SCIENCE · INVESTIGATE AND EXPLAIN

Identify the charge and voltage constraints of capacitor networks

Capacitor combination rules follow different common quantities from resistor rules. Across parallel capacitors, the potential difference is the same and plate charges add: Qtotal=(C1+C2)V. In a series pair with an initially neutral isolated intermediate conductor, the two capacitor charge magnitudes are equal. Their voltage drops add, so V=Q/C1+Q/C2 and 1/Ceq=1/C1+1/C2.

Learning outcome: Compare equal and unequal capacitances in both topologies. Retain individual voltages, charges and energy before summing. Then describe what extra datum would be needed if the isolated midpoint began with nonzero net charge.

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

For C1=2 µF, C2=4 µF and source 6 V, series Ceq=4/3 µF and common plate charge is 8 µC. Voltages are 4 and 2 V; energies 16 and 8 µJ total 24 µJ. In parallel Ceq=6 µF, charges are 12 and 24 µC, and total energy is 108 µJ.

Open the experiment

Predict

The equal series-capacitor charge magnitudes rely here on:

Your prediction and reason

Experiment

Compare equal and unequal capacitances in both topologies. Retain individual voltages, charges and energy before summing. Then describe what extra datum would be needed if the isolated midpoint began with nonzero net charge.

Compare series and parallel charge and energy

Worked teaching model · synthetic values · Two ideal linear capacitors across an ideal maintained 6 V source, compared in two distinct topologies. In series the isolated intermediate conductor initially has zero net charge. No leakage, breakdown, ESR or physical handling.

First capacitance: 2 µF Second capacitance: 4 µF

Reset keeps your prediction, captures and practice history.

Series equivalent capacitance versus second element

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

Equivalent numerical coordinates for Series equivalent capacitance versus second element

isolated neutral midpoint

second capacitance (µF)	equivalent capacitance (µF)
2	1
2.1	1.02439
2.2	1.04762
2.3	1.06977
2.4	1.09091
2.5	1.11111
2.6	1.13043
2.7	1.14894
2.8	1.16667
2.9	1.18367
3	1.2
3.1	1.21569
3.2	1.23077
3.3	1.24528
3.4	1.25926
3.5	1.27273
3.6	1.28571
3.7	1.29825
3.8	1.31034
3.9	1.32203
4	1.33333
4.1	1.34426
4.2	1.35484
4.3	1.36508
4.4	1.375
4.5	1.38462
4.6	1.39394
4.7	1.40299
4.8	1.41176
4.9	1.42029
5	1.42857
5.1	1.43662
5.2	1.44444
5.3	1.45205
5.4	1.45946
5.5	1.46667
5.6	1.47368
5.7	1.48052
5.8	1.48718
5.9	1.49367
6	1.5

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
Series equivalent capacitance	1.33333	µF
Series plate-charge magnitude	8	µC
First series capacitor voltage	4	V
Second series capacitor voltage	2	V
Series total energy	0.000024	J
Parallel equivalent capacitance	6	µF
First parallel plate charge	12	µC
Second parallel plate charge	24	µC
Parallel total energy	0.000108	J
Series voltage residual	0	V

Series capacitors share charge magnitude under the neutral-midpoint premise; parallel capacitors share voltage.

For series Q/C1+Q/C2=V, so 1/Ceq=1/C1+1/C2. For parallel Qtotal=(C1+C2)V. These rules contrast with resistor combinations because different conserved/common quantities enter. Preexisting net charge on the isolated midpoint changes the equal-charge premise. The comparison uses two separately charged configurations, not an unmodeled sudden reconnection with missing dissipated energy.

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

Series capacitors share charge magnitude under the neutral-midpoint premise; parallel capacitors share voltage.

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 neutral intermediate-node premise matters. A preexisting net charge trapped there would change the equal-charge relation and require a separate charge balance. In the stated series case, the smaller capacitance takes the larger voltage because V_i=Q/C_i. In parallel, the larger capacitance instead carries more charge because every branch has the same voltage.

Check evidence and the limits of the model

Energy is the sum of each half-CV² term. For a maintained source voltage this is also half-CeqV², but that does not make the two topologies equivalent. They represent separately prepared configurations. Suddenly reconnecting already charged capacitors would require a transient/dissipation account; a final electrostatic energy comparison alone cannot explain the path or missing energy. All topology drawings are virtual, not capacitor-handling or wiring instructions.

Worked original case

Investigation

Changed-case practice

Try the changed case before feedback

Capacitors 3 and 6 µF are in series across 9 V with neutral midpoint. Find common charge magnitude in µC.

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

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Account for electrical energy across a full schedule Next: Close a dielectric-change energy ledger under two constraints

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