SCIENCE · INVESTIGATE AND EXPLAIN

Separate resistance geometry from a power claim

Resistance R describes a particular component, whereas resistivity ρ describes a material under specified conditions. For a uniform wire, R=ρL/A. Doubling its length doubles resistance; doubling its cross-sectional area halves resistance. Doubling diameter instead quadruples area for a circular wire, so diameter and area changes must not be confused.

Learning outcome: Compare lengths 1 and 2 m at area 0.20 mm². Record R and both powers, naming the source condition for every ratio. Then halve area and compare the result with doubling length.

Starting recap

Distinguish charge flow from energy transfer. Identify nodes, choose consistent current/voltage signs and state source and initial conditions before applying algebra.

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Try

With ρ=1.0×10⁻⁶ Ω m, L=1 m and A=0.20 mm²=2.0×10⁻⁷ m², R=5 Ω. At 2 V the current is 0.40 A and power 0.80 W. Under the separate fixed 0.20 A condition, power is 0.20 W.

Open the experiment

Predict

If length doubles, do the fixed-voltage and fixed-current power predictions have the same sign of change?

Your prediction and reason

Experiment

Compare lengths 1 and 2 m at area 0.20 mm². Record R and both powers, naming the source condition for every ratio. Then halve area and compare the result with doubling length.

Separate material resistivity, geometry and power constraints

Worked teaching model · synthetic values · Hypothetical constant resistivity 1.0×10⁻⁶ Ω m at fixed temperature. Uniform wire; ideal 2 V source. A separate 0.20 A fixed-current comparison is explicitly a different source boundary. No physical heating or short-circuit activity.

Wire length: 1 m Cross section: 0.2 mm²

Reset keeps your prediction, captures and practice history.

Circuit connections: Ideal 2 V source and one uniform resistive wire in a closed loop; separate fixed-current comparison is not a second simultaneous source.

Axes rescale with settings; compare numerical values, not screen angles or apparent lengths. Solid/points show the selected model relationship or observations; the dashed comparison is explained below.

Calculated evidence ledger — synthetic teaching model
Quantity	Value	Unit
Wire resistance	5	Ω
Current at fixed 2 V	0.4	A
Power at fixed 2 V	0.8	W
Power at fixed 0.20 A	0.2	W

Accessible plot coordinates

Primary plotted points
wire length (m)	resistance (Ω)
0.5	2.5
0.75	3.75
1	5
1.25	6.25
1.5	7.5
1.75	8.75
2	10

Resistance depends on material and geometry; the effect on power depends on what the source holds fixed.

Use R=ρL/A, P=IV, P=V²/R at fixed voltage and P=I²R at fixed current. Material resistivity is held constant here; real metal heating often increases it. These are two different source conditions, not simultaneous constraints.

Resistance depends on material and geometry; the effect on power depends on what the source holds fixed.

Observation to attach to a capture

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

Understand

Establish the model and evidence

The model holds temperature and resistivity fixed. Real metallic components often become more resistive as they warm. An ohmic idealization treats voltage proportional to current over the stated operating range. A specimen may look approximately ohmic in a narrow temperature-controlled test but depart from that behavior when heating changes its material state. The browser does not simulate those thermal changes.

Electrical power is P=IV. For an ohmic element this can also be written I²R or V²/R, but deciding which expression predicts a change requires a source condition. Increasing resistance at fixed current raises power; increasing it at fixed voltage lowers power. Neither statement is a universal rule that larger resistance always heats more or less.

The two power entries deliberately compare different source conditions: a 2 V voltage source and a separate 0.20 A current source. They are not simultaneous constraints on the same circuit. No physical heating trial is requested. Actual component ratings, temperature, insulation and supervised electrical safety cannot be inferred from a small numerical wattage in an ideal model.

Worked evidence

With ρ=1.0×10⁻⁶ Ω m, L=1 m and A=0.20 mm²=2.0×10⁻⁷ m², R=5 Ω. At 2 V the current is 0.40 A and power 0.80 W. Under the separate fixed 0.20 A condition, power is 0.20 W.

Predict and investigate

If length doubles, do the fixed-voltage and fixed-current power predictions have the same sign of change?

Compare lengths 1 and 2 m at area 0.20 mm². Record R and both powers, naming the source condition for every ratio. Then halve area and compare the result with doubling length.

Transfer to changed conditions

A 12 Ω resistor is replaced by 24 Ω. Compare power ratios at fixed 6 V and at fixed 0.10 A.

Compare the worked transfer

At 6 V, power falls from 3 W to 1.5 W. At 0.10 A, it rises from 0.12 W to 0.24 W. The conditions describe different source-controlled comparisons.

Explain the boundary

How would unmeasured heating undermine an inference that resistivity stayed constant?

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. Original cases and computed/synthetic evidence, not actual laboratory measurements. These activities are virtual only: do not reproduce wiring, short-circuit, battery charging, heating or charged-capacitor experiments. Any real practical work requires qualified supervision and an independently approved safety procedure.

Apply

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

How would unmeasured heating undermine an inference that resistivity stayed constant?

Your explanation and limitations

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Previous: Explain current without consuming charge Next: Test a constant-resistance claim against curved data

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