On 10/04/19 08:00, Hugh Byrne wrote:
Today the car wouldn't start at a friend's house who had no jumper cables.
He had some electrical wire though.
We were just starting to jury rig something when his wife showed up with
cables in her trunk.

If we did jury rig electrical wires, we were unsure of what size would
work.

How can we determine what size electrical wire would work to jump a typical
sedan in an emergency?

I would start with a 5 AWG if less than 10 ft length.
Climb to 4 AWG size at least for a 12 ft length.

I made up 12 ft cables with 0 AWG size that always work.

The problem with small size jumper cables is not the
overheating, but the voltage drop. If your source is low
of power, it drops alone to 8-9 volts or less under strain,
and adding a drop of 3 volts gives you only 5-6 volts at the
receiving side ... useless.

Resistance of a cable:

R = rho * length / section

where rho is the electrical resistivity. If rho(copper)
is given as

rho(copper) = 1,68 * 10^(-8) ohm*meter

the length should be in meters and the section in squaremeters.
Cable can be made with aluminium (less weight):

rho(aluminium) = 2,65 * 10^(-8) omh*meter

For copper calculated values:

https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes

The suggested 5 AWG for 10 feet length would read as:

0,3133 mΩ/ft * 10 ft = 3,133 mΩ per cable

You need to sum up both cables resistance, that si

3,133 * 2 = 6,266 mΩ = 0,006266 ohms

From Ohm's law, the voltage drop would be:

V = R * I = 0,006266 * I

At I = 100 A you have only a 0,63 voltage drop.
But if I climbs to 500 A you have up to 3,15 volts.


The suggested 4 AWG for 12 feet length would read as:

0,2485 mΩ/ft * 12 ft = 2,982 mΩ per cable

almost as above with the smaller, but shorter cable.


My 0 AWG 12 feet cables read as:

0.09827 mΩ/ft * 12 ft = 1,179 mΩ per cable

and take up to:

I = V / R = 1 / 0,001179 = 848 amperes

before dropping 1 volt.

The trick with the Li-ion power bank cables is that
they are very short, a couple of feet.