# 3.5 Alternating current versus direct current  (Page 3/10)

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## Power losses are less for high-voltage transmission

(a) What current is needed to transmit 100 MW of power at 200 kV? (b) What is the power dissipated by the transmission lines if they have a resistance of $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega$ ? (c) What percentage of the power is lost in the transmission lines?

Strategy

We are given ${P}_{\text{ave}}=\text{100 MW}$ , ${V}_{\text{rms}}=\text{200 kV}$ , and the resistance of the lines is $R=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega$ . Using these givens, we can find the current flowing (from $P=\text{IV}$ ) and then the power dissipated in the lines ( $P={I}^{2}R$ ), and we take the ratio to the total power transmitted.

Solution

To find the current, we rearrange the relationship ${P}_{\text{ave}}={I}_{\text{rms}}{V}_{\text{rms}}$ and substitute known values. This gives

${I}_{\text{rms}}=\frac{{P}_{\text{ave}}}{{V}_{\text{rms}}}=\frac{\text{100}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{W}}{\text{200}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V}}=\text{500 A}.$

Solution

Knowing the current and given the resistance of the lines, the power dissipated in them is found from ${P}_{\text{ave}}={I}_{\text{rms}}^{2}R$ . Substituting the known values gives

${P}_{\text{ave}}={I}_{\text{rms}}^{2}R=\left(\text{500 A}{\right)}^{2}\left(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \right)=\text{250 kW}.$

Solution

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

$\text{% loss=}\frac{\text{250 kW}}{\text{100 MW}}×\text{100}=0\text{.}\text{250 %}.$

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.

It is widely recognized that high voltages pose greater hazards than low voltages. But, in fact, some high voltages, such as those associated with common static electricity, can be harmless. So it is not voltage alone that determines a hazard. It is not so widely recognized that AC shocks are often more harmful than similar DC shocks. Thomas Edison thought that AC shocks were more harmful and set up a DC power-distribution system in New York City in the late 1800s. There were bitter fights, in particular between Edison and George Westinghouse and Nikola Tesla, who were advocating the use of AC in early power-distribution systems. AC has prevailed largely due to transformers and lower power losses with high-voltage transmission.

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