Why the grid runs at 400,000 volts.
Push the same 500 MW down the same wire — only the voltage changes. Since current falls as voltage rises (P = VI), and resistive loss goes as the square of current, a 10× step up in voltage leaves just 1/100th the loss.
power station
to city / distribution
Transmission voltage V
V ↑ 10× ⇒ loss ↓ 100×
11 kVdistribution ↔ ultra-high-voltage transmission765 kV
Try
This models pure Joule heating (I²R loss) in the conductor itself — it leaves out corona discharge and transformer core losses, which are real but comparatively small. P = 500 MW is representative of one large power-station unit; R = 15 Ω represents one specific, fixed wire — a ~300 km line at roughly 0.05 Ω/km, a plausible resistance for a large overhead conductor — held constant on purpose, so the only thing changing across the slider is voltage, matching the "same wire" framing above. Since I = P/V, loss = I²R = R(P/V)², which is why doubling voltage always quarters the loss, independent of P or R. The ~5% line on the gauge is a common informal ceiling engineers use for "economical" line loss, not a hard regulatory limit; at low, distribution-class voltages, 500 MW over 300 km would draw more current than any real conductor of this resistance could carry without failing — which is exactly why bulk power is never actually sent that way. This is also why the grid uses step-up transformers to reach transmission voltage for the long haul, then step-down transformers to bring it back down near the load, where I²R no longer matters because the distances are short.