Line Loss Calculator: Voltage Drop and Power Loss

An electrical line loss calculator estimates voltage drop and power dissipated as heat in a cable from its length, current, conductor resistance, and circuit type. Enter the one-way cable length, conductor resistance per unit length, current, and voltage; the calculator returns voltage drop, percentage drop, and resistive power loss. These are planning estimates, not a way to choose a safe wire size or protective device.

Formula

The calculator uses a simplified resistive line model. It treats the entered resistance as the resistance of one conductor in ohms per kilometre (Ω/km), and the entered length as the one-way route length in metres. A circuit factor accounts for the path of current:

• Voltage drop: ΔV = I × R × (L / 1000) × k
• Percentage drop: ΔV% = 100 × ΔV / V
• Resistive power loss: P_loss = I² × R × (L / 1000) × k

Here, ΔV is voltage drop in volts (V), ΔV% is the drop as a percentage of supply voltage, P_loss is power dissipated in watts (W), I is current in amperes (A), R is conductor resistance in ohms per kilometre, L is one-way length in metres, V is supply voltage in volts, and k is the selected circuit factor. For a two-wire DC or single-phase AC circuit, k = 2 because current travels out and back through two conductors. For a balanced three-phase circuit, the simplified line-to-line voltage-drop factor is √3, approximately 1.732.

The model assumes equal conductors and a resistive load path. It does not include reactance, power factor, temperature corrections, connectors, or changes in load. The three-phase result is an approximation for balanced conditions, not an analysis of an unbalanced system.

Worked Example

Using the defaults—30 m one-way length, 1.15 Ω/km conductor resistance, 20 A current, 230 V supply, and DC or single-phase circuit—the factor k is 2.

1. Convert route length to kilometres: 30 / 1000 = 0.03 km.
2. Calculate voltage drop: 20 × 1.15 × 0.03 × 2 = 1.38 V.
3. Calculate percentage drop: 100 × 1.38 / 230 = 0.6%, shown as 0.60%.
4. Calculate line power loss: 20² × 1.15 × 0.03 × 2 = 27.60 W.

The calculator therefore estimates a 1.38 V drop, a 0.60% drop, and 27.60 W of resistive loss under those assumptions.

What Line Loss Means

Conductors have electrical resistance. When current flows, some electrical energy is converted to heat in the conductors rather than reaching the load. The voltage measured at the load is consequently lower than the voltage at the source. The amount depends chiefly on current, conductor resistance, and the length of the current path.

“Line loss” can also describe losses across an entire electricity network, including transformers and many distribution lines. This calculator covers only estimated conductor loss for a specified cable run. It does not calculate utility-system losses, energy use over time, or a complete installation’s efficiency.

How to Enter the Inputs

• Length: Enter the one-way distance from source to load, in metres. Do not double it; the selected factor already accounts for the return conductor in DC and single-phase circuits.
• Resistance: Use the conductor’s resistance in Ω/km, from reliable cable data for the relevant conductor material, cross-section, and operating conditions. Do not enter total circuit resistance here.
• Current: Enter the expected current in amperes. For a changing load, a single current value cannot describe every operating condition.
• Voltage: Enter the supply voltage used as the reference for the percentage calculation. For three-phase circuits, use line-to-line voltage when interpreting the displayed percentage.
• Circuit type: Choose DC or single-phase for a two-conductor path, or three-phase for the balanced-system approximation.

For example, if cable data lists 1.15 Ω/km and the route is 30 m one way, the resistance for one conductor along that route is 1.15 × 0.03 = 0.0345 Ω. In a two-conductor circuit, the round-trip resistance in this simplified model is 0.069 Ω.

Reading the Results and Limits

Voltage drop in volts is the estimated difference between source and load voltage under the modeled load. The percentage expresses that difference relative to the entered voltage. Power loss is the estimated rate of heat generation in the conductors at that current; it is not energy. To estimate energy over a steady operating period, multiply watts by hours to get watt-hours, while recognizing that real loads and current may vary.

Resistance changes with conductor temperature, and AC circuits can also have reactance. Long runs, larger conductors, unusual cable layouts, non-unity power factor, and unbalanced three-phase loads may require a more detailed calculation. The calculator’s rounded display is useful for an estimate, not evidence that the underlying inputs or assumptions are exact.

Common Mistakes and Safety

• Entering the round-trip length and also using a factor of 2, which counts the return path twice.
• Mixing metres with a resistance value in Ω/km without converting length to kilometres.
• Using a cable’s total circuit resistance as though it were resistance per kilometre for one conductor.
• Applying the three-phase factor to a single-phase circuit, or using phase-to-neutral voltage where line-to-line voltage is needed for a three-phase percentage.
• Assuming a small calculated drop proves that a cable is safe or correctly sized.

Voltage-drop arithmetic alone cannot determine a safe conductor size. Cable selection and installation also depend on factors such as current-carrying capacity, insulation, installation method, ambient temperature, grouping, fault conditions, and applicable electrical rules. Have a qualified electrical professional assess an actual installation. Do not work on energized wiring to collect measurements.

Frequently Asked Questions

How do I calculate electrical line loss?
For the simplified model, multiply current by conductor resistance per kilometre, route length in kilometres, and the circuit factor to estimate voltage drop. Multiply current squared by the same resistance and path terms to estimate power loss.

Do I enter cable length one way or round trip?
Enter one-way route length. The circuit factor includes the return conductor for DC and single-phase circuits. If you instead use a measured total loop resistance, this calculator’s resistance input and factor are not the right representation.

What is an acceptable voltage drop?
There is no universal answer for every circuit and installation. Acceptability depends on the equipment, system, design requirements, and applicable rules. Treat this result as an estimate and confirm the requirements with a qualified professional.

Does line loss equal voltage drop?
No. Voltage drop is a change in electrical potential measured in volts. Power loss is the rate at which energy is dissipated, measured in watts. They are related, but they are different quantities.

Can I use this for three-phase power?
Use the three-phase option only as a simplified estimate for a balanced system, with line-to-line voltage for the percentage. Significant imbalance, cable reactance, or other non-ideal conditions call for a fuller engineering calculation.

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in len | One-way cable length | m | 30 | 0 | 100000
in r | Conductor resistance | ohm/km | 1.15 | 0 | 100000
in i | Load current | A | 20 | 0 | 100000
in v | Supply voltage | V | 230 | 0.001 | 1000000
sel ph | Circuit type | DC or single-phase=2, Three-phase=1.732
out drop | Voltage drop | V | i * r * (len / 1000) * ph | 2
out pct | Voltage drop | % | 100 * drop / v | 2
out loss | Line power loss | W | i ^ 2 * r * (len / 1000) * ph | 2