# joule heating - transmitting power at higher voltages reduces resistive loss?

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

• Your second calculation falls on its face immediately. $\dfrac{W}{V}\neq \Omega$ Dec 29, 2013 at 3:48
• thanks, yeah i realized that half way through. i sort of got so tangled in my own math mistakes and naivete I gave up and posted with obvious bone head mistakes. Dec 29, 2013 at 4:02

Wire 1:

V = 1000V

P = 500W

I = P/V = 500W / 1000V = 0.5A

R_wire = 2 Ohm

Loss in wire 1 = I^2*R = 0.5A^2 * 2 Ohm = 0.5W

Wire 2:

V = 10000V

P = 500W

I = P/V = 500W / 10000V = 0.05A

R_wire = 2 Ohm

Loss in wire 2 = I^2*R = 0.05A^2 * 2 Ohm = 0.005W

UPDATE:

To work out the exact loss, assuming voltage is measured at start of wire:

simulate this circuit – Schematic created using CircuitLab

Total current, I = V1 / (R1 + R2)

Load power, P2 = 500W = I^2 * R2 = (V1^2 R2) / (R1 + R2)^2

This equation can be rearranged into a quadratic form and R2 solved for. Once R2 is found, the loss in the wire can be found by I^2 * R1

To find it using the P = VI formula, calculate V2 as I * R2, the loss is given by (V1 - V2) * I.

• why does Joule's Law (I^2)*R give the same result as voltage * current? is V*I the exact same thing mathematically as I^2*R? thanks! Dec 30, 2013 at 7:30
• Yes it will, but you have to use the voltage between the start and end of the wires, not to ground. The above calculation assume 1000v / 10000v is the voltage at the load end. I'll update... Dec 30, 2013 at 12:00
• And P=I^2*R only applies to resistors, while V*I is always equal to power. Dec 30, 2013 at 18:19