University Physics II · topic UPII-07.05
Charging an RC circuit
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.
Charging an RC circuit: predict a finite change
Charging an RC circuit is the transient that follows closing a switch on a series emf, resistor and capacitor: charge and capacitor voltage climb exponentially towards their final values while the current decays, all on the single time scale τ = RC. Use the supplied topic model Y_out = 12*(1-exp(-t/max(r*c,0.01))), where Y_out is v_C AT CURSOR (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. t is Cursor time t, measured in ms. Keep r (Resistance R) = 40 kΩ; c (Capacitance C) = 2.5 μF fixed. For this exercise use only 42.5 ms ≤ t ≤ 357.5 ms 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 = 111.8 ms and 155.9 ms. 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.
v_C AT CURSOR against t
Predictions from the supplied Charging an RC circuit 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): v_C AT CURSOR / V
Horizontal axis (x): t: Cursor time t / ms
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Cursor time t / ms | v_C AT CURSOR / V |
|---|---|
| 42.5 | 4.154762578441833 |
| 121.25 | 8.430567392139112 |
| 200 | 10.375976601160648 |
| 278.75 | 11.261100491386413 |
| 357.5 | 11.66381489070934 |
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 = 111.8 ms, substitute into Y_out = 12*(1-exp(-t/max(r*c,0.01))): Y₀ = 8.0768 V.
- At t = 155.9 ms, the same fixed-parameter model gives Y₁ = 9.47584 V.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.39904 V; the input change is Δt = 44.1 ms.
- Average sensitivity is ΔY_out/Δt = 0.0317243 V per ms. 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 = 8.0768 V, Y₁ = 9.47584 V; ΔY_out = 1.39904 V; average sensitivity = 0.0317243 V per ms.
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.
Charging an RC circuit: infer an input with uncertainty
Charging an RC circuit is the transient that follows closing a switch on a series emf, resistor and capacitor: charge and capacitor voltage climb exponentially towards their final values while the current decays, all on the single time scale τ = RC. Use the supplied topic model Y_out = 12*(1-exp(-t/max(r*c,0.01))), where Y_out is v_C AT CURSOR (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. t is Cursor time t, measured in ms. Keep r (Resistance R) = 40 kΩ; c (Capacitance C) = 2.5 μF fixed. For this exercise use only 42.5 ms ≤ t ≤ 357.5 ms 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 = 10.851558 V, with a bounded reading uncertainty ±0.12184 V. 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 0.0315 ms, 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.
v_C AT CURSOR against t
Predictions from the supplied Charging an RC circuit 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): v_C AT CURSOR / V
Horizontal axis (x): t: Cursor time t / ms
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Cursor time t / ms | v_C AT CURSOR / V |
|---|---|
| 42.5 | 4.154762578441833 |
| 121.25 | 8.430567392139112 |
| 200 | 10.375976601160648 |
| 278.75 | 11.261100491386413 |
| 357.5 | 11.66381489070934 |
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(42.5) = 4.15476 V and Y_out(357.5) = 11.6638 V. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 42.5 + (10.8516 − 4.15476)(357.5 − 42.5)/(11.6638 − 4.15476) = 323.426 ms. 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 = 234.65 ms; substituting back gives Y_out = 10.851558 V.
- The interpolation error in the input is |t_line − t| = 88.7763 ms. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 10.729718 V and 10.973398 V. Sort the roots even if the response decreases: 224.567 ms ≤ t ≤ 245.865 ms.
- The lower and upper input deviations are 10.0833 ms and 11.2152 ms. 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 ≈ 234.65 ms; compatible interval [224.567, 245.865] ms. Endpoint interpolation gives 323.426 ms.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.