A 300 MW power station produces electricity at 80 kV which is then supplied to consumers along cabes of total resistance 5.0 ohms. What percentage of the power is lost in the cables.
P = VI = V2/R
The Attempt at a Solution
I said the total power was PT = 300 MW
Power dissipated in the cables: Pd = V2/R = (80 kV)2/(5.0 ohms) = 1280 MW
This didn't make any sense because that power dissipated exceeds the power produced. So here's another attempt:
Power remaining: PR = 300 MW
Power dissipated: PD = V2/R = (80 kV)2/(5.0 ohms) = 1280 MW
Total Power: PT = PR + PD = (1280 + 300) MW = 1580 MW
Percentage Lost: PD/PT = 1280/1580 = 81.0%
I'm not sure if this is right but it's my best guess. Any help is appreciated. Thanks.