SCIENCE · INVESTIGATE AND EXPLAIN

Retain initial current and the inductive energy cross term

An ideal linear inductor stores a current state rather than a capacitor voltage state. With a maintained 4 V source and resistance 2 Ω, the equation is L di/dt+2i=4. The independently prepared initial current i0 may oppose the source; no physical preparation or energized-switch procedure is requested.

Learning outcome: Compare i0=−1,2,4 A for the same L. Record initial slope, current at one second, source work, resistor conversion and stored-energy change. Explain why positive dissipative conversion does not require source power to be positive at every instant.

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

For L=1 H and i0=−1 A, τ=0.5 s and i(1)=2−3e^(−2)≈1.593994 A. Its slope is 6e^(−2)≈0.812012 A/s. The passive inductor voltage is about 0.812012 V and the resistor drop about 3.187988 V, adding to the 4 V source.

Open the experiment

Predict

For L=1 H and initial current −1 A, what is current at one second?

Your prediction and reason

Experiment

Compare i0=−1,2,4 A for the same L. Record initial slope, current at one second, source work, resistor conversion and stored-energy change. Explain why positive dissipative conversion does not require source power to be positive at every instant.

Solve an inductive transient with signed initial current

Worked teaching model · synthetic values · Ideal maintained 4 V source, 2 ohm resistor and linear inductor; analyze t=1 s and plot 0–3 s. Initial signed current is independently prepared. Passive voltage convention vL=L di/dt. No core saturation, physical opening of an energized inductor or handling instructions.

Inductance: 1 H Initial current: -1 A

Reset keeps your prediction, captures and practice history.

Signed current approaches the source-resistor equilibrium

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

Equivalent numerical coordinates for Signed current approaches the source-resistor equilibrium

general initial RL solution

time (s)	current (A)
0	-1
0.0375	-0.7832305
0.075	-0.5821239
0.1125	-0.3955487
0.15	-0.2224547
0.1875	-0.06186784
0.225	0.08711555
0.2625	0.2253339
0.3	0.3535651
0.3375	0.4725307
0.375	0.5829003
0.4125	0.685295
0.45	0.780291
0.4875	0.8684229
0.525	0.9501868
0.5625	1.026043
0.6	1.096417
0.6375	1.161707
0.675	1.222279
0.7125	1.278475
0.75	1.33061
0.7875	1.378977
0.825	1.42385
0.8625	1.465481
0.9	1.504103
0.9375	1.539935
0.975	1.573178
1.0125	1.604018
1.05	1.632631
1.0875	1.659176
1.125	1.683802
1.1625	1.70665
1.2	1.727846
1.2375	1.747511
1.275	1.765755
1.3125	1.782681
1.35	1.798383
1.3875	1.812952
1.425	1.826467
1.4625	1.839006
1.5	1.850639
1.5375	1.861431
1.575	1.871444
1.6125	1.880733
1.65	1.88935
1.6875	1.897346
1.725	1.904763
1.7625	1.911645
1.8	1.918029
1.8375	1.923952
1.875	1.929447
1.9125	1.934545
1.95	1.939274
1.9875	1.943662
2.025	1.947733
2.0625	1.95151
2.1	1.955013
2.1375	1.958264
2.175	1.96128
2.2125	1.964077
2.25	1.966673
2.2875	1.969081
2.325	1.971315
2.3625	1.973388
2.4	1.975311
2.4375	1.977095
2.475	1.97875
2.5125	1.980285
2.55	1.98171
2.5875	1.983031
2.625	1.984257
2.6625	1.985395
2.7	1.98645
2.7375	1.987429
2.775	1.988338
2.8125	1.98918
2.85	1.989962
2.8875	1.990687
2.925	1.99136
2.9625	1.991985
3	1.992564

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
Inductive time constant	0.5	s
Current at one second	1.593994	A
Current slope at one second	0.8120117	A/s
Inductor passive voltage	0.8120117	V
Source work through one second	2.812012	J
Resistor conversion through one second	2.041603	J
Stored-energy change	0.7704087	J
Energy residual	-2.22045e-16	J

Inductor current retains its initial state; a negative initial current can oppose the source before crossing zero.

L di/dt+2i=4 gives i=2+(iinitial−2)e^(−2t/L). The passive inductor voltage is L di/dt, while its induced emf uses the opposite orientation. Multiplying the differential equation by i gives source power = resistive conversion + d(Li²/2)/dt. The displayed finite-time integrals retain the cross term in i²; omitting it breaks energy closure.

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

Inductor current retains its initial state; a negative initial current can oppose the source before crossing zero.

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

The equilibrium is 2 A and τ=L/2. Subtract equilibrium and solve to obtain i(t)=2+(i0−2)e^(−t/τ). Passive inductor voltage is L di/dt. The induced-emf orientation is opposite, so exchanging the two conventions without changing signs would corrupt Kirchhoff’s equation.

Test conservation, uncertainty and limits

Multiply the differential equation by i: 4i=2i²+d(Li²/2)/dt. Integrating the square includes its constant, cross and exponential-square terms. Omitting the cross term does not merely change precision; it violates the energy account. A negative initial current may first reduce stored magnetic energy, cross zero, and then rebuild it with the other current polarity.

Worked original case

Investigation

Changed-case practice

Try the changed case before feedback

For L=2 H and initial current 4 A, find current at t=ln 2 seconds in amperes.

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.

Apply

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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

Continue

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Previous: Match the state instead of restarting the transient Next: Fit an offset and reserve a fourth record

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