SCIENCE · INVESTIGATE AND EXPLAIN

Use a nonzero initial voltage in an RC transient

A capacitor need not start uncharged. This packet begins with 100 µF at 2 V, then an ideal switch connects it through R to a maintained 6 V source. With finite resistor-limited current, capacitor voltage cannot jump at the switching instant. The initial resistor drop is 6−2=4 V, so initial current is 4/R, not 6/R.

Learning outcome: Compare t=0,τ and a long-time limit for two resistances. Retain initial and final states before evaluating an intermediate value. Verify source work equals energy change plus heat, and explain why a larger R changes timescale but not the ideal final stored energy.

Starting recap

Use charge continuity, energy per charge, consistent SI units and an explicit circuit topology. Exponential responses are supplied algebraic models; initial state, source constraint and measurement loading must be declared.

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Try

At R=2 kΩ, τ=.2 s. After .2 s, Vc=6−4/e≈4.528482 V and I=(4/2000)/e≈.735759 mA. Initial current is 2 mA. Final capacitor energy gain is 1.6 mJ, source work 2.4 mJ and total resistor heat .8 mJ.

Open the experiment

Predict

For this initially 2 V capacitor and 6 V source, initial current is:

Your prediction and reason

Experiment

Compare t=0,τ and a long-time limit for two resistances. Retain initial and final states before evaluating an intermediate value. Verify source work equals energy change plus heat, and explain why a larger R changes timescale but not the ideal final stored energy.

Resolve a nonzero initial capacitor voltage after switching

Worked teaching model · synthetic values · Ideal 100 µF capacitor initially at 2 V, connected through R to an ideal 6 V source at t=0. Supplied algebraic law Vc=6−4 exp(−t/RC); no calculus proof required. Positive current enters the positive capacitor plate. No real switching or capacitor handling.

Series resistance: 2 kΩ Elapsed time after switch: 0.2 s

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Nonzero initial voltage approaches the maintained source

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

Equivalent numerical coordinates for Nonzero initial voltage approaches the maintained source

supplied exponential response

time (s)	capacitor voltage (V)
0	2
0.025	2.47001
0.05	2.8848
0.075	3.25084
0.1	3.57388
0.125	3.85895
0.15	4.11053
0.175	4.33255
0.2	4.52848
0.225	4.70139
0.25	4.85398
0.275	4.98864
0.3	5.10748
0.325	5.21235
0.35	5.3049
0.375	5.38658
0.4	5.45866
0.425	5.52227
0.45	5.5784
0.475	5.62794
0.5	5.67166
0.525	5.71024
0.55	5.74429
0.575	5.77434
0.6	5.80085
0.625	5.82425
0.65	5.8449
0.675	5.86313
0.7	5.87921
0.725	5.8934
0.75	5.90593
0.775	5.91698
0.8	5.92674
0.825	5.93535
0.85	5.94294
0.875	5.94965
0.9	5.95556
0.925	5.96079
0.95	5.96539
0.975	5.96946
1	5.97305

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
Time constant	0.2	s
Capacitor voltage	4.52848	V
Charge on positive plate	0.000452848	C
Charging current	0.000735759	A
Resistor power	0.00108268	W
Source work since switching	0.00151709	J
Capacitor energy gain	0.000825358	J
Resistor heat since switching	0.000691732	J
Energy residual	0	J
Initial current	0.002	A
Final capacitor voltage limit	6	V

The initial current depends on source minus initial capacitor voltage, not source voltage alone.

At t=0 the capacitor is already at 2 V; its voltage cannot jump with finite resistor-limited current. After one RC interval the remaining 4 V gap is multiplied by exp(−1), about .368. At long times current approaches zero and voltage approaches 6 V. The supplied heat expression closes the integrated energy account without requiring calculus; it is not obtained by multiplying final power by elapsed time.

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 current depends on source minus initial capacitor voltage, not source voltage alone.

Observation to attach to a capture

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

Understand

Define topology, recipient and initial conditions

Derive the quantitative response

The supplied response is Vc(t)=6−4exp(−t/RC), with current I(t)=(4/R)exp(−t/RC). The time constant RC has seconds as its unit. After one time constant, the remaining gap to the final voltage is exp(−1)≈.368 times its initial value. The actual capacitor voltage is therefore about 4.52848 V, not necessarily 63.2% of the final voltage when the initial state is nonzero. The exponential law is supplied for algebraic evaluation; a calculus proof is not required.

Check evidence and the limits of the model

Energy must be integrated over the changing process rather than estimated from final power times time. Here source work since switching is 6C[Vc(t)−2], capacitor energy change is C[Vc(t)²−4]/2, and their difference is resistor heat C(4²)[1−exp(−2t/RC)]/2. These supplied expressions can be checked algebraically. At t=0 every accumulated change is zero; at long time current approaches zero while finite heat has accumulated. Disconnecting or reversing the source would require a separately stated circuit and initial condition.

Worked original case

Investigation

Changed-case practice

Try the changed case before feedback

For the stated circuit with R=4 kΩ, find the time constant in seconds.

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 evidence boundary

Original numerical cases and prose checked 30 October 2026 against the relevant primary explanation. All records are synthetic and all diagrams mathematical. No practical wiring, battery testing, capacitor handling, mains measurement or equipment safety rating is requested. Public feedback does not verify independent mastery; retain first predictions, contrasting recipes and criterion-based critique in the separate course evidence workspace.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Which independent observation could reject this model rather than merely refit its parameters?

Your explanation and limitations

Continue

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Skipping retains experiments and awards no mastery.

Previous: Close a dielectric-change energy ledger under two constraints Next: Infer a decay time separately from a voltage offset

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