Why the voltage at the end of the cable is lower than at the supply
Calculating voltage drop over the run, and why the cable size that satisfies current may not satisfy voltage.
The engineering question
The circuit is correctly sized for current but the load is still underpowered at the far end. How much drop is acceptable, and how do you reduce it?
Worked example
The figures below were produced by running the Voltage Drop Calculator with the same inputs — during the build, from its authoritative engine.
| Voltage drop | 5.2704 V |
|---|---|
| Voltage drop | 2.2915 % |
| Voltage at the load | 224.7296 V |
| Conductor resistance (one way) | 0.0824 Ω |
| Assessment | Drop 2.3% is within the 3% target. |
- Single-phase current travels out and back, so k = 2 over the loop.
- Loop resistance is the one-way conductor resistance doubled over the route length.
- Compare the resulting percentage with the design limit — commonly 3 % for a branch circuit.
Sanity check
Halving the length halves the drop. If your result does not behave that way, the length has been entered as the loop rather than one-way.
Tools for this problem
These are the EngiMetric calculators that compute the numbers in this topic. Each one shows its formula, its units and its limits rather than returning a bare result.
References & standards
- IEC 60364-5-52 — voltage drop recommendations
- BS 7671 — current-carrying capacity and voltage drop
- NEC Chapter 9 Table 8 — conductor resistance
Related problems
Sizing a cable for both current and voltage drop
Two independent constraints that must both be satisfied, and recognising which one actually chose the size.
Why a conduit cannot be filled completely, and how full is too full
Conduit fill percentages by conductor count, the reason they exist, and choosing the next conduit when the first does not fit.
Also: Construction & QS