Cable Size Calculator — IEC 60364

Purpose

Fixes which tabulated data the calculator reads. Ratings and protective conductor rules are not interchangeable between standards, so the whole chain follows one dataset.

Engineering explanation

Current capacity and derating come from IEC 60364-5-52 Annex B, protective conductor sizing from IEC 60364-5-54, and conductor resistance from IEC 60228 class 2 stranded.

Reference conditions are 30 °C in air, 20 °C in soil, soil thermal resistivity 2.5 K·m/W. Derating factors multiply the tabulated rating relative to those conditions rather than replacing it.

Formulas

Tables

Notes

• Tabulated values shipped here are a working dataset flagged unverified in the JSON. Check them against your licensed copy of the standard before production use.
• National rules may be more restrictive than IEC 60364, particularly for voltage drop limits.

Examples

• A design verified at 30 °C IEC reference ambient needs rechecking against a national standard assuming 40 °C, because the base rating differs.

References

• IEC 60364-5-52
• IEC 60364-5-54
• IEC 60364-4-41
• IEC 60228

Purpose

Sets how load power converts to current, how many conductors count as loaded, and the constant in the voltage drop equation.

Engineering explanation

Three-phase current uses the phase-to-phase voltage and the √3 factor. Single-phase and DC use phase-to-neutral or pole-to-pole voltage, and their voltage drop includes the return conductor, hence the factor of two.

A three-phase circuit is rated on three loaded conductors, single-phase on two. A balanced neutral carries no significant current and is not counted as loaded.

Formulas

Tables

Notes

• For three-phase motors enter the line current, not the delta winding current.
• Above 15 percent third harmonic, IEC 60364-5-52 Annex E treats the neutral as loaded. Use the harmonic field under derating.

Examples

• 700 kW at 415 V, pf 0.9: 700000 ÷ (1.732 × 415 × 0.9) = 1082 A, which requires parallel cables.

References

• IEC 60364-5-52 Clause 523, Annex E

Purpose

Sets the design current for a given load power and the base for percentage voltage drop.

Engineering explanation

Enter phase-to-phase voltage for three-phase, phase-to-neutral for single-phase, pole-to-pole for DC. The calculator applies the convention from the phase arrangement.

Choose Other for any value from 1 V to 1000 V. Above 1000 V AC this calculator does not apply, since HV cable sizing uses different rating tables and screen earthing rules.

Formulas

Tables

Notes

• Percentage drop is referenced to the nominal voltage entered here, not a measured supply voltage.
• Low-voltage DC is dominated by voltage drop. A 24 V circuit at 3 percent allows only 0.72 V.

Examples

• 20 kW draws 87 A at 230 V single-phase but 32 A at 415 V three-phase, several sizes apart.

References

• IEC 60038
• IEC 60364-1 Clause 131

Purpose

Every downstream calculation starts from the design current, so this field accepts whatever unit the nameplate uses and converts it.

Engineering explanation

In amperes, enter the per-phase line current. If each of three phases carries 100 A, enter 100 A, not 300 A.

kW and W need a power factor, because the conductor carries apparent current regardless of useful work. kVA and VA need none for the current calculation.

Horsepower is shaft output, so converting to electrical input needs power factor and efficiency. One horsepower is 745.7 W.

Formulas

Tables

Notes

• This is the continuous load current, not starting current. Motor starting is a separate assessment.
• Apply diversity before entering the rating. The calculator sizes for the current it is given.

Examples

• 50 hp at 415 V, pf 0.85, efficiency 0.92: (50 × 745.7) ÷ (1.732 × 415 × 0.85 × 0.92) = 66.3 A.

References

• IEC 60364-5-52 Clause 523.1
• IEC 60364-1 Clause 132.3

Purpose

Converts real power to current and, in specified mode, sets the phase angle for voltage drop.

Engineering explanation

Lower power factor means more current for the same real power, so a larger conductor. It also changes how reactance contributes to voltage drop.

Reactance grows relative to resistance as size increases, so power factor matters most on large cables. On 4 mm² it is negligible; on 300 mm² the two terms are comparable.

Formulas

Tables

Notes

• Indicative values for preliminary design only. Use nameplate data for the final calculation.
• Correction capacitors reduce current upstream of themselves, not between capacitor and load.

Examples

• 30 kW at 415 V draws 41.7 A at unity but 49.1 A at 0.85, which can move the size up one step.

References

• IEC 60364-5-52 Annex G

Purpose

Usually the binding constraint on long circuits. Current capacity decides short runs; beyond a few tens of metres voltage drop decides the size.

Engineering explanation

IEC 60364-5-52 Annex G recommends 4 percent from origin to point of use, split in practice between distribution and final circuits.

Excessive drop reduces motor torque, dims lighting and stresses electronic supplies. Motors are most sensitive, since torque falls roughly with the square of voltage.

Formulas

Tables

Notes

• Drop accumulates through the distribution chain. Budget per segment rather than applying the full allowance at each stage.
• Choose Other for any limit from 0.01 to 100 percent where equipment demands it.

Examples

• At 415 V with a 1 percent limit the budget is 4.15 V. Over 40 m at 1082 A that needs parallel large conductors.

References

• IEC 60364-5-52 Clause 525, Annex G

Purpose

Voltage drop and power loss are directly proportional to length, so route accuracy matters as much as conductor size.

Engineering explanation

Enter the one-way length. The return path is added automatically for single-phase and DC, so do not double the figure.

Use the installed route length, not straight-line distance. Risers, tray routing, terminations and slack all add length, and underestimating it is a common cause of a compliant calculation with a non-compliant installation.

Formulas

Tables

Notes

• For distributed loads along a run, the effective length is shorter than the full route. Assess load by load rather than assuming all current reaches the far end.

Examples

• A 40 m single-phase circuit has 80 m of conductor in the loop. Enter 40; the factor of two is applied for you.

References

• IEC 60364-5-52 Clause 525

Purpose

Sets how conservative the result is. Both defaults are the conservative choice and can be relaxed when operating conditions are known.

Engineering explanation

Copper resistance rises about 0.393 percent per kelvin, so a conductor at 40 °C rather than 90 °C has roughly 17 percent less resistance. Calculate mode estimates the operating temperature from the load ratio; Maximum assumes the insulation limit.

Worst case power factor uses the impedance magnitude, as though cable and load angles coincide. Specified uses the true angle and yields a lower drop.

Formulas

Tables

Notes

• Temperature coefficients at 20 °C: copper 0.00393 per K, aluminium 0.00403 per K.
• Use specified power factor only where it is known and stable at full load, such as a motor at rated output.

Examples

• A 90 °C cable at 70 percent load in 30 °C air runs near 30 + 60 × 0.49 = 59 °C, well below the maximum assumption.

References

• IEC 60364-5-52 Annex G
• IEC 60228 Clause 4

Purpose

Determines how hot the conductor may run. A higher permitted temperature allows a greater rise above ambient, so more current on the same size.

Engineering explanation

PVC is limited to 70 °C conductor temperature under IEC 60364; XLPE and EPR reach 90 °C. That headroom is worth roughly 20 to 25 percent more capacity, often a full size step.

Insulation also sets the adiabatic k factor for the short-circuit check. XLPE tolerates 250 °C under fault, PVC only 160 °C up to 300 mm².

The code labels follow the reference implementation. Engineering behaviour depends only on temperature class and material family, both carried in the JSON dataset.

Formulas

Tables

Notes

• Halogen-free grades marked HF produce less corrosive smoke but do not change the current rating.
• A 90 °C cable on 70 °C rated terminals must be sized on the lower limit. Check switchgear terminal ratings.

Examples

• Moving a 50 mm² four-core copper cable from PVC to XLPE can avoid stepping up to 70 mm², provided terminations are rated for 90 °C.

References

• IEC 60364-5-52 Table 52.1
• IEC 60364-5-54 Table 54.3
• IEC 60502-1

Purpose

Determines the number of loaded conductors for the rating table and which rating and reactance dataset applies.

Engineering explanation

Multi-core cables carry all cores in one sheath, so mutual heating lowers the rating, while the close spacing gives low reactance.

Single-core cables can be spaced for better cooling and a higher rating, but the larger loop area raises reactance. Trefoil is the compromise between the two.

Types marked mains combine earth and neutral in one PEN conductor. That is permitted only upstream of the point of separation, never in a final circuit downstream of it.

Tables

Notes

• A balanced three-phase neutral is not a loaded conductor. Harmonic current changes that.
• A PEN conductor must never be switched or interrupted, and requires a minimum cross-section under IEC 60364-5-54.

Examples

• For a 1082 A three-phase submain, single-core 4x1C+E in trefoil with parallel groups is usually more economical and easier to install than very large multi-core cable.

References

• IEC 60364-5-52 Clause 521
• IEC 60364-5-54 Clause 543.4

Purpose

Flexible cables use class 5 conductors with many fine strands, giving slightly higher resistance per square millimetre and a different rating basis.

Engineering explanation

IEC 60228 class 5 conductors have a higher maximum DC resistance than class 2 at the same nominal area, because the fine strands pack less efficiently and each strand carries a thin oxide layer.

Flexible cables are intended for appliance connections, portable equipment and short movable runs. They are not intended for fixed distribution wiring, where their lower mechanical robustness and higher cost are not justified.

Formulas

Tables

Notes

• Flexible cable coiled on a drum or reel needs additional derating, since the coiled section cannot dissipate heat.
• Fine-stranded conductors require terminals rated for them. Screw terminals designed for class 2 can crush class 5 strands.

Examples

• A 2.5 mm² class 5 flexible cord has slightly higher resistance than 2.5 mm² class 2, so voltage drop on a long extension is marginally worse.

References

• IEC 60228 Clauses 5 and 6
• IEC 60245
• IEC 60227

Purpose

Material sets resistivity, which drives both current capacity and voltage drop. Aluminium needs a larger cross-section for the same performance but weighs and costs less.

Engineering explanation

Aluminium resistivity is about 1.64 times that of copper, so an aluminium conductor needs roughly 1.6 times the cross-sectional area to match a copper one. In practice that is one to two standard size steps.

Aluminium is around half the weight of copper for the same current capacity, which matters for long runs, heavy submains and cable support design. Its drawbacks are creep at terminations and a tenacious surface oxide, both of which demand terminals rated for aluminium and correctly applied torque.

Formulas

Tables

Notes

• Aluminium below 16 mm² is not offered here. Small aluminium conductors are mechanically unreliable at terminations.
• Never join copper and aluminium directly. Galvanic corrosion at the joint raises resistance over time. Use bimetallic lugs.

Examples

• Replacing 185 mm² copper with aluminium typically requires 300 mm² to achieve a comparable rating and voltage drop.

References

• IEC 60228 Clause 4
• IEC 60364-5-54 Table 54.3

Purpose

Auto finds the smallest standard size that satisfies every enabled check. A manual size lets you verify a predetermined cable against the same checks.

Engineering explanation

In Auto mode the calculator walks the size list upward and returns the first size that passes all of the enabled criteria: current capacity after derating, voltage drop, short-circuit withstand and earth fault loop impedance. Because each criterion is monotonic in size, the first pass is also the smallest pass.

With a manual size, all checks still run and the results panel reports which ones fail. That is the mode to use when auditing an existing installation or confirming a design someone else produced.

Formulas

Tables

Notes

• Auto returns the smallest compliant size, not necessarily the most economical whole-life choice. A larger conductor reduces losses and may pay back over time.
• If Auto reports no compliant size, use parallel cables, raise the voltage drop allowance if permitted, or improve the installation method.

Examples

• For 100 A over 30 m at 415 V with a 4 percent limit, current capacity governs and 25 mm² copper is selected. Extend the run to 200 m and voltage drop pushes the same circuit to 70 mm².

References

• IEC 60364-5-52 Clause 523
• IEC 60364-4-43 Clause 433

Purpose

Above roughly 400 A a single cable becomes impractical to handle, bend and terminate. Parallel cables split the current so each conductor stays a manageable size.

Engineering explanation

Each parallel path carries the design current divided by the number of paths, and the effective impedance is the single-cable impedance divided by the same number. Both the current capacity and the voltage drop improve proportionally.

That proportional gain assumes equal sharing, which requires identical length, identical cross-section, identical construction and identical installation for every path. Unequal length is the usual failure mode: the shorter path carries more current and overheats while the longer one is underused.

Parallel cables also count as multiple circuits for grouping derating. Three parallel groups on one tray are derated as three circuits, which partially offsets the capacity gained.

Formulas

Tables

Notes

• A minimum of 4 mm² per parallel conductor is a common wiring-rule requirement, to keep mechanical strength adequate.
• The short-circuit check is applied per conductor by default, with no assumed fault current split. That is the conservative treatment.
• Parallel cables are generally not permitted for installations exposed to direct sunlight, because the thermal behaviour is not covered by the standard rating tables.

Examples

• A 1082 A load with three parallel groups gives 361 A per group, which a 185 mm² copper conductor can carry in a well-ventilated installation.

References

• IEC 60364-5-52 Clause 523.7
• IEC 60364-4-43 Clause 434.4

Purpose

The protective conductor must carry earth fault current for long enough for the protective device to operate, without exceeding its own temperature limit, and must keep loop impedance low enough for that device to actually trip.

Engineering explanation

IEC 60364-5-54 offers two routes. The simplified route uses Table 54.7, which relates the protective conductor size to the line conductor size. The calculated route applies the adiabatic equation with the actual fault current and clearing time, and generally yields a smaller conductor.

For multi-core cables only Auto is available, because the earth core is part of the cable construction and its size is set by the manufacturer. For single-core cables the earth is a separate cable and can be sized independently, so both Auto and manual selection are offered.

In Auto mode the calculator starts from the Table 54.7 minimum and increases the size until the enabled fault rating and loop impedance checks pass. If no listed size passes, it falls back to matching the active size.

Formulas

Tables

Notes

• Aluminium protective conductors with copper line conductors are not offered, because mixed-material terminations at the earth bar are a reliability risk.
• Parallel protective conductors must be the same length, follow the same route, and be mechanically protected and separated from the live conductors, otherwise the assumed equal fault current split does not hold.
• Where the protective conductor is not part of the cable, IEC 60364-5-54 sets absolute minima of 2.5 mm² with mechanical protection and 4 mm² without.

Examples

• A 240 mm² copper line conductor gives a Table 54.7 minimum of 120 mm². The adiabatic route at 10 kA for 0.1 s needs only √(10000² × 0.1) ÷ 143 = 22 mm², so the table governs.

References

• IEC 60364-5-54 Clause 543, Tables 54.2, 54.3, 54.7

Purpose

Installation method is the single largest influence on current capacity, because it determines how readily heat leaves the cable. The same conductor can differ by more than 30 percent in rating between the best and worst method.

Engineering explanation

A cable spaced clear of a surface on a ladder is cooled on all sides and rates highest. A cable enclosed in conduit or surrounded by thermal insulation traps its own heat and rates lowest. Buried cables sit in between and depend on soil thermal resistivity.

Select the worst-case section of the route, meaning the section with the lowest rating. A cable that runs 40 m on open tray and 2 m through thermal insulation is limited by the insulated section if that section is long enough to matter thermally.

Each method in the picker maps to a reference method in IEC 60364-5-52 Annex A, which in turn selects a rating table in Annex B. The detailed per-method engineering notes are in the installation method reference section below.

Formulas

Tables

Notes

• Relative ratings above are directional guidance for design decisions. The calculation uses the tabulated values in the JSON dataset, not these ratios.
• Changing the installation method resets the cable support and spacing options, because the valid choices depend on the method.

Examples

• Moving a circuit from conduit in a wall to spaced tray can lift the rating enough to drop one conductor size, often cheaper than upsizing the cable.

References

• IEC 60364-5-52 Annex A, Annex B

Purpose

Formation trades current capacity against reactance. It applies only to single-core cables, since multi-core geometry is fixed by the cable itself.

Engineering explanation

Trefoil places the three phases in a touching triangle. The tight symmetric spacing minimises the loop area, giving the lowest reactance and the best magnetic balance, but the cables heat each other so the rating is lower.

Laid flat spreads the cables in a plane, improving cooling and raising the rating, but the wider and asymmetric spacing increases reactance and produces slightly unequal impedance between phases.

For large conductors where reactance dominates the voltage drop, trefoil usually wins overall even though its tabulated rating is lower. For smaller conductors where resistance dominates, laid flat is often better.

Formulas

Tables

Notes

• Trefoil groups must be cleated at intervals to hold formation, since fault current produces large mechanical forces between the cables.
• Single-core cables passing through a ferrous wall or gland plate must pass through the same aperture, or induced eddy currents will heat the steel.

Examples

• On a 240 mm² single-core group at 415 V and 0.85 power factor, trefoil typically gives a lower total voltage drop than laid flat despite the lower tabulated rating, because the reactive term dominates.

References

• IEC 60364-5-52 Annex B
• IEC 60287-1-1

Purpose

Tabulated ratings assume one circuit at reference ambient temperature. Real installations are hotter, more crowded, or thermally obstructed, and every departure reduces the safe current.

Engineering explanation

Factors multiply. Four modest factors of 0.9 combine to 0.66, so a cable tabulated at 100 A carries only 66 A. Grouping and ambient temperature together account for most real-world derating.

Ambient temperature derating reflects the smaller permitted rise: a 90 °C cable in 30 °C air has 60 K of headroom, but only 40 K in 50 °C air. Grouping derating reflects mutual heating, which is why spacing cables recovers capacity.

Underground installations add soil thermal resistivity and burial depth. Thermal insulation is the most severe case, since a cable completely surrounded by insulation can lose half its rating.

Formulas

Tables

Notes

• Values above are the working dataset and are flagged unverified in derating.json. Confirm against your licensed copy of the standard.
• Count every cable in the installation for grouping, not just the circuit being designed. Heat from adjacent circuits affects this one.
• Above 15 percent third harmonic the neutral becomes loaded and IEC 60364-5-52 Annex E applies an additional factor.

Examples

• Three circuits bunched in 40 °C air on XLPE: 0.70 × 0.91 = 0.64. A cable tabulated at 200 A is limited to 128 A.

References

• IEC 60364-5-52 Clause 523.5 to 523.9, Tables B.52.14 to B.52.21, Annex E

Purpose

Conduit fill limits exist for two reasons: cables must be installable without damage, and a crowded conduit traps heat and undermines the current rating.

Engineering explanation

IEC practice sizes conduit by space factor, the ratio of total cable cross-sectional area to the internal area of the conduit. A common limit is 45 percent for three or more cables, with more generous allowances for one or two cables where drawing-in is easier.

Use the overall cable diameter including insulation and sheath, not the conductor size. Bends make drawing-in harder, so a run with several bends warrants a larger conduit than the space factor alone suggests.

Formulas

Tables

Notes

• The conduit option appears only for installation methods that involve an enclosure.
• Space factor governs installability. It does not replace grouping derating, which still applies to cables inside the conduit.
• Draw-in boxes are typically required every two right-angle bends, or at intervals on long straight runs.

Examples

• Four cables of 12 mm overall diameter: Σd² = 576 mm². At 45 percent, D ≥ √(576 ÷ 0.45) = 35.8 mm, so a 40 mm conduit.

References

• IEC 61386-1
• IEC 60364-5-52 Clause 522.8

Purpose

During a short circuit the conductor may carry many times its rated current. The check confirms the insulation does not reach a damaging temperature before the device operates.

Engineering explanation

The adiabatic equation assumes all fault energy heats the conductor with no loss to the surroundings, which is valid for clearing times up to about five seconds. It is conservative for longer faults.

The k factor combines the conductor specific heat, resistivity and the permitted initial and final insulation temperatures. XLPE gives a higher k than PVC because it tolerates 250 °C rather than 160 °C under fault.

For current-limiting devices the manufacturer I²t let-through is the correct input and is usually far lower than the prospective fault current would suggest, which permits a smaller conductor.

Formulas

Tables

Notes

• Enter the prospective fault current at the origin of the circuit, which is the highest value the cable will see.
• For current-limiting devices, use the manufacturer I²t let-through rather than the prospective current and a nominal time. The result is usually far less onerous.
• The adiabatic assumption stops being valid beyond about five seconds. For longer clearing times consult the cable manufacturer.

Examples

• 10 kA cleared in 0.1 s on XLPE copper: S = √(10000² × 0.1) ÷ 143 = 22.1 mm², so 25 mm² is the minimum. The same fault on PVC needs √(10000² × 0.1) ÷ 115 = 27.5 mm², so 35 mm².

References

• IEC 60364-5-54 Clause 543.1.2, Table 54.3
• IEC 60364-4-43 Clause 434.5.2

Purpose

Fault protection depends on the device tripping fast enough to limit touch voltage duration. Too much loop impedance means too little fault current, and the device operates slowly or not at all.

Engineering explanation

The loop comprises the source impedance, the line conductor impedance and the protective conductor impedance. Because the protective conductor is often half the line size, it typically contributes more than half the cable portion of the loop.

The criterion is that the fault current must reach the current that operates the device within the required time. Rearranged, that sets a maximum permitted loop impedance for each device type and rating.

Conductors are hot during a fault, so their resistance is higher than at 20 °C. Applying a factor of about 1.2 to the cold resistance is standard conservative practice.

Formulas

Tables

Notes

• C_min accounts for the permitted voltage tolerance at the origin and is commonly taken as 0.95.
• An RCD provides fault protection where loop impedance is too high for the overcurrent device to operate in time, but it is not a substitute for a correctly sized protective conductor.
• On a TT system the loop includes the earth electrode resistance, which usually dominates and generally requires RCD protection.

Examples

• A 32 A type C MCB needs 320 A to trip instantaneously. At 230 V with C_min 0.95, the maximum loop impedance is (230 × 0.95) ÷ 320 = 0.68 Ω.

References

• IEC 60364-4-41 Clauses 411.3.2, 411.4, Table 41.1
• IEC 60364-6 Clause 6.4.3

Purpose

The device rating sits between the design current and the derated cable capacity. It also sets the fault current needed for the loop impedance check and the clearing time for the short circuit check.

Engineering explanation

The coordination rule is that the design current must not exceed the device rating, and the device rating must not exceed the derated cable capacity. A device larger than the cable can carry allows a sustained overload the cable cannot survive.

Device type determines the instantaneous trip threshold. A type B MCB trips at five times rating, type C at ten, type D at twenty. Higher thresholds tolerate motor inrush but demand lower loop impedance to operate in time.

Fuses have inverse time-current curves without a sharp instantaneous threshold, so the operating current for a target disconnection time must be read from the manufacturer curve.

Formulas

Tables

Notes

• The 1.45 factor allows for the conventional tripping current of the device being above its nominal rating.
• A device selected for motor starting inrush may fail the loop impedance check even though it is correct for overload. Both checks must pass.

Examples

• For a 66 A motor load, an 80 A device requires a cable with at least 80 A of derated capacity, not merely 66 A.

References

• IEC 60364-4-43 Clauses 433.1, 434
• IEC 60898-1
• IEC 60947-2

Purpose

This is the thermal limit of the cable. Exceeding it continuously degrades the insulation, shortening cable life and eventually causing failure.

Engineering explanation

The tabulated rating is a laboratory figure for a single circuit at reference conditions. Multiplying it by the combined derating factor gives the installed capacity, which is the value the design current and device rating must respect.

Physically the limit is thermal equilibrium: the conductor heats by I²R and loses heat to the surroundings. The rating is the current at which the conductor settles exactly at its permitted maximum temperature.

Because loss rises with the square of current, a modest overload produces a disproportionate temperature rise. Twenty percent over rating is roughly forty-four percent more heat generated.

Formulas

Tables

Notes

• The margin reported here is thermal only. A cable with ample current margin can still fail the voltage drop check.
• For parallel cables the capacity shown is the total across all paths, on the assumption of equal sharing.

Examples

• A cable tabulated at 200 A with a combined derating factor of 0.64 has an installed capacity of 128 A, so a 125 A device is the largest that coordinates.

References

• IEC 60364-5-52 Clause 523, Annex B
• IEC 60364-4-43 Clause 433

Purpose

Voltage drop determines whether the load actually receives usable voltage. It is often the criterion that governs conductor size on long runs.

Engineering explanation

Current flowing through conductor impedance produces a voltage drop. The resistive component always contributes; the reactive component matters increasingly as conductor size grows and as power factor falls.

Three-phase drop uses the √3 factor and is calculated as the line-to-line value. Single-phase and DC include both outgoing and return conductors, hence the factor of two.

In worst-case mode the impedance magnitude is used directly, which is equivalent to assuming the load and cable phase angles coincide. In specified mode the true load angle is applied, which lowers the calculated drop.

Formulas

Tables

Notes

• The division by 1000 converts the tabulated Ω/km to Ω for a length in metres.
• For parallel cables the effective impedance is divided by the number of paths before the drop is calculated.
• Motor starting drop is a separate calculation using the locked-rotor current, and is not covered by this result.

Examples

• 1082 A over 40 m at 415 V through three parallel 185 mm² groups with Z of 0.12 Ω/km: ΔV = (1.732 × 1082 × 0.04 ÷ 1000) × (0.12 ÷ 3) × 1000 = 3.0 V, which is 0.72 percent and inside a 1 percent limit.

References

• IEC 60364-5-52 Clause 525, Annex G

Purpose

Resistance drives both the resistive voltage drop and the I²R power loss, and it is strongly temperature dependent.

Engineering explanation

Tabulated DC resistance at 20 °C comes from IEC 60228. It is corrected to operating temperature with the linear coefficient, then adjusted upward for skin and proximity effects to give the AC value.

Skin effect pushes current toward the conductor surface at power frequency, raising effective resistance. The effect is negligible below about 50 mm² but reaches several percent on very large conductors.

DC circuits have no skin or proximity effect, so the DC resistance is used directly. That makes DC resistance slightly lower than the AC value for the same conductor.

Formulas

Tables

Notes

• These are the IEC 60228 class 2 maxima at 20 °C, flagged unverified in impedance.json. Confirm against your licensed copy.
• Circular conductor data is used throughout in preference to shaped conductor data, because it is the more conservative choice.

Examples

• 95 mm² copper at 20 °C is 0.193 Ω/km. At 75 °C it rises to 0.193 × [1 + 0.00393 × 55] = 0.235 Ω/km, an increase of 22 percent.

References

• IEC 60228 Clause 4, Table 2
• IEC 60287-1-1 Clause 2.1

Purpose

Reactance adds to the voltage drop and, unlike resistance, does not fall as conductor size increases. On large cables it eventually dominates.

Engineering explanation

Reactance arises from the magnetic field around the conductors and depends on geometry rather than cross-section: the spacing between conductors and the conductor diameter, not the amount of metal.

Because resistance falls roughly inversely with area while reactance stays near 0.07 to 0.10 Ω/km, upsizing beyond about 300 mm² buys little voltage drop improvement. Parallel cables are the effective remedy, since they divide both terms.

DC circuits have no reactance in steady state, so only resistance contributes to their voltage drop.

Formulas

Tables

Notes

• At 60 Hz, multiply 50 Hz reactance by 1.2.
• Reactance values in impedance.json are flagged unverified and depend on the actual cable construction. Use manufacturer data for critical designs.

Examples

• On 300 mm² copper at 90 °C, resistance is about 0.080 Ω/km and trefoil reactance about 0.080 Ω/km. The two contribute equally, so geometry matters as much as metal.

References

• IEC 60287-1-1
• IEC 60364-5-52 Annex G

Purpose

Cable loss is a permanent operating cost and a source of heat in the building. On heavily loaded circuits running continuously, upsizing the conductor can pay for itself.

Engineering explanation

Loss is I²R per conductor, summed over the current-carrying conductors. Three-phase circuits have three such conductors; single-phase and DC have two, counting the return path.

Because the relationship is quadratic in current and inverse in area, doubling the conductor area halves the loss for the same current. Efficiency here is transmission efficiency of the cable only, not of the load.

Annual cost follows from loss, operating hours and tariff. For a submain running near capacity around the clock, the difference between adjacent sizes can be a significant sum over the installation life.

Formulas

Tables

Notes

• Loss is computed at the same conductor temperature used for voltage drop, so the advanced temperature option affects this figure too.
• Cable loss becomes a building cooling load in air-conditioned spaces, so its true cost exceeds the energy cost alone.

Examples

• 400 A three-phase over 100 m of 185 mm² copper at 0.12 Ω/km: 3 × 400² × 0.12 × 0.1 = 5760 W. At 4000 hours and 0.15 per kWh that is roughly 3456 per year in energy.

References

• IEC 60364-8-1 — Energy efficiency
• IEC 60287-3-2 — Economic optimisation of cable size