What is Voltage Drop in Electrical Cables?
Whenever electric current flows through an electrical cable, energy is dissipated as heat due to the internal resistance of the metallic conductor. This resistance causes a reduction in electrical potential along the length of the wire, meaning the voltage delivered to the load is lower than the voltage supplied at the source origin.
For example, in a 230 V single-phase circuit delivering 40 A over a 50-meter run of 10 mm² copper cable, the total conductor resistance creates a voltage drop of approximately 6.33 V (2.75%). The load receives 223.67 V, safely within BS 7671 permissible limits.
That 6.33 V figure is the cold cable. Conductors gain about 0.39% of resistance per °C, so at the 70 °C thermoplastic design temperature the same circuit drops 7.57 V (3.29%), and the tabulated mV/A/m route lands near 9.2 V. Set the conductor temperature in Advanced Options to match how you intend to prove the circuit.
This calculator checks results against BS 7671:2018+A4:2026 (UK) by default and can switch to IEC 60364-5-52 Annex G international guidance via the standards selector in the input panel. Both quote 3% for lighting and 5% for other uses from a public LV supply; IEC adds a 6%/8% allowance for private supplies and up to +0.5% on runs over 100 m.
The Voltage Drop Formulas & Equations
DC Circuits (Two-Wire Loop)
V_drop = 2 × I × L × (ρ / A)Accounts for total round-trip conductor resistance across positive and return conductors, where ρ is resistivity and A is cross-sectional area.
Single-Phase AC Circuits (1-Φ)
V_drop = 2 × I × L × (r cos φ + x sin φ)Combines cable active resistance (r) and reactive reactance (x) adjusted for the load displacement power factor (cos φ).
Three-Phase AC Circuits (3-Φ Balanced)
V_drop = √3 × I × L × (r cos φ + x sin φ)The square root of 3 (≈ 1.732) represents line-to-line voltage across three balanced 120° phase-shifted conductors.
Conductor Temperature Correction
ρ_T = ρ_20 × [1 + α × (T - 20)]Accounts for positive thermal coefficient of resistance (α = 0.00393 for copper), meaning hotter cables suffer greater voltage drop.
How to Calculate Voltage Drop: Step-by-Step
- 1
Select System Type & Nominal Voltage
Choose between DC, Single-Phase AC (e.g. 230 V), or Three-Phase AC (e.g. 400 V line-to-line).
- 2
Enter Load Current and One-Way Run Length
Input the maximum design current in Amperes (A) and the total physical length of the cable route in meters (m).
- 3
Specify Conductor Cross-Section & Material
Select conductor cross-sectional area (e.g. 2.5 mm², 6 mm², 10 mm², 16 mm²) and material (Copper or Aluminum).
- 4
Review Results Against the Selected Standard
Verify voltage drop against the limits in force — BS 7671 Reg 525.1 (3% lighting / 5% other uses from the origin) or IEC 60364-5-52 Annex G, which shares those ceilings on a public supply and adds the 6%/8% private-supply and >100 m allowances.
BS 7671 & IEC 60364 Permissible Voltage Drop Limits
Under the UK Wiring Regulations (BS 7671:2018+A4:2026, Regulation 525.1 with the limits tabulated in Appendix 4 Table 4Ab) and the international IEC 60364-5-52 Annex G guidance (Table G.52.1), the voltage drop between the origin of the installation and any fixed equipment must not exceed:
| Installation Type & Standard | Lighting Circuits (Max %) | Other Uses: Sockets, Power, Heating (Max %) |
|---|---|---|
| Public LV Supply — BS 7671 (UK) | 3.0% (6.9 V at 230 V) | 5.0% (11.5 V at 230 V) |
| Public LV Supply — IEC 60364-5-52 Annex G, Table G.52.1 | 3.0% (6.9 V at 230 V) | 5.0% (11.5 V at 230 V) |
| Private LV Supply (generator, on-site transformer, solar PV) | 6.0% (13.8 V at 230 V) | 8.0% (18.4 V at 230 V) |
| Three-phase 400 V final circuits (both standards) | 3.0% (12 V) | 5.0% (20 V) |
| Long runs — IEC Annex G note | Beyond 100 m of main wiring the ceilings may be raised by 0.005% per extra metre, capped at +0.5% | |
These are the installation allowances only. The distributor's own drop sits outside them: EN 50160 lets a 230 V public supply run between −6% and +10% of nominal at the meter, so a circuit calculated at exactly 5% can still arrive at the equipment hungrier than expected. Where a board is fed by a submain the budget is shared — a common split is about 0.5% for the consumer mains, 1.5–2% for the submain, and the remainder for the final circuit.
How to Reduce Voltage Drop in Long Cable Runs
- Upsize Conductor Cross-Section: Increasing from 6 mm² to 10 mm² or 16 mm² halves conductor resistance, drastically reducing voltage loss and heat build-up.
- Shorten Cable Routing: Optimize cable paths through containment and cable trays to eliminate unnecessary run length.
- Switch to Higher Voltages or Three-Phase: For large industrial loads, distributing load across three phases reduces current per conductor by a factor of 1.732.
- Use High-Conductivity Copper: Electrolytic tough-pitch copper (0.0172 Ω·mm²/m at 20 °C) has about 39% lower resistivity than the same size of EC-grade aluminum (0.0282 Ω·mm²/m) — equivalently, aluminum drops ~64% more voltage per metre.
- Size to the hot cable, not the cold one: 0.39% more resistance per °C means a PVC conductor at its 70 °C design temperature carries roughly a fifth more volt drop than the same cable at 20 °C, and the BS 7671 mV/A/m tables already assume the hot value.
Frequently Asked Questions
What is the maximum permitted voltage drop in the UK under BS 7671?
BS 7671:2018+A4:2026 Regulation 525.1 (with the limits tabulated in Appendix 4, Table 4Ab) permits a maximum voltage drop from the origin of the installation of 3% of the nominal voltage for lighting circuits — 6.9 V at 230 V, 12 V at 400 V — and 5% for all other circuits — 11.5 V at 230 V, 20 V at 400 V. Where the installation is fed from a private LV supply (generator, transformer, solar PV), Table 4Ab doubles the allowance to 6% for lighting and 8% for other uses. The DNO-side drop from the transformer to your meter is not included: EN 50160 separately permits the supply itself to sit between −6% and +10% of 230 V.
Why does voltage drop occur in electrical cables?
Every metallic conductor has an internal electrical resistance determined by its material resistivity, length, and cross-sectional area. When electric current flows through this resistance, energy is dissipated as heat (I²R loss), which causes the electrical potential (voltage) at the end of the cable to be lower than at the supply origin.
How do you calculate 3-phase voltage drop compared to single-phase?
In a single-phase circuit, current flows out on the live conductor and returns on the neutral, so the multiplier is 2 (round trip). In a balanced 3-phase circuit, the 120° phase angle displacement between conductors reduces the effective line-to-line impedance multiplier to √3 (approximately 1.732).
How do I reduce excessive voltage drop in a long cable run?
The most effective way to reduce voltage drop is to increase the conductor cross-sectional area (e.g., upsizing from 6 mm² to 10 mm² or 16 mm²), which directly reduces resistance. Other methods include optimizing cable routing to reduce length, balancing loads across three phases, or using copper conductors instead of aluminum.
Does temperature affect voltage drop in electrical cables?
Yes. Metals have a positive temperature coefficient of resistance. As conductor temperature rises due to ambient heat or load current, resistivity increases according to ρ_T = ρ_20[1 + α(T - 20)], with α = 0.00393 /°C for copper and 0.00403 /°C for aluminum (IEC 60287-1-1). A copper conductor at the 70 °C PVC design temperature therefore has 19.6% more resistance than the same conductor cold at 20 °C — on the 230 V / 40 A / 50 m / 10 mm² example that is the difference between 6.33 V (2.75%) and 7.57 V (3.29%).
Why is my answer different from the BS 7671 mV/A/m tables?
Three assumptions differ. (1) Temperature — the Appendix 4 tables quote mV/A/m at the conductor's maximum operating temperature (70 °C for thermoplastic, 90 °C for thermosetting), while this calculator uses the temperature you enter, 20 °C by default, i.e. a cold cable. (2) Cable data — the tables use the maximum d.c. resistance permitted for the size, a few per cent above the nominal resistivity used here, and for cables up to 16 mm² they publish the resistive drop only (inductance ignored). (3) Power factor — above 16 mm² the tabulated impedance assumes cos φ ≈ 0.8, so set the same power factor and switch cable reactance on to compare like with like. Worked check: 10 mm² thermoplastic copper is tabulated at 4.6 mV/A/m, which gives 4.6 × 40 × 50 / 1000 = 9.2 V on the example run; this tool returns 6.33 V cold at 20 °C and 7.57 V when you set 70 °C. For a lightly loaded cable Appendix 4 also permits a Ct correction: mV/A/m × [230 + t_p − (C_a² − I_b²/I_t²)(t_p − 30)] / (230 + t_p).
How much voltage drop does a 7 kW EV charger need?
A 7 kW single-phase charger draws 32 A for hours at a time, so volt drop matters more here than on a shower or cooker that runs for ten minutes. Taking the 70 °C design temperature of thermoplastic cable, 6 mm² copper loses about 2.2 V per 10 m at 32 A (roughly 1% of 230 V), 10 mm² about 1.3 V per 10 m, and 2.5 mm² about 5.3 V per 10 m. That is why 6 mm² is normally the longest run you would accept for a budget of 3% (about 30 m) and 10 mm² is the default beyond it, while 2.5 mm² runs out of the 5% ceiling (11.5 V) at around 20 m. Check the chargepoint manual too: many specify a maximum loop impedance or minimum conductor size so the 6 mA DC smoothing and the RCD type still work.
Should I enter the whole ring-main length for a ring final circuit?
No — a ring feeds the load from both directions, so the two paths are effectively in parallel and carry half the current each. Take a 100 m ring of 2.5 mm² copper at 30 A with the hot-cable resistance (8.23 mΩ/m): each 50 m path carries 15 A, so the drop to the furthest socket is 15 × 0.4115 Ω ≈ 6.2 V (2.7%). A radial over the same 50 m distance would drop 24.7 V — four times as much. To reproduce the ring figure in this calculator, enter one eighth of the total ring length as the one-way run (12.5 m above), and keep margin: real rings unbalance when most of the load sits on one side, which is why long rings are commonly run in 4 mm² instead of 2.5 mm².
Is voltage drop the same thing as power wasted in the cable?
They are related but not identical. The heat in the conductors is I²R, which is what this calculator reports as "Power Loss"; the volts the load loses is I × Z projected onto the supply phasor, which is why the power factor enters the drop but never the loss. For a resistive load the two line up: a 5% drop (11.5 V) at 40 A is about 460 W being turned into heat inside the wall or ceiling — which is why a cable that is merely "legal" on volt drop is often still the wrong choice thermally.