Voltage drop
The key idea
Every conductor has impedance, and current through impedance costs voltage. The farther the power travels and the more current it carries, the less voltage arrives at the end.
The idea
The voltage printed on the network diagram is a promise made at the source. What a customer actually receives is that promise minus everything lost along the way.
Each meter of cable has resistance and reactance — an impedance. When load current flows through it, a little of the voltage is used up pushing the current along. The drop follows from three things, multiplied together:
- How much current flows. Twice the load, twice the drop.
- How far it flows. Twice the length, twice the impedance, twice the drop.
- What it flows through. A small cable has more ohms per kilometer than a large one.
At the source end, everything is fine. It is the customer at the end of a long, heavily loaded feeder who receives 0.93 pu and watches the lights dim when the neighbor's motor starts.
Try it
at the load: 382 V · 0.955 pu · drop 18 V
400 V feeder, 0.32 Ω/km cable, power factor 0.9. Push either slider and watch the voltage at the load fall — and hover the plot to read the voltage anywhere along the run.
Why it matters
- Voltage limits are contractual and physical. Networks are typically required to deliver within ±5% or ±10% of nominal. Motors overheat and electronics reset below that.
- It decides conductor sizes. On long rural feeders, voltage drop — not heating — is usually the constraint that forces a larger cable.
- Reactive power makes it worse. The drop depends on the current the conductor carries, and a poor power factor means more current for the same work. Fixing power factor at the load is often the cheapest voltage fix.
- It is the question a power flow answers. For a real meshed network with dozens of loads, the simple formula stops being enough — the power flow is the full version of this widget.
The math, if you want itOptional — the page reads completely without it
For a three-phase feeder with current I, length L, and cable impedance R + jX per unit length, at load power factor cos φ:
the approximate drop
ΔV ≈ √3 · I · L · ( R cos φ + X sin φ )
The R cos φ term dominates on small LV cables (mostly resistance); the X sin φ term matters on larger conductors and overhead lines — which is why reactive-heavy loads sag the voltage even on lightly loaded feeders.
See it in Phasor
Run a power flow in Phasor and the voltage heatmap paints this page's story across your whole network at once: green at the source, fading toward the weak ends, with every bus outside its band flagged in the violations list.