In a 120 volt application, you are permitted a 5% voltage drop. This isn't much as 5% of 120 volts is only about 6 volts. No big deal when running a 1200 watt portable hair dryer. If you simply size the wire to now do the same thing on 12 volts, you no longer have a 5% voltage drop. At the same current you still have a 6 volt drop with the 10X larger wire but you now lost 50% of your power in the wire. Take a hint from the pro..
Well, it's the 5% that's the issue, not the 6 V. The same would go for the 12 VDC circuits: 5% of 12 V is 0.6 V and not 6 V resulting in the same reduction in DELIVERED power, of course, not power losses---a very important difference! In fact (EE 091, not even 101), the lower the current the lower the losses (that's the electric power dissipated to heat) IN THE WIRE (R*I^2), right? The 10% losses in an inverter are the true, ACTUAL losses (converting electricity into heat).
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If your wire is ten times the thickness, it will have 1/100 the resistance---not the 1/10... :)
Well, it's the 5% that's the issue, not the 6 V. The same would go for the 12 VDC circuits: 5% of 12 V is 0.6 V and not 6 V resulting in the same reduction in DELIVERED power, of course, not power losses---a very important difference! In fact (EE 091, not even 101), the lower the current the lower the losses (that's the electric power dissipated to heat) IN THE WIRE (R*I^2), right? The 10% losses in an inverter are the true, ACTUAL losses (converting electricity into heat).
There is no magic here, is there?