electrical engineering hopeful here.
Can someone explain with math how transmitting power at higher voltages reduces resistive loss?
I know Joule's Law
Power is proportional to (I^2)*R
lets say we have two identical direct current power lines - 1000 feet and resistance is 2 ohms.
One wire will run at 1,000 volts, and the other 10,000 volts. if both have a load of 500 watts, how can it be shown that the higher voltage line with experience less heating loss?
my botched attempt to solve this - I know half of every step i make is probably wrong:
wire 1: 1,000v load = 500w, so 500w/1000v = .5a = appliance amps 1000v/.5a = 2,000 ohms 2000 + 2 = 2002 ohm serial resistance. 1000v/2002r = .49a circuit amps (less current
wire 2: 10,000v
load = 500w, so 500w/10,000v = .05 ohms = appliance 2 resistance.
.05 + 2 = 2.05 ohm serial circuit resistance.
10,000/2.05 ohm = 4878a * 10,000v = 48,780,000 watts