Choose a conductor using two explicit constraints

Translate a design brief into measurable requirements

A hypothetical lightweight device needs a 2 m total conducting path carrying a steady 2 A. Its teaching brief allows at most 0.100 V drop along that path and at most 10.0 g of bare metal. Both limits must be met. The allowed menu contains copper and aluminum; the area grid runs from 0.5 to 4.0 mm² in 0.5 mm² steps.

These are assigned design conditions, not an electrical installation standard. The model holds temperature at 20 °C and excludes insulation, connectors and mechanical supports. A passing result answers this brief only. Start by predicting whether the material with lower resistivity must also be the best choice for a mass-limited task.

The reference constants are resistivities 1.68 × 10⁻⁸ Ω m for copper and 2.65 × 10⁻⁸ Ω m for aluminum, and densities 8.96 g/cm³ and 2.70 g/cm³, respectively. Sources appear beside the ledger: the OpenStax resistivity table and Royal Society of Chemistry entries for copper and aluminum. These selected properties are reference inputs, not fresh measurements of a particular manufactured wire.

Compare equivalent geometries first

At area 1.0 mm², copper has R = 0.0336 Ω and a 0.0672 V drop, meeting the voltage requirement. Its metal volume is 2.0 cm³ and its mass is 17.92 g, exceeding the mass requirement. The lower-resistivity material fails the combined brief at this geometry.

Switch only the material to aluminum. Its mass is 5.40 g, but R = 0.053 Ω gives a 0.106 V drop, just beyond the allowed 0.100 V. The lighter material also fails at this geometry, for a different reason. Report which requirement fails; saying only “bad design” hides the information needed for a useful revision.

Changing both material and area simultaneously can find a candidate, but it obscures which change caused which outcome. Save the equal-area comparison first. Then vary area for one material while holding length, current and temperature fixed.

Make a revision and check both constraints

Increase aluminum area to 1.5 mm². Resistance falls to approximately 0.035333 Ω, voltage drop to 0.070667 V, and power converted to heat to 0.141333 W. Volume rises to 3.0 cm³, producing a bare-metal mass of 8.10 g. This candidate meets both assigned limits.

The voltage margin is approximately 0.100 − 0.070667 = 0.029333 V; the mass margin is 10.0 − 8.10 = 1.90 g. A margin is distance from a constraint under the model, not an allowance that can automatically be spent on uncertain temperature, contact resistance or fittings. Those omitted quantities need their own accounting.

Increase aluminum area again to 2.0 mm². The voltage improves further, but mass becomes 10.80 g, failing the mass limit. More of a conducting material can improve one requirement while worsening another. A design decision therefore needs a brief and evidence, not a single adjective such as “best conductor.”

Try copper at 0.5 mm²: its mass is 8.96 g, but the voltage drop is 0.1344 V. Moving to 1.0 mm² repairs voltage at the cost of excess mass. On the offered grid, only the 1.5 mm² aluminum candidate meets both conditions. Investigate the neighboring failures rather than treating a green result as a complete explanation.

Keep units and chart meanings explicit

For the fixed 2 m path, volume is 2S cm³ when the numerical area S is entered in mm². One way to check this is to convert 2 m into 200 cm and 1 mm² into 0.01 cm²: 200 × 0.01 = 2 cm³. Multiplying by density gives mass in grams. Density is mass per volume; it is not resistivity, even though both are sometimes denoted with the Greek letter rho. The ledger names them separately.

The chart uses percent of each requirement’s limit. A voltage value of 70.667% means 0.070667/0.100 × 100. A mass value of 81% means 8.10/10.0 × 100. Values at or below 100% meet their respective limit. These percentages are not probabilities, accuracy scores or interchangeable physical quantities. Do not add them into an unexplained overall grade.

The area grid also matters. It is a menu of supported candidates, not a claim that every mathematically possible area is manufactured or available. A design could fail on one menu yet be feasible with a different set of options. Conversely, a continuous mathematical solution does not establish manufacturing tolerances or practical suitability.

State what would change the recommendation

The calculated heat power is not a prediction that the wire stays cool. The model maintains temperature; an actual installation requires an account of heat removal, environment and permitted operating conditions. Contact resistance and connections can change voltage loss. Insulation and fittings add mass. Strength, fatigue, corrosion, alloy grade and joining methods can affect whether a candidate is usable.

Cost and resource impacts need additional evidence too. Bare-metal mass alone cannot establish extraction impact, manufacturing energy, service life or recyclability. A supported decision brief identifies the comparison boundary and lists the information that could overturn the provisional recommendation. This exercise is completed entirely with calculations; it does not ask you to construct or energize a physical circuit.

Transfer: Keep the 1.5 mm² aluminum candidate and both numerical limits, but double the required current from 2 to 4 A. Without changing its area, predict mass, voltage drop and heat power. Decide whether the previous pass still applies, and explain why a mass-only review would miss the failure.

Check the conductor-design transfer

Mass remains 8.10 g and resistance stays approximately 0.035333 Ω under the fixed-temperature assumption. Voltage doubles to approximately 0.141333 V and now fails the 0.100 V limit. Power increases by a factor of four to approximately 0.565333 W. The current changed; a prior passing calculation is not evidence for the new operating condition.