SCIENCE · INVESTIGATE AND EXPLAIN

Solve an RC initial-value problem with energy return

The common zero-charge formula is only one initial-value problem. Take a maintained 6 V source, resistance 2000 Ω and capacitance 500 µF, so RC=1 s. Initial voltage Vi can be below, equal to or above 6 V, including a negative polarity. Positive current enters the capacitor positive plate.

Learning outcome: Compare Vi=−2,6,10 V at the same time. Before revealing values, predict current sign, source-work sign and whether field energy rises or falls. Retain the full finite-time energy ledger rather than labeling every source as always delivering energy.

Starting recap

Review charge continuity, definite integrals, first-order differential equations, initial conditions and signed source work. Distinguish displayed time from physical time.

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Try

With Vi=10 V at t=1 s, V=6+4/e≈7.471518 V and current is −2/e mA≈−0.735759 mA. Source work is about −0.00758545 J: energy has returned to the source. Resistor conversion remains positive, about 0.00345866 J, and the capacitor loses the sum of those two outgoing amounts.

Open the experiment

Predict

For Vi=10 V at t=1 s, what is the signed current?

Your prediction and reason

Experiment

Compare Vi=−2,6,10 V at the same time. Before revealing values, predict current sign, source-work sign and whether field energy rises or falls. Retain the full finite-time energy ledger rather than labeling every source as always delivering energy.

Solve RC relaxation from an arbitrary initial voltage

Worked teaching model · synthetic values · Ideal maintained 6 V source, 2000 ohm resistor and 500 microfarad capacitor, time constant 1 s. Signed current enters the capacitor positive plate. Initial voltage is prescribed, including opposite polarity or above-source states. No breakdown, parasitic, switching impulse or physical apparatus model.

Initial capacitor voltage: 10 V Elapsed time: 1 s

Reset keeps your prediction, captures and practice history.

General initial-state relaxation

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: analytic RC solution (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for General initial-state relaxation

analytic RC solution

time (s)	capacitor voltage (V)
0	10
0.05	9.804918
0.1	9.61935
0.15	9.442832
0.2	9.274923
0.25	9.115203
0.3	8.963273
0.35	8.818752
0.4	8.68128
0.45	8.550513
0.5	8.426123
0.55	8.307799
0.6	8.195247
0.65	8.088183
0.7	7.986341
0.75	7.889466
0.8	7.797316
0.85	7.70966
0.9	7.626279
0.95	7.546964
1	7.471518
1.05	7.399751
1.1	7.331484
1.15	7.266547
1.2	7.204777
1.25	7.146019
1.3	7.090127
1.35	7.036961
1.4	6.986388
1.45	6.938281
1.5	6.892521
1.55	6.848992
1.6	6.807586
1.65	6.7682
1.7	6.730734
1.75	6.695096
1.8	6.661196
1.85	6.628949
1.9	6.598274
1.95	6.569096
2	6.541341
2.05	6.51494
2.1	6.489826
2.15	6.465937
2.2	6.443213
2.25	6.421597
2.3	6.401035
2.35	6.381477
2.4	6.362872
2.45	6.345174
2.5	6.32834
2.55	6.312327
2.6	6.297094
2.65	6.282605
2.7	6.268822
2.75	6.255711
2.8	6.24324
2.85	6.231377
2.9	6.220093
2.95	6.209359
3	6.199148
3.05	6.189436
3.1	6.180197
3.15	6.171409
3.2	6.163049
3.25	6.155097
3.3	6.147533
3.35	6.140337
3.4	6.133493
3.45	6.126983
3.5	6.12079
3.55	6.114899
3.6	6.109295
3.65	6.103965
3.7	6.098894
3.75	6.094071
3.8	6.089483
3.85	6.085119
3.9	6.080968
3.95	6.077019
4	6.073263

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Capacitor voltage	7.471518	V
Signed capacitor current	-0.0007357589	A
Stored-energy change	-0.01104411	J
Source work	-0.007585447	J
Resistor conversion	0.003458659	J
Energy residual	-4.33681e-19	J
Initial voltage slope	-4	V/s

The initial deviation from the final voltage, not the final voltage alone, decays exponentially.

Kirchhoff and I=C dV/dt give RC dV/dt+V=6. Since RC=1 s, V=6+(Vinitial−6)e^(−t). Integrating 6 I and I²R gives source work and nonnegative resistor conversion. An initially overcharged capacitor sends energy back into the maintained source while still dissipating heat. At Vinitial=6 there is no transient. Source work alone is not equal to stored-energy change.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

The initial deviation from the final voltage, not the final voltage alone, decays exponentially.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

State the current, source and measurement model

Derive the integral or initial-value solution

Kirchhoff and I=C dV/dt give RC dV/dt+V=6. Subtract the equilibrium voltage and separate variables to obtain V(t)=6+(Vi−6)e^(−t). Current is (6−Vi)e^(−t)/2000. The sign is part of the solution: an initially overcharged capacitor drives current back into the ideal maintained source.

Test conservation, uncertainty and limits

Integrate source power to get Ws=6C[V(t)−Vi]. Resistor conversion is C(Vi−6)²[1−e^(−2t)]/2, always nonnegative. Stored-energy change is C[V(t)²−Vi²]/2, and Ws=ΔU+Qresistor. This identity distinguishes negative source work from negative dissipation and includes the exact no-transient state Vi=6 V.

Worked original case

Investigation

Changed-case practice

Try the changed case before feedback

For Vi=−2 V at t=ln 2 seconds, find capacitor voltage in volts.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original derivations and synthetic evidence checked 31 October 2026 against the primary explanation. No live circuit construction, energized switching, component rating or electrical handling instruction is requested. Preserve first predictions, critique, revision and fresh assigned work separately from public practice. Software agreement does not certify measurement technique or learning.

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Which initial state, calibration or constitutive assumption is needed for this conclusion, and what fresh evidence could test it?

Your explanation and limitations

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Previous: Solve node voltages and audit delivered power Next: Match the state instead of restarting the transient

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