Electronics
How To Calculate Voltage Drop
Voltage drop is the loss along a cable as current flows through its resistance. It is proportional to the length of the run, the current, and the resistance of the conductor, and it is counted twice because the current must return.
Quick Answer
V_drop = 2 x length in km x resistance per km x current
- L
- One-way cable length, in kilometres for the resistance figure used
- I
- Current in amperes
- R
- Conductor resistance in ohms per kilometre
- V_drop
- Volts lost along the run, counted over both conductors
A 30 metre run of cable rated 7.41 ohms per kilometre carrying 20 amperes has a loop resistance of 0.4446 ohms. That drops 8.892 volts, which is 3.8660869565 percent of a 230 volt supply, leaving 221.108 volts at the load. Doubling the run to 50 metres raises the drop to 14.82 volts.
What Is Voltage Drop?
Voltage drop is the reduction in voltage along a cable caused by the current flowing through the cable's own resistance. Every conductor has some resistance, and pushing current through it produces a voltage loss given by Ohm's law. The load at the far end therefore sees less than the supply voltage, and the difference is the drop.
The calculation counts the distance twice because current has to travel out to the load and back to the source. A 30 metre run has 60 metres of conductor in the circuit, and that doubling is the factor of two at the front of the formula. Forgetting it halves the answer and is the single most common error in the calculation.
Resistance is normally tabulated per kilometre of cable, which is why the length enters the formula in kilometres. A 30 metre run is 0.03 kilometres, and multiplying by a resistance of 7.41 ohms per kilometre gives 0.2223 ohms per conductor. Doubling for the return path gives a loop resistance of 0.4446 ohms.
Multiplying the loop resistance by the current gives the drop in volts. Twenty amperes through 0.4446 ohms loses 8.892 volts. That is the absolute figure; the figure that usually matters for compliance is the drop expressed as a percentage of the nominal supply voltage.
Expressing the drop as a percentage makes it independent of the supply voltage and allows a single limit to be applied everywhere. Dividing 8.892 volts by 230 volts and multiplying by 100 gives 3.8660869565 percent. Wiring regulations commonly set a limit somewhere between three and five percent for the whole installation.
The drop grows in direct proportion to the length, so doubling the run doubles the loss. At 50 metres the loop resistance becomes 0.741 ohms, the drop becomes 14.82 volts and the percentage rises to 6.4434782609, which is well past the usual limit. Length is the parameter with the biggest practical leverage.
The drop also grows in direct proportion to the current, so a load that draws twice as much loses twice as many volts. That is why a circuit sized for a small load can fail when a larger appliance is plugged in: the cable has not changed, but the current has, and the drop scales with it.
Conductor resistance falls as the cross-sectional area rises. Moving from a cable rated 7.41 ohms per kilometre to one rated 4.61 ohms per kilometre cuts the loop resistance to 0.2766 ohms and the drop to 5.532 volts, or 2.4052173913 percent. Choosing a heavier cable is the standard remedy for an excessive drop.
Three-phase circuits behave slightly differently because the return current is shared between the other phases. The drop in a balanced three-phase circuit is smaller than in a single-phase circuit carrying the same current, and the formula picks up a factor of the square root of three rather than two. The single-phase form used here is the conservative one.
The voltage at the load is the supply voltage minus the drop. Subtracting 8.892 from 230 leaves 221.108 volts, and it is that figure a device actually receives. Equipment rated for a narrow voltage window may malfunction well before the drop reaches a regulatory limit, so checking the delivered voltage is worth doing separately.
Formula
V_drop = 2 L I R
The factor of two accounts for the outbound and return conductors. Length is in kilometres when resistance is given in ohms per kilometre.
| Symbol | Meaning | Unit | Notes |
|---|---|---|---|
| L | One-way length | kilometres | In kilometres, to match the resistance units. |
| I | Current | amperes | The load current in amperes. |
| R | Resistance per kilometre | ohms per kilometre | From the cable datasheet. |
| V | Voltage drop | volts | The volts lost over the whole loop. |
percent = V_drop / V_supply x 100
Converts the absolute drop into the form regulations are written in, so a single limit applies at any supply voltage.
| Symbol | Meaning | Unit | Notes |
|---|---|---|---|
| V_drop | Voltage drop | volts | The loss computed from length, current and resistance. |
| V_supply | Nominal supply voltage | volts | 230 volts for a typical single-phase supply. |
| pct | Percentage drop | percent | Usually limited to a few percent of the supply. |
V_load = V_supply - V_drop
What the equipment actually receives. A load can be outside its own tolerance band even when the percentage drop looks acceptable.
| Symbol | Meaning | Unit | Notes |
|---|---|---|---|
| V_supply | Supply voltage | volts | Measured at the origin of the circuit. |
| V_drop | Voltage drop | volts | The loss along the run. |
| V_load | Voltage at the load | volts | The delivered voltage. |
How To Calculate Voltage Drop
- 1
Convert the one-way length into kilometres
Cable resistance is quoted per kilometre, so a 30 metre run becomes 0.03 kilometres. Keeping the length in metres while using a per-kilometre resistance understates the drop by a factor of a thousand.
- 2
Double the length for the return path
Current flows out along one conductor and back along another, so the loop is twice the one-way length. This factor of two is the most frequently forgotten part of the formula.
- 3
Multiply by the resistance per kilometre
0.03 kilometres at 7.41 ohms per kilometre is 0.2223 ohms per conductor, or 0.4446 ohms around the loop. The resistance figure comes from the cable datasheet for the conductor size and material.
- 4
Multiply by the current to get the drop in volts
Twenty amperes through 0.4446 ohms gives 8.892 volts. The drop scales linearly with current, so a heavier load loses proportionally more.
- 5
Convert to a percentage and subtract from the supply
8.892 divided by 230 and multiplied by 100 is 3.8660869565 percent, and subtracting the drop from 230 leaves 221.108 volts at the load. Both figures matter: one for compliance, one for the equipment.
Examples
Example 1: A 30 metre run at 20 amperes
- Supply
- 230 volts
- Current
- 20 amperes
- One-way length
- 30 metres
- Cable resistance
- 7.41 ohms per kilometre
| Step | Calculation | Result |
|---|---|---|
| Loop resistance | 2 x 0.03 x 7.41 | 0.4446 |
| Voltage drop | 0.4446 x 20 | 8.892 |
| Percentage drop | 8.892 / 230 x 100 | 3.8660869565 |
| Voltage at the load | 230 - 8.892 | 221.108 |
Result: The drop is 8.892 volts, or 3.8660869565 percent of the supply, which leaves 221.108 volts at the load.
Example 2: The same load on a 50 metre run
- Supply
- 230 volts
- Current
- 20 amperes
- One-way length
- 50 metres
- Cable resistance
- 7.41 ohms per kilometre
| Step | Calculation | Result |
|---|---|---|
| Loop resistance | 2 x 0.05 x 7.41 | 0.741 |
| Voltage drop | 0.741 x 20 | 14.82 |
| Percentage drop | 14.82 / 230 x 100 | 6.4434782609 |
Result: Stretching the run to 50 metres raises the drop to 14.82 volts, which is 6.4434782609 percent of the 230 volt supply.
Example 3: A heavier cable on the original run
- Supply
- 230 volts
- Current
- 20 amperes
- One-way length
- 30 metres
- Cable resistance
- 4.61 ohms per kilometre
| Step | Calculation | Result |
|---|---|---|
| Loop resistance | 2 x 0.03 x 4.61 | 0.2766 |
| Voltage drop | 0.2766 x 20 | 5.532 |
| Percentage drop | 5.532 / 230 x 100 | 2.4052173913 |
Result: Switching to a cable rated 4.61 ohms per kilometre cuts the drop to 5.532 volts, or 2.4052173913 percent of the supply.
Calculator
Voltage drop in volts
8.892
- Drop as a percent of supply
- 3.8661
- Voltage at the load
- 221.108
- Loop resistance in ohms
- 0.4446
Values update as you type. This calculator covers the single scenario its formula assumes — see Common Mistakes for what it leaves out.
Prefer a full-width tool? Open the Voltage Drop calculator page.
Common Mistakes
Forgetting the return conductor
The factor of two at the front of the formula counts the outbound and return paths. Leaving it out halves the drop and can make a non-compliant circuit look acceptable.
Mixing metres with a per-kilometre resistance
Resistance tables are quoted per kilometre, so the length must be converted before multiplying. Using 30 instead of 0.03 inflates the drop by a thousand.
Applying a single-phase factor to a three-phase circuit
A balanced three-phase circuit shares the return current between phases and uses the square root of three rather than two. Applying the single-phase factor overstates the drop on a three-phase run.
Checking only the percentage and ignoring the delivered voltage
A drop that passes the regulatory limit can still leave equipment below its own tolerance band. Both the percentage and the voltage at the load deserve a look.
Assuming the drop is independent of load
The drop is proportional to current, so a circuit sized for a small load can fail when a heavier appliance is connected. The cable has not changed; the current has.
Using the conductor's nominal size instead of its resistance
Two cables of the same cross-sectional area can have different resistances depending on the conductor material and construction. Use the datasheet figure rather than assuming a standard value.
FAQ
Why does the formula multiply the length by two?
Because the current has to travel out to the load and back to the source, so it passes through twice the one-way length of conductor. In a single-phase circuit that doubling is exactly what the factor of two represents.
What is an acceptable voltage drop?
Wiring regulations commonly limit the drop to a few percent of the nominal supply, often three percent for a final circuit and five percent for the whole installation from the origin. The exact figure depends on the standard in force and the type of circuit.
How do I reduce voltage drop?
Use a heavier cable, shorten the run, or reduce the current. Since the drop is proportional to length, current and resistance, any of those three levers works, and the choice usually comes down to cost and practicality.
Does voltage drop waste energy?
Yes. The lost volts multiplied by the current give the power dissipated as heat in the cable, so a large drop is both a performance problem and a running cost. A drop of 8.892 volts at 20 amperes wastes about 178 watts along the run.
How is the calculation different for three-phase?
A balanced three-phase circuit shares the return current between the other two phases, so the loop factor is the square root of three rather than two. That makes the drop smaller for the same conductor and current, which is why three-phase distribution is preferred for long runs.
What resistance per kilometre should I use?
Take it from the cable manufacturer's datasheet, because resistance depends on the conductor material, its cross-sectional area and its construction. Published tables for standard copper and aluminium sizes cover most domestic and commercial work.
References
- [1]Wikipedia, Voltage drop — https://en.wikipedia.org/wiki/Voltage_drop
- [2]Wikipedia, Ohm's law — https://en.wikipedia.org/wiki/Ohm%27s_law
- [3]Wikipedia, Electrical resistance and conductance — https://en.wikipedia.org/wiki/Electrical_resistance_and_conductance