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Simulation · Electronics

Voltage drop calculator

Set the supply, the current, the cable length and the wire size. Watch the voltage fade along the cable, see the heat it wastes, and get the smallest size that keeps the drop under 3 or 5 percent.

Change the length, current or wire size
SupplyBattery, DC
12 V
10 A
10 m
4 mm²
Voltage at the load11.14 V
Voltage drop
Heat in the cable
Size for the limit
Longest run at this size

Current makes a round trip

The current flows out on one wire and back on the other, so a load 10 m away has 20 m of wire in its circuit. Both wires drop voltage.

Vdrop = I × 2L × ρ ÷ A

Volts matter less at high voltage

The same cable and current lose the same volts at 12 V and at 230 V, but as a share of the supply the loss is 19 times bigger at 12 V.

% drop = Vdrop ÷ Vsupply

The lost volts become heat

Every volt the cable drops, times the current, is power warming the cable instead of running the load.

Ploss = I2 × Rloop

How do you calculate voltage drop in a cable?

Multiply the current by the resistance of both wires: drop = I × 2L × ρ ÷ A, where L is the one-way length, A the conductor area and ρ the resistivity (0.0172 Ω·mm²/m for copper). A 10 A load at the end of 10 m of 4 mm² copper loses 0.86 V. On a 230 V circuit that is 0.4%, nothing to worry about. On a 12 V solar system it is 7.2%, and the charge controller sees only 11.14 V.

That is the whole problem in one example: the volts lost depend only on current, length and wire size, but whether they matter depends on the supply voltage. Most limits are 3 percent for lighting and sensitive loads and 5 percent for everything else. The simulation draws the cable with its voltage marked along the run, shows the heat it wastes and picks the thinnest standard size that meets your limit, in mm² as used in India and Europe or in AWG. For a worked 12 V solar example, read what wire size you need for 12 V solar over 10 metres.

Key results from the simulation

  • Low voltage needs thick wire: 10 A over 10 m needs 9.58 mm² of copper for a 3 percent drop at 12 V, so a 10 mm² cable. At 24 V the same 10 A needs half that, 4.79 mm², so 6 mm².
  • Length doubles: the current goes out and comes back, so a 10 m run has 20 m of wire in the loop. Forgetting this halves the calculated drop.
  • Aluminium needs about 1.6 times the area: its resistivity is 1.64 times copper's. A 40 m, 32 A feed at 230 V drops 3.1% on 10 mm² aluminium and 1.9% on 10 mm² copper.
  • Hot wire loses more: copper's resistance rises about 0.39 percent per degree. The 12 V solar example drops 7.2% at 20 °C and 8.6% with the conductor at 70 °C.

Six setups worth trying

Each button loads the setup into the simulation. Lengths are one way, from the supply to the load. Numbers are for conductors at 20 °C unless stated.

12 V solar panel, 10 m from the charge controller

DC, 10 A, 4 mm² copper

Two 100 W panels in parallel deliver about 10 A at 12 V nominal. Run that through 10 m of 4 mm² cable and the loop resistance is 86.2 mΩ, so 0.86 V disappears and 8.6 W warms the cable. That is a 7.2% drop, more than twice the 3 percent target.

The calculator's answer is 10 mm², which drops 0.34 V (2.9%). The cheaper fix is to raise the voltage: wire the panels in series into an MPPT controller and the current halves, so the drop in volts halves and the supply doubles. The 12 V solar wire size post works through both options.

Drop 0.86 V (7.2%) At the controller 11.14 V Heat 8.6 W Size for 3% 10 mm²

A 12 V LED strip 5 m from its power supply

DC, 6 A, 1.5 mm² copper

A 72 W strip draws 6 A. On 1.5 mm² wire over 5 m the drop is 0.69 V (5.7%), and the far end of the strip runs visibly dimmer and often yellower. LED strips are worse than lamps here, because their current limiting resistors leave little headroom.

The limit-meeting size is 4 mm². The other fix is feeding the strip from both ends, which halves the current in each lead. The existing post on what gauge wire for 12 V covers strips and fridges in AWG.

Drop 0.69 V (5.7%) At the strip 11.31 V Size for 3% 4 mm² Max run on 1.5 mm² 2.61 m

A 3 kW geyser on 2.5 mm² wire, 25 m from the board

Single-phase AC, 13 A

A 3 kW water heater on 230 V draws 13 A. Through 25 m of 2.5 mm² copper the drop is 4.5 V, which is 1.9%, inside the 3 percent limit. The heater still sees 225.5 V. Yet the cable turns 58.3 W into heat, spread over 50 m of conductor.

At mains voltage, drop rarely decides the size for normal house runs. Current carrying capacity does: 2.5 mm² PVC cable is good for roughly 27 A clipped to a wall, and less when bunched in conduit or buried in insulation. Long runs to an outhouse or a borewell pump are where drop starts to matter.

Drop 4.5 V (1.9%) At the geyser 225.5 V Heat in cable 58.3 W Max run for 3% 38.5 m

Aluminium versus copper on a 40 m feed

Single-phase AC, 32 A, 10 mm²

Aluminium is cheaper and lighter, and it is common for service cables and long feeders in India. Its resistivity is 1.64 times copper's, so the same 10 mm² cable over 40 m at 32 A drops 7.2 V (3.1%) instead of 4.4 V (1.9%) for copper, and wastes 232 W instead of 141 W.

The rule of thumb is to go one standard size up when you switch to aluminium. Both pass the 5 percent limit here, but only copper passes 3 percent. Aluminium also needs terminals rated for it, because it creeps under pressure and its oxide layer insulates.

Aluminium drop 3.1% Copper drop 1.9% Resistivity ratio 1.64× Al heat 232 W

A 15 A circuit on 14 AWG, 30 m (about 100 ft) away

Single-phase AC, 120 V

14 AWG copper is 1.63 mm across, 2.08 mm², and has 8.29 Ω per kilometre at 20 °C. A full 15 A load 30 m away drops 7.5 V, which is 6.2% of 120 V. The load sees 112.5 V.

US guidance (an informational note in the National Electrical Code) suggests 3 percent for a branch circuit. That needs 10 AWG for this run. Lower supply voltage is why long US runs are upsized more often than the same runs at 230 V.

Drop 7.5 V (6.2%) At the load 112.5 V Size for 3% 10 AWG 14 AWG area 2.08 mm²

A 1200 W inverter on a 12 V battery, 2 m of cable

DC, 100 A, 25 mm² copper

At 12 V a 1200 W inverter pulls about 100 A. Even 2 m of 25 mm² cable, a thick welding-style lead, drops 0.28 V (2.3%) and wastes 27.6 W. Each metre of each wire dissipates 6.9 W.

Short and fat is the rule for battery leads. Inverters cut out at low input voltage, and a sagging battery plus cable drop trips that alarm long before the battery is flat. This is also why bigger systems move to 24 V or 48 V.

Drop 0.28 V (2.3%) Heat 27.6 W Size for 3% 25 mm² Max run at 25 mm² 2.61 m

Copper cable: drop for 10 A over 10 m, and the longest run for 3 percent

SizeΩ/km per wireDrop at 10 A, 10 mDrop at 12 VHeatMax run, 12 VMax run, 230 V
1.5 mm²11.492.30 V19.2%23.0 W1.57 m30 m
2.5 mm²6.901.38 V11.5%13.8 W2.61 m50 m
4 mm²4.310.86 V7.2%8.6 W4.18 m80 m
6 mm²2.870.57 V4.8%5.7 W6.26 m120 m
10 mm²1.720.34 V2.9%3.4 W10.4 m200 m
16 mm²1.080.22 V1.8%2.2 W16.7 m320 m
25 mm²0.690.14 V1.1%1.4 W26.1 m500 m

Resistance from the annealed copper standard, 1/58 Ω·mm²/m at 20 °C. IEC 60228 allows real stranded conductors up to about 7 percent more (for example 7.41 Ω/km maximum for 2.5 mm² against 6.90 here), so add a margin when you are close to the limit. Max run is one way, for 10 A and a 3 percent drop.

How the calculator works

Each conductor has resistance R = ρL ÷ A. For copper the simulation uses ρ = 1/58 = 0.017241 Ω·mm²/m at 20 °C, the International Annealed Copper Standard. For aluminium it uses 0.028264 Ω·mm²/m (61 percent of copper's conductivity), the value in Indian Standard IS 10810 Part 2. The "hot wire" setting raises both with temperature coefficients of 0.00393 and 0.00403 per °C.

AWG sizes are converted to area with the standard formula d = 0.127 mm × 92^((36 − n) ÷ 39), so 12 AWG is 2.05 mm across and 3.31 mm². At 75 °C that gives 1.93 Ω per 1000 ft, close to the 1.98 Ω/kft the US electrical code lists for stranded 12 AWG.

The circuit has two conductors in series with the load, so the loop resistance is 2LR' and the drop is I times that. The percent drop is the drop divided by the supply voltage. The heat is I² times the loop resistance. The size for the limit is the area A = 2LIρ ÷ (limit × V), rounded up to the next standard mm² or AWG size.

The 3 percent and 5 percent limits are the IEC 60364-5-52 Annex G values for installations fed from a public low voltage network (3 percent for lighting, 5 percent for other uses), the same figures the US code's informational notes recommend. The red verdict also checks a rough current rating so a cable that is merely within the drop limit but too thin to carry the current is flagged.

What the model leaves out

  • Reactance and power factor: for AC the model treats the cable as a pure resistance at unity power factor. Above about 16 mm², or with motors, the cable reactance (about 0.08 mΩ per metre) and a power factor below 1 add a little more drop.
  • Three-phase: a balanced three-phase circuit drops √3 × I × L × R' instead of 2 × I × L × R'. The simulation is single-phase and DC only.
  • Real conductor resistance: stranded cable is allowed to be a few percent above the ideal value. See the table note.
  • Inrush and motor starting: motors draw several times their running current when they start, so the drop at that moment is much larger.
  • Connections: every crimp, fuse holder, switch and terminal adds resistance. On 12 V systems a single poor joint can add as much drop as metres of cable.
  • Current capacity: the ampacity check is a rough guide for PVC cable on a wall. Bunching, conduit, insulation and hot roofs reduce what a cable can carry. Use your wiring code and the cable maker's tables for that.

Common misconceptions

"Voltage drop depends on the voltage." The drop in volts does not: 10 A through 0.086 Ω loses 0.86 V whether the supply is 12 V or 230 V. Only the percentage changes, which is why low-voltage systems need much thicker wire.

"If the wire does not get hot, it is big enough." Ampacity and voltage drop are different limits. A 12 V run can be cool to the touch and still deliver 10 percent less voltage than the load needs.

"Use the one-way length." The current goes there and back. Unless the return is a vehicle chassis, use twice the distance.

Frequently asked questions

What is the formula for voltage drop?

For DC or single-phase AC: voltage drop = 2 × length × current × resistivity ÷ area. With length in metres, area in mm² and copper resistivity 0.0172 Ω·mm²/m, 10 A over 10 m of 4 mm² copper drops 0.86 V.

What is an acceptable voltage drop?

IEC 60364-5-52 gives 3 percent for lighting and 5 percent for other uses on installations fed from the public network. Low-voltage DC systems usually aim for 3 percent or less, because the loads and chargers have little headroom.

What size wire do I need for 12V over 10 metres?

For 10 A and a 3 percent limit you need at least 9.58 mm² of copper, so 10 mm² (about 7 AWG). For a 5 percent limit, 6 mm² is enough. Halving the current, or doubling the voltage, halves the area you need.

Does aluminium wire have more voltage drop than copper?

Yes. Aluminium's resistivity is about 1.64 times copper's, so for the same drop it needs about 1.6 times the cross-section, usually one standard size larger.

Is voltage drop calculated on one way or round trip length?

Round trip. Current flows to the load on one conductor and back on the other, so the resistance of both counts. Formulas that use the one-way length include a factor of 2 for this.

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