Halve the current, quarter the loss

Move the same power at twice the voltage and you need half the current — and the heat the cable makes falls to a quarter, not a half. That square is why traction packs are 400 volts and not 12.

Reading time: 7 min read
Tier: Free, and stays free
Physics review: Awaiting physics review

Rests on

Electrical power P = VI
Ohm's law V = IR
Joule heating P = I²R
Resistivity and its temperature coefficient

This lesson assumes

Not required. Read it here, or go down into it and come back.

Where the energy goes — A vehicle never makes energy and never destroys it. It moves energy between forms, and everything it cannot use becomes heat. That is the whole subject.

A car has run on 12 volts for seventy years. An electric one runs its traction system at 400, and increasingly at 800. Nobody moved to high voltage because it is safer or cheaper — it is neither. They moved because of one squared term.

The claim you will see everywhere, and why it is wrong

Search for why electric cars use high voltage and you will find, over and over, some version of this:

Halve the current and you halve the losses.

That sentence has appeared on this website, and it is wrong. It is not wrong by a rounding error either — it understates the effect by a factor of two, and in doing so it hides the entire reason the industry went to the trouble.

Halve the current and you quarter the losses. Here is why, and here is what it is worth.

Two equations, and they are not the same equation

Power delivered: P = VI. Watts are volts times amps. A 100 kW motor can be fed 100 kW as a small current at a high voltage or a large current at a low one — the load does not care which, as long as the watts arrive.

Power wasted in the cable: P = I²R. This is , and only the current appears in it. Not the supply voltage. The cable does not know or care what voltage the system runs at; it only knows how many amps are being pushed through it and how much it has.

Put those side by side and the whole design follows. Voltage buys you power for free. Current costs you heat, and it costs you heat at the square.

The same 100 kW, at three voltages

I = P ÷ V

At 12 V: 100,000 ÷ 12 = 8,333 A

At 400 V: 100,000 ÷ 400 = 250 A

At 800 V: 100,000 ÷ 800 = 125 A

What that costs in a real cable

Take a run of 35 mm² copper, two metres out to the motor and two metres back, which is four metres of conductor.

Cable resistance and the heat it makes

R = ρL ÷ A, where ρ is resistivity, L is length, A is cross-section

ρ (copper, 20 °C) = 1.68 × 10⁻⁸ Ω·m, L = 4 m, A = 35 mm² = 35 × 10⁻⁶ m²

R = (1.68 × 10⁻⁸ × 4) ÷ (35 × 10⁻⁶) = 1.92 × 10⁻³ Ω, or 1.92 mΩ

Loss = I²R

At 400 V and 250 A: 250² × 0.00192 = 62,500 × 0.00192 = 120 W

At 800 V and 125 A: 125² × 0.00192 = 15,625 × 0.00192 = 30 W

Half the current. A quarter of the heat. 120 watts down to 30, from the same cable moving the same 100 kilowatts to the same motor.

Predict, then run

You are about to halve the current through a cable, with its resistance unchanged. Before you touch the slider — what happens to the heat it makes?

Nothing — the cable carries what it carries

Halved

Quartered

Doubled

You learn more from being wrong on purpose than right by accident.

The 12-volt version, which is where it stops being an argument

Run that same cable at 12 volts and the arithmetic turns violent.

8,333 amps through the same 35 mm² cable

Loss = I²R = 8,333² × 0.00192 = 69,440,000 × 0.00192 = 133,000 W

That is 133 kW of heat — more than the 100 kW you were trying to deliver.

So size the copper to hold the loss at the same 120 W instead:

Required resistance R = P ÷ I² = 120 ÷ 8,333² = 1.73 × 10⁻⁶ Ω

Required area A = ρL ÷ R = (1.68 × 10⁻⁸ × 4) ÷ (1.73 × 10⁻⁶) = 0.0389 m² = 38,900 mm²

That is a round bar about 220 mm across.

Its mass, at copper's 8960 kg/m³: 4 m × 0.0389 m² × 8960 = 1,390 kg

Fourteen hundred kilograms of copper, per cable, to do at 12 volts what 35 mm² does at 400. The cable would weigh more than the car it was fitted to. That is not an engineering trade-off; it is a closed door, and the square is what closes it.

The same square, in a fault you will actually meet

Most "electrical faults" on an ordinary car are not electrical in any interesting sense. They are connections. Corrosion, a loose earth, a spread terminal, a chafed strand count — all of which do exactly one thing: add a small resistance in series with something that needs current.

Take a starter motor drawing 300 A through an earth strap that has picked up 5 milliohms of corrosion.

Five milliohms, at starter current

Voltage lost at the joint: V = IR = 300 × 0.005 = 1.5 V

On a 12 V system, that is over 12% of the supply gone before the starter sees it.

Heat made at the joint: P = I²R = 300² × 0.005 = 90,000 × 0.005 = 450 W

Four hundred and fifty watts, in a joint the size of a coin. That is a domestic fan heater on its low setting, concentrated in a terminal — which is why bad earths are found by touch as often as by meter, and why they get worse rather than staying still: heat drives oxidation, oxidation adds resistance, more resistance makes more heat.

Why you measure the drop and not the resistance

Put an ohmmeter across that strap and it will very likely read 0.00 Ω. Not because the fault is not there, but because a hand-held meter's own test leads carry a few tens of milliohms and its resolution is no better. You are trying to see five milliohms through an instrument that cannot resolve twenty.

So you measure the across the joint while the circuit is doing its work. At 300 A the same five milliohms produces 1.5 volts, which any meter reads without difficulty.

Most workshops carry a rule of thumb — on the order of 0.2 V across a single ground connection and 0.5 V across a whole high-current circuit, measured under load. Treat those as a starting point, not gospel, and use the manufacturer's figure where one is published.

And it gets worse as it gets hot

Copper's resistance rises with temperature by about 0.4% per kelvin. A cable running 100 K above ambient has roughly 39% more resistance than the figure on the datasheet, so it makes about 39% more heat at the same current.

For a properly sized cable that settles somewhere — losses rise, temperature rises, the cable sheds more heat to the air, and it finds an equilibrium. For a bad joint it often does not settle, because the mechanism that adds resistance is corrosion rather than temperature, and heat accelerates it. That is the difference between a warm cable and a melted connector.

What the electric car does with all this

An 800-volt pack does the same job as a 400-volt one with half the current, so:

Conduction losses fall to a quarter in the cables, the busbars and the connectors.
Copper mass falls, which matters because every kilogram of cable is a kilogram of range.
DC fast charging goes faster at the same cable current limit, since power is volts times amps and the amps are what the cable and the connector are limited by.

It is not free. Higher voltage needs power semiconductors that can block it, which in practice means silicon carbide rather than the silicon IGBTs a 400-volt inverter can use, and it needs wider clearance and creepage distances everywhere the insulation has to hold off a fault. Both cost money. The industry paid it anyway, because the square is on the other side of the ledger.

What you now know

That P = VI and P = I²R are two different equations doing two different jobs. That voltage is how you deliver power and current is how you waste it. That the waste goes as the square, so halving the current quarters the loss — and that this one fact is why a traction pack is 400 volts, why a bad earth burns, and why you find that bad earth by measuring volts under load rather than ohms at rest.