Voltage Drop
Part of Power Transmission
How to calculate and control the voltage lost along a wire run, so that loads receive adequate voltage to operate correctly.
Why This Matters
Voltage drop is the invisible thief of electrical systems. You wire a circuit, connect a motor, and the motor runs — but sluggishly, overheating, drawing more current than it should, and wearing out prematurely. The problem is not the motor. The problem is that the voltage arriving at the motor terminals is significantly less than the voltage at the source. The resistance of the wire between them consumed some of the electrical pressure the motor needed.
This matters practically because many loads are sensitive to voltage. Motors operate below rated torque and efficiency at reduced voltage. Incandescent lamps are dimmer. Electronics may malfunction. LED drivers may flicker. Heating elements produce less heat. In every case, the load performs worse than intended, and the excess current drawn by a motor straining to maintain speed actually increases the I²R losses in the wire — a self-reinforcing problem.
Calculating voltage drop before running wire saves you from these problems. It is a simple calculation, requires only basic arithmetic, and tells you definitively whether a given wire gauge will work for a given run.
The Basic Formula
Voltage drop across a conductor follows Ohm’s Law:
V_drop = I × R
I = current through the wire (amps)
R = resistance of the wire (ohms)
For a circuit with two conductors (current flows out on one wire, returns on the other), the total wire length is twice the one-way run length, so:
R_total = 2 × L × r
L = one-way distance (meters)
r = resistance per meter for the wire gauge (Ω/m)
V_drop = I × R_total = I × 2 × L × r
The received voltage at the load is:
V_load = V_source - V_drop
And percentage voltage drop:
% drop = (V_drop / V_source) × 100
Worked Example: Lighting Circuit
You have a 120V AC system and want to run a lighting circuit to a barn 60 meters away. Total load is 200W of lamps.
Step 1 — Current:
I = P / V = 200W / 120V = 1.67A
Step 2 — Try 14 AWG copper (r = 0.0084 Ω/m):
R_total = 2 × 60m × 0.0084 = 1.008Ω
V_drop = 1.67A × 1.008Ω = 1.68V
% drop = (1.68 / 120) × 100 = 1.4%
1.4% drop — excellent. The lamps receive 118.3V, essentially full brightness. 14 AWG is fine for this run.
Worked Example: Motor Circuit
Now a more demanding case: a 1-horsepower (750W) workshop motor on a 120V circuit, 80 meters from the panel.
Step 1 — Current (at full load):
I = 750W / 120V = 6.25A
(Real motors draw somewhat more due to power factor; use 8A as a safe estimate)
Step 2 — Try 14 AWG:
R_total = 2 × 80m × 0.0084 = 1.344Ω
V_drop = 8A × 1.344 = 10.75V
% drop = (10.75 / 120) × 100 = 8.96%
Nearly 9% — unacceptable for a motor. The motor receives 109V instead of 120V. It will run hot, develop less torque, and age faster.
Step 3 — Try 10 AWG (r = 0.0033 Ω/m):
R_total = 2 × 80m × 0.0033 = 0.528Ω
V_drop = 8A × 0.528 = 4.22V
% drop = (4.22 / 120) × 100 = 3.5%
3.5% — acceptable. The motor receives 115.8V, within the typical ±5% tolerance.
Voltage Drop Limits by Load Type
Different loads tolerate different amounts of voltage drop:
| Load Type | Maximum Acceptable Voltage Drop | Notes |
|---|---|---|
| Lighting circuits | 3-5% | LEDs are often more tolerant than incandescents |
| General outlets | 5% | Typical residential standard |
| Motors | 3-5% | Sensitive to undervoltage; larger motors need tighter limits |
| Sensitive electronics | 2-3% | Regulators help but can’t compensate for severe drops |
| Heating elements | 8-10% | Power output proportional to V², but usually tolerable |
| Battery charging | 2-3% | Chargers need adequate input voltage to function |
For a settlement grid, design for 3% maximum on any individual circuit. This leaves margin for additional loads added later and for voltage variations at the source.
The Quick Reference Table
For 120V AC circuits, the wire gauge needed to stay under 3% voltage drop:
| Current | 10m run | 25m run | 50m run | 100m run | 200m run |
|---|---|---|---|---|---|
| 2A | 18 AWG | 18 AWG | 16 AWG | 14 AWG | 12 AWG |
| 5A | 18 AWG | 16 AWG | 14 AWG | 12 AWG | 8 AWG |
| 10A | 16 AWG | 14 AWG | 12 AWG | 10 AWG | 6 AWG |
| 15A | 16 AWG | 14 AWG | 10 AWG | 8 AWG | 4 AWG |
| 20A | 14 AWG | 12 AWG | 10 AWG | 6 AWG | 4 AWG |
| 30A | 12 AWG | 10 AWG | 8 AWG | 6 AWG | 2 AWG |
For 240V AC circuits: because voltage is doubled, the same wire carries the same wattage at half the current. Wire gauges from this table remain correct if you halve the current column. A 2,400W load at 240V draws 10A — use the 10A row.
For 12V DC systems (common in off-grid setups): divide all run distances by 10. A 50-meter run at 12V 10A behaves like a 500-meter run at 120V — the percentage drop is the same, but because 12V is your only voltage level, there is no stepping back up. Use the table, but treat each entry as applying to distances one-tenth as long.
Voltage Drop at Low Voltage Is a Bigger Problem
This bears emphasis because many off-grid systems operate at 12V or 24V.
The percentage drop from a given wire over a given distance and current is fixed by I×R / V_source. For the same wattage, lower-voltage systems have proportionally higher currents (I = P/V), and the I² factor means losses scale as the square of the voltage ratio.
Comparison: same 500W load, same wire, same distance
| System voltage | Current | Wire (10 AWG, 40m total) | V_drop | % drop |
|---|---|---|---|---|
| 12V DC | 41.7A | 0.132Ω | 5.5V | 46% |
| 48V DC | 10.4A | 0.132Ω | 1.37V | 2.9% |
| 120V AC | 4.17A | 0.132Ω | 0.55V | 0.46% |
| 240V AC | 2.08A | 0.132Ω | 0.27V | 0.11% |
A 12V system with any significant load over any significant distance is nearly impossible to make efficient. This is why 12V battery banks are only practical for loads within a few meters — lighting in a single room, a nearby radio, a water pump next to the battery. As soon as distance increases, stepping up to 48V, 120V, or 240V becomes mandatory.
Measuring Voltage Drop in an Existing Installation
To check an installed circuit:
- Connect the full load (run all lights, start the motor)
- Measure voltage at the source — at the panel terminals or transformer secondary
- Measure voltage at the load — at the receptacle, motor terminals, or lamp holder
- The difference is your actual voltage drop
Record measurements during maximum expected load conditions, not with a partial or no load. Voltage drop only appears when current flows.
If measured drop is too high, your options are:
- Replace wire with a larger gauge. Most effective but labor-intensive if the wire is run through walls or conduit.
- Reduce the load on the circuit. Move some loads to a separate circuit.
- Increase the source voltage. If your transformer output is adjustable (some wound transformers can be tapped at different turns), increasing output voltage by the drop amount compensates exactly. A 5V drop on a 120V circuit means setting the transformer output to 125V so the load receives 120V.
- Add a voltage booster. A small autotransformer or voltage regulator at the load end can compensate for drop, but this adds complexity and cost.
- Shorten the run. Move the source closer to the load, or add a subpanel at an intermediate point.
Voltage Drop in Transmission Lines vs. Branch Circuits
Voltage drop analysis applies at two scales:
Transmission lines (from generator to settlement distribution point) carry the total grid load. Here, the entire grid voltage budget is at stake. A 5% drop on the transmission line means every building in the settlement starts at 5% below nominal. All their internal drops stack on top of that. Design transmission line voltage drop to be 2% or less.
Branch circuits (from the distribution panel to individual loads) carry only one circuit’s worth of current. A 3% drop here is acceptable because it doesn’t affect other circuits. Design branch circuits for 3-5% depending on load type.
Total system voltage drop at any load point equals the sum of all drops in series from generator to load. If your generator output is 125V, transmission line drops to 121V (3.2% drop), and the building’s internal circuit drops another 3.5V (2.9% drop), the load receives 117.5V — a total drop of 5.9% from nominal. Still within tolerance, but any additional drop in a later circuit extension will push into problem territory.
This cascading effect is why building an electrical system requires calculating drops at each stage, not just the last one.