Question 20.10: (a) What current is needed to transmit 100 MW of power at 20...

(a)
To find the current, we rearrange the relationship P_{ ave }=I_{ rms } V_{ rms } and substitute known values. This gives

I_{ rms }=\frac{P_{\text {ave }}}{V_{ rms }}=\frac{100 \times 10^{6} W }{200 \times 10^{3} V }=500 A.                        (20.52)

(b)
Knowing the current and given the resistance of the lines, the power dissipated in them is found from P_{ ave }=I_{ rms }^{2} R. Substituting the known values gives

P_{ ave }=I_{ rms }^{2} R=(500 A )^{2}(1.00 \Omega)=250 kW.                      (20.53)

(c)
The percent loss is the ratio of this lost power to the total or input power, multiplied by 100:

\% \text { loss }=\frac{250 kW }{100 MW } \times 100=0.250 \%.                  (20.54)

Discussion
One-fourth of a percent is an acceptable loss. Note that if 100 MW of power had been transmitted at 25 kV, then a current of 4000 A would have been needed. This would result in a power loss in the lines of 16.0 MW, or 16.0% rather than 0.250%. The lower the voltage, the more current is needed, and the greater the power loss in the fixed-resistance transmission lines. Of course, lower-resistance lines can be built, but this requires larger and more expensive wires. If superconducting lines could be economically produced, there would be no loss in the transmission lines at all. But, as we shall see in a later chapter, there is a limit to current in superconductors, too. In short, high voltages are more economical for transmitting power, and AC voltage is much easier to raise and lower, so that AC is used in most large-scale power distribution systems.