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): BATTERY WORK εq / mJ
Horizontal axis (x): t: Cursor time t / s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Cursor time t / s | BATTERY WORK εq / mJ |
|---|---|
| 0.1275 | 6.787982818569638 |
| 0.36374999999999996 | 12.063910899485876 |
| 0.5999999999999999 | 13.683066215502759 |
| 0.8362499999999999 | 14.179976692138833 |
| 1.0724999999999998 | 14.332475973306074 |
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(t) = 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(0.1275) = 6.78798 mJ and Y_out(1.0725) = 14.3325 mJ. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 0.1275 + (13.9737 − 6.78798)(1.0725 − 0.1275)/(14.3325 − 6.78798) = 1.02756 s. 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 t = 0.70395 s; substituting back gives Y_out = 13.973662 mJ.
- The interpolation error in the input is |t_line − t| = 0.323606 s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 13.91984 mJ and 14.027484 mJ. Sort the roots even if the response decreases: 0.680173 s ≤ t ≤ 0.730941 s.
- The lower and upper input deviations are 0.0237774 s and 0.0269905 s. 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 t ≈ 0.70395 s; compatible interval [0.680173, 0.730941] s. Endpoint interpolation gives 1.02756 s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.