Voltage Drop Calculator

Resistive AC approximation at 20°C, including power factor but excluding cable reactance.

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1 Inputs

Minimum: 0.1

Minimum: 0.1

Minimum: 0.1

Minimum: 1

Minimum: 0.1 · Maximum: 1

2 Results

Voltage drop
4.2 V
Voltage drop %
1.83%
Voltage at load
225.8 V
Conductor resistance
2.917 Ω/km
ΔV ≈ K × I × L × R × cosφ / 1000
Resistive AC approximation at 20°C, including power factor but excluding cable reactance. It does not select conductor size or establish compliance. Verify temperature, reactance, installation method and applicable design requirements with an electrical engineer.

Worked example

Example values are illustrative. Replace them with your measured data.

InputValue
Load current32 A
One-way run length25 m
Conductor cross-section6 mm²
System voltage230 V
PhaseSingle phase (K=2)
ConductorCopper (ρ 17.5)
Power factor (resistive estimate)0.9

Voltage drop: 4.2 V

Related tools

About this calculator

How it works

Conductor resistance (Ω/km) = resistivity constant ÷ cross-section, using 17.5 for copper or 28 for aluminium. Voltage drop = K × current × one-way length × resistance ÷ 1,000, where K = 2 for single-phase and √3 for three-phase. The tool reports the drop in volts and as a percentage of system voltage, and the voltage left at the load.

Worked example (illustrative inputs)

32 A over a 25 m run on 6 mm² copper, single-phase at 230 V. R = 17.5 ÷ 6 ≈ 2.917 Ω/km. Drop = 2 × 32 × 25 × 2.917 ÷ 1,000 ≈ 4.67 V, or 2.03%. That leaves about 225.3 V at the load, within 3%.

How to read the result

The tool checks against the common guidance of 3% or less on a branch circuit and 5% or less in total for feeder plus branch. Above 5% it advises a larger conductor. Doubling the cross-section roughly halves the drop. This is a resistive estimate at about 20 °C. Reactance on long or large cables and hot conductors push the real drop higher.

When to use it

Use it for a quick check when sizing a cable, before a detailed calculation to your local wiring rules.

À propos de ce calculateur

Traduction (brouillon) — à relire.

Fonctionnement

Résistance du conducteur (Ω/km) = constante de résistivité ÷ section, avec 17,5 pour le cuivre et 28 pour l'aluminium. Chute de tension = K × courant × longueur aller × résistance ÷ 1 000, avec K = 2 en monophasé et √3 en triphasé. L'outil donne la chute en volts et en pourcentage de la tension, et la tension restante à la charge.

Exemple chiffré (valeurs d'illustration)

32 A sur 25 m en cuivre de 6 mm², monophasé 230 V. R = 17,5 ÷ 6 ≈ 2,917 Ω/km. Chute = 2 × 32 × 25 × 2,917 ÷ 1 000 ≈ 4,67 V, soit 2,03 %. Il reste environ 225,3 V à la charge, sous les 3 %.

Lire le résultat

L'outil compare à la recommandation courante : 3 % au plus sur un circuit terminal et 5 % au plus au total (distribution + terminal). Au-delà de 5 %, il conseille d'augmenter la section. Doubler la section divise environ la chute par deux. C'est une estimation résistive à environ 20 °C. La réactance des câbles longs ou de forte section et l'échauffement des conducteurs augmentent la chute réelle.

Quand l'utiliser

À utiliser pour une vérification rapide lors du choix d'un câble, avant un calcul détaillé selon les règles d'installation locales.

حول هذه الحاسبة

ترجمة (مسودة) — للمراجعة.

طريقة الحساب

مقاومة الموصل (أوم/كم) = ثابت المقاومية ÷ المقطع، بقيمة 17.5 للنحاس و28 للألمنيوم. هبوط الجهد = K × التيار × الطول في اتجاه واحد × المقاومة ÷ 1000، حيث K = 2 للأحادي الطور و√3 للثلاثي الطور. وتعرض الأداة الهبوط بالفولت وكنسبة من جهد النظام، والجهد المتبقي عند الحمل.

مثال محسوب (قيم توضيحية)

32 أمبير على 25 م بنحاس 6 مم²، أحادي الطور 230 فولت. R = 17.5 ÷ 6 ≈ 2.917 أوم/كم. الهبوط = 2 × 32 × 25 × 2.917 ÷ 1000 ≈ 4.67 فولت، أي 2.03%. يتبقى نحو 225.3 فولت عند الحمل، ضمن حد 3%.

قراءة النتيجة

تقارن الأداة بالإرشاد الشائع: 3% كحد أقصى للدائرة الفرعية و5% كحد أقصى للمجموع (المغذي والدائرة الفرعية). فوق 5% تنصح بزيادة المقطع. مضاعفة المقطع تخفض الهبوط إلى النصف تقريباً. هذا تقدير مقاومي عند نحو 20 درجة مئوية، والممانعة في الكابلات الطويلة أو الكبيرة وسخونة الموصلات ترفع الهبوط الفعلي.

متى تستخدمه

استخدمها كفحص سريع عند اختيار مقطع الكابل، قبل حساب تفصيلي وفق قواعد التمديدات المحلية.

In one line

Resistive AC approximation at 20°C, including power factor but excluding cable reactance. It does not select conductor size or establish compliance. Verify temperature, reactance, installation method and applicable design requirements with an electrical engineer.

Worked example

EXAMPLE

Illustrative values (the calculator defaults), not data from a real site.

Inputs
Load current32 A
One-way run length25 m
Conductor cross-section6 mm²
System voltage230 V
PhaseSingle phase (K=2)
ConductorCopper (ρ 17.5)
Power factor (resistive estimate)0.9
Results
Voltage drop4.67 V
Voltage drop %2.03%
Voltage at load225.3 V
Conductor resistance2.917 Ω/km
VerdictOK

Frequently asked questions

What voltage drop is considered acceptable?

The common design target is no more than 3% on a branch circuit and 5% combined for feeder plus branch, a guideline referenced in the NEC. Exceeding this dims lighting, causes motors to run hotter and lose torque, and wastes energy as heat in the cable.

How do I reduce voltage drop?

Increasing the conductor's cross-sectional area is usually the biggest lever, followed by shortening the run, switching from aluminum to copper, raising the system voltage, or balancing load across phases. Doubling the conductor area roughly halves the voltage drop for the same current and length.

Does this calculation account for power factor and reactance?

This is a simplified resistive, DC, calculation using the conductor's resistivity; it doesn't separately model the inductive reactance component of cable impedance, which becomes more significant on larger conductors, longer runs, or lower power factors. For critical or long runs, a full AC impedance calculation gives a more precise figure.

Should I enter the one-way or round-trip cable length?

Enter the one-way run length from source to load — the calculator's K factor, 2 for single-phase or √3 for three-phase, already accounts for the return path in the formula, so doubling the length yourself would overstate the drop.

Why does copper versus aluminum matter so much?

Copper has a lower resistivity than aluminum, so a copper conductor of the same cross-section carries the same current with less voltage drop — the tool's resistivity constant, 17.5 for copper versus 28 for aluminum, reflects that difference directly in the result.

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