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?

**Answer 1**

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

**Answer 2**

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.

**Answer 3**

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

**Answer 4**

38000