# how much less power in kW is wasted?

A power station delivers 530 kW of power at 12,000 V to a factory through wires with total resistance 2.5 . How much less power is wasted if the electricity is delivered at 50,000 V rather than 12,000 V?

P(transmission) = I * V (solve for I)

I(transmission) = 530,000 Watts / 12,000 V = 44.167 Amps

P(loss) @12kV = (I^2) * R

(44.167 Amps)^2 * 2.5 Ohms = 4876.8 Watts (Power Loss @ 12kV)

P(transmission) = I * V (solve for I)

I(transmission) = 530,000 Watts / 50,000 V = 10.6 Amps

P(loss) @50kV = (I^2) * R

(10.6 Amps)^2 * 2.5 Ohms = 280.9 Watts (Power Loss @ 50kV)

Answer = (Power Loss @12kV) – (Power Loss @ 50kV)
= 4876.8 Watts – 280.9 Watts
= 4595.9 Watts (less power wasted if delivered at 50kV)

4595.9 Watts

Source(s): I have an Elelectrical Engineering Degree

P = I^2 * R
at 12,000 V
I = P/V = 530/12=44.166666666666666666666666666667
P = 44.16 ^2 * 2.5
=
4876.736111

at 50,000 V
I = P/V = 530/50=10.6
P = 10.6^2 * 2.5
=280.9

4876.736111 -280.9 = 4595.83611

Source(s): I am working on my civil engineering degree.

P = I^2R

watts = amps *volts

530,000 watts = 12,000 V * I
I = 44.16 amps

530,000 watts = 50,000 V *I
I = 10.6 amps

P of 12,000V = (44.16)^2 * 2.5 = 4,875.3
P of 50,000V = (10.6)^2 *2.5 = 280.9