750 watts is slightly less than 63 amperes. So, the question is whether the car
chassis can supply 63 amperes.
Let's pick a 4 meters long steel object. If it weighs 500 kg (I assume a car
chassis weighs 500 kg, correct me if this is incorrect), its volume for density
8000 kg/m3 (different types of steel have different density, but 8000 kg/m3 is a rough approximation) is is 0.0625 m3 and its cross sectional area is then 0.015625 m2.
The electrical resistivity of carbon steel (NOT stainless steel) is 1.43e-7
ohm*m. This means the resistance of 0.015625 m2 cross section is 9.1520e-6
ohm/m, and for 4 meter long object it's 3.6608e-5 ohm. For a current of 63
amperes, it's 0.0023063 volt drop. In other words, 2.3 millivolts. Do you think
this is a problem? I don't think it is a problem.
The loss of power in watts is 0.145 watts, compared to 750 watts. Quite
Now, how much copper do you need to achieve the same resistance? For 4 meters
long wire, you need 1.59e-8 ohm*m / A * 4 m = 3.6608e-5 ohm, solving the
equation gives A = 0.0017373 m2. In other words, 1737 mm2. Such a 4 meter long
copper wire would weigh 62.3 kg! My 5 meter long jumper cables are just 50 mm2.
I don't know if you consider 62.3 kg copper wire a problem, but I do. There's no way you can match the resistivity of the chassis with copper!
In summary: almost always you want to use the amount of metal in the car
chassis instead of installing separate wires. Even though steel does not have optimal resistivity, the sheer amount of metal in the car chassis overcomes the slightly less than optimal resistivity.
Edit: what is the current carrying capacity of the car's chassis? If you
consider 0.1 volt drop ok, it's 0.1 V / 3.6608e-5 ohm = 2732 amperes. This
would make the car chassis dispose of heat at 273 watts. I'm pretty sure you
could perhaps even double the voltage drop to 0.2 V, meaning 5463 amperes and
1.1 kilowatt heat disposal. I'm certain the car chassis can dispose of that
amount of heat.