Halve the current, quarter the loss

Low-voltage vehicle systems have long operated around 12 volts, while many electric traction systems use hundreds of volts. Higher voltage brings safety, insulation and component costs, but it can reduce current for a chosen power.

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

Electrical power: P = VI. Watts are volts times amps. In this simplified DC comparison, 100 kW can be represented by a smaller current at higher voltage or a larger current at lower voltage. Real motors, inverters and insulation are designed for particular operating ranges.

Power wasted in the cable: P = I²R. This is Joule heating, 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 resistance it has.

Put those side by side and the design pressure becomes clear. Raising voltage is not free, but reducing current can sharply reduce conductor heating because current is squared.

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.

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

Commit

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.

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

That calculation is deliberately self-contradictory: a 12 V source cannot both hold 12 V at the load and lose 133 kW in that cable while delivering 100 kW. The enormous result means the assumed current and conductor cannot coexist; voltage would collapse and the conductor would overheat.

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

About fourteen hundred kilograms of copper for the complete four-metre out-and-return conductor assumed here, to hold conductor loss to 120 W at 12 V. That simplified comparison ignores terminals and cooling, but it shows why a traction system cannot sensibly deliver 100 kW at 12 V.

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.

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, concentrated in a small terminal. That is enough to create dangerous temperatures quickly. Do not search for it by touch: use an appropriate voltage-drop or thermal test while following the vehicle's service procedure. Heating can damage surfaces and insulation and may accelerate some corrosion mechanisms, worsening the connection.

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 only when the same power, resistance and voltage-at-load assumptions hold.
• Required conductor cross-section can fall for a chosen loss and thermal limit, reducing mass and packaging burden.
• 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 appropriately rated power semiconductors and insulation coordination, including suitable clearance and creepage distances. Silicon-carbide devices can improve switching and conduction performance and are common in 800 V systems, but they are not a physical requirement; silicon devices can also be designed for these voltages. Cost, cooling, switching frequency and efficiency determine the choice.

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.