SCIENCE · INVESTIGATE AND EXPLAIN

Preserve inductor current when the closed path changes

The first virtual topology is a series loop containing a 6 V source, 2 ohm resistor and inductor L, initially with zero current.

Learning outcome: Retain a node/path sketch for both topologies, predict which state is continuous and compute both one-sided voltages. Compare current and energy decay rates. Recompute the changed equivalent-resistance packet and explain why an open interruption is outside this finite model.

Starting recap

Differentiate linked flux, preserve circuit state and test energy and independent-calibration constraints.

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Try

For L = 2 H and ts = 2 s, the charging time constant is 1 s and Is ≈ 2.593994 A. Passive inductor voltage changes from +0.812012 V to −10.375977 V while current remains Is. One discharge time constant later, current is ≈ 0.954277 A and energy is exp(−2) of its event value.

Open the experiment

Predict

Which quantity is continuous through the stipulated finite-voltage switching event?

Your prediction and reason

Experiment

Retain a node/path sketch for both topologies, predict which state is continuous and compute both one-sided voltages. Compare current and energy decay rates. Recompute the changed equivalent-resistance packet and explain why an open interruption is outside this finite model.

Carry current through a closed-path topology change

Worked teaching model · synthetic values · Initially zero current charges through 6 V and 2 ohm, then a stipulated ideal switch transfers the inductor to a closed 4 ohm discharge path. No open-circuit interruption, arc or parasitic model. Current is continuous under finite voltage.

Inductance: 2 H Charging duration: 2 s

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State and topology reference

This original schematic defines one closed loop; it is not an apparatus wiring instruction. Exact element connections and event conditions are stated below. Component sizes are schematic.

Driven topology: source, resistor and one inductor are in series; E = R I + L dI/dt.
For the switch packet only, the post-event topology removes the source and closes the same inductor through a 4 ohm resistor. The drawn source is not retained in that second topology.
Positive passive inductor voltage is the drop in the current direction. Current continuity does not imply voltage continuity.

Event-aligned current

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Event-aligned current: numerical coordinates

time relative to event (s)	current (A)
-2	0
-1.94167	0.169994
-1.88333	0.330355
-1.825	0.481629
-1.76667	0.624331
-1.70833	0.758947
-1.65	0.885936
-1.59167	1.00573
-1.53333	1.11873
-1.475	1.22533
-1.41667	1.32589
-1.35833	1.42076
-1.3	1.51024
-1.24167	1.59466
-1.18333	1.67429
-1.125	1.74941
-1.06667	1.82028
-1.00833	1.88713
-0.95	1.95019
-0.891667	2.00967
-0.833333	2.06579
-0.775	2.11873
-0.716667	2.16866
-0.658333	2.21577
-0.6	2.26021
-0.541667	2.30213
-0.483333	2.34167
-0.425	2.37898
-0.366667	2.41417
-0.308333	2.44736
-0.25	2.47868
-0.191667	2.50822
-0.133333	2.53609
-0.075	2.56237
-0.0166667	2.58717
0.0416667	2.38659
0.1	2.12378
0.158333	1.88992
0.216667	1.6818
0.275	1.4966
0.333333	1.3318
0.391667	1.18515
0.45	1.05464
0.508333	0.938504
0.566667	0.835158
0.625	0.743192
0.683333	0.661353
0.741667	0.588526
0.8	0.523718
0.858333	0.466047
0.916667	0.414727
0.975	0.369058
1.03333	0.328418
1.09167	0.292253
1.15	0.260071
1.20833	0.231432
1.26667	0.205947
1.325	0.183269
1.38333	0.163088
1.44167	0.145129
1.5	0.129147

Calculated values under the declared model
Quantity	Value	Unit
Charging time constant	1	s
Discharge time constant	0.5	s
Current continuous at event	2.59399	A
Passive inductor voltage just before event	0.812012	V
Passive inductor voltage just after event	-10.376	V
Current one discharge time constant later	0.954277	A
Energy ratio after one discharge time constant	0.135335	ratio

Current is continuous through this finite-voltage change of closed path, while its derivative and the inductor voltage jump.

The charging state becomes the discharge initial condition. Passive inductor voltage is L dI/dt, so it is positive during charging and negative while returning stored energy to the resistor. An ideal open interruption of nonzero current is not modeled; no real switching procedure is requested.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Current is continuous through this finite-voltage change of closed path, while its derivative and the inductor voltage jump.

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 two circuits, not just a switch word

The first virtual topology is a series loop containing a 6 V source, 2 ohm resistor and inductor L, initially with zero current. At a stipulated event, the source is removed from that loop and the inductor is connected into a closed 4 ohm resistive discharge path. This is a mathematical topology change with a guaranteed current path, not an instruction to open a real energized inductor.

Transfer the correct state across the event

Before the event, current is 3[1 − exp(−t/(L/2))] A. Its value Is at event time ts becomes the discharge initial condition. Finite inductor voltage implies continuous current because integrating vL = L dI/dt across a vanishingly short interval gives no finite current jump. Voltage and dI/dt may jump. After the event, current is Is exp[−(t − ts)/(L/4)].

Predict voltage polarity and energy

Just before the event, passive vL = 6 − 2Is is positive. Just after, vL = −4Is is negative: the inductor is now delivering energy to the resistor. After one discharge time constant current is Is/e, while stored energy is reduced by exp(−2), not exp(−1). At infinite ideal discharge time, the full initial magnetic energy has become resistor heat.

Reduce a changed resistor network carefully

In a separate source-free topology, a 3 ohm resistor in series with a parallel pair of 6 and 3 ohm gives an equivalent resistance of 5 ohm. An inductor of 2 H therefore has tau = 0.4 s, not the sum of three separate time constants. The equivalent network must describe the actual path seen by the one inductor; a resistor disconnected by the switch cannot be left in the formula.

Distinguish finite-state modeling from ideal singularity

An ideal open circuit demanding zero current immediately after a nonzero inductor current would require an unbounded voltage impulse in this lumped model. Real systems have parasitic capacitance, insulation limits and other mechanisms; this lesson does not model or direct those operations. The safe virtual problem instead supplies a finite closed discharge path and checks its conservation law.

Worked original case

Investigation

Changed-case practice

Try a new practice case

For a 3 H inductor discharging through a 6 ohm equivalent resistance, what is tau in seconds?

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and boundaries

Original teaching checked 31 October 2026 against current official Unit 13 and the relevant OpenStax discussion. No source exercises or diagrams are reproduced. All records and apparatus are virtual, supplied or generated; no real wiring, switching, magnetic braking or electrical measurement is requested. Public feedback and retained work do not establish independent mastery, supervised practical participation or AP authorization.

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

Which orientation, initial state, symmetry or independent calibration assumption would change this induction inference?

Your explanation and limitations

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