The same 100 MW is sent down the same 1.00 ohm line at 25 kV instead of 200 kV. What fraction of the power is now lost?

1. First find the current: I = P / V = (100 x 10^6 W) / (25 x 10^3 V) = 4000 A. 2. Then the line loss: P_loss = I squared R = (4000 A) squared (1.00 ohm) = 16.0 x 10^6 W. 3. That is 16.0 MW lost out of the 100 MW sent. 4. As a percentage, (16.0 MW / 100 MW) x 100 = 16.0 percent. 5. Dropping the voltage by a factor of eight raised the current by eight and the loss by sixty four. 6. 0.250 percent is the loss at 200 kV, so choosing it ignores the change of voltage entirely. 7. 1.60 percent scales the loss linearly with voltage instead of with its square. _Source: OpenStax College Physics (CC BY 4.0), Ch 20 "Electric Current, Resistance, and Ohm's Law", section Why Use AC for Power Distribution?_