University Physics II · topic UPII-07.06
Discharging 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.
Discharging an RC circuit: predict a finite change
Discharging an RC circuit is what happens when a charged capacitor is connected across a resistor with no source in the loop: the stored charge drives its own current and dies away exponentially with time constant τ = RC. Use the supplied topic model Y_out = v0/r, where Y_out is INITIAL CURRENT V₀/R (mA). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Resistance R, measured in kΩ. Keep v0 (Starting voltage V₀) = 9 V; c (Capacitance C) = 20 μF fixed. For this exercise use only 26.375 kΩ ≤ r ≤ 73.625 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 r = 36.77 kΩ and 43.385 kΩ. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
INITIAL CURRENT V₀/R against r
Predictions from the supplied Discharging 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): INITIAL CURRENT V₀/R / mA
Horizontal axis (x): r: Resistance R / kΩ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Resistance R / kΩ | INITIAL CURRENT V₀/R / mA |
|---|---|
| 26.375 | 0.3412322274881517 |
| 38.1875 | 0.23567921440261866 |
| 50 | 0.18 |
| 61.8125 | 0.14560161779575329 |
| 73.625 | 0.12224108658743633 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r = 36.77 kΩ, substitute into Y_out = v0/r: Y₀ = 0.244765 mA.
- At r = 43.385 kΩ, the same fixed-parameter model gives Y₁ = 0.207445 mA.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0373198 mA; the input change is Δr = 6.615 kΩ.
- Average sensitivity is ΔY_out/Δr = -0.00564169 mA per kΩ. 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 = 0.244765 mA, Y₁ = 0.207445 mA; ΔY_out = -0.0373198 mA; average sensitivity = -0.00564169 mA 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.
Discharging an RC circuit: infer an input with uncertainty
Discharging an RC circuit is what happens when a charged capacitor is connected across a resistor with no source in the loop: the stored charge drives its own current and dies away exponentially with time constant τ = RC. Use the supplied topic model Y_out = v0/r, where Y_out is INITIAL CURRENT V₀/R (mA). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Resistance R, measured in kΩ. Keep v0 (Starting voltage V₀) = 9 V; c (Capacitance C) = 20 μF fixed. For this exercise use only 26.375 kΩ ≤ r ≤ 73.625 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 = 0.16305086 mA, with a bounded reading uncertainty ±0.0061215 mA. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.004725 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.
INITIAL CURRENT V₀/R against r
Predictions from the supplied Discharging 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): INITIAL CURRENT V₀/R / mA
Horizontal axis (x): r: Resistance R / kΩ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Resistance R / kΩ | INITIAL CURRENT V₀/R / mA |
|---|---|
| 26.375 | 0.3412322274881517 |
| 38.1875 | 0.23567921440261866 |
| 50 | 0.18 |
| 61.8125 | 0.14560161779575329 |
| 73.625 | 0.12224108658743633 |
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(r) = 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(26.375) = 0.341232 mA and Y_out(73.625) = 0.122241 mA. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 26.375 + (0.163051 − 0.341232)(73.625 − 26.375)/(0.122241 − 0.341232) = 64.8198 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 r = 55.1975 kΩ; substituting back gives Y_out = 0.16305086 mA.
- The interpolation error in the input is |r_line − r| = 9.62229 kΩ. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.15692936 mA and 0.16917236 mA. Sort the roots even if the response decreases: 53.2002 kΩ ≤ r ≤ 57.3506 kΩ.
- The lower and upper input deviations are 1.99732 kΩ and 2.15314 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 r ≈ 55.1975 kΩ; compatible interval [53.2002, 57.3506] kΩ. Endpoint interpolation gives 64.8198 kΩ.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.