Silicon, Circuits, and the Digital Revolution
At 120 V, this power would require a current of I = P / V = 50,000 kW / 120 V = 400 kA. A 10-gauge wire (5 mm2 cross-sectional area) can handle 30 A before overheating -- thus more than 13,000 such wires would be required or a wire of diameter 15 cm, both physically unreasonable!
To deliver a particular amount of power, the current may be decreased by increasing the voltage, since P = I V. For example, boosting the voltage to 80 kV results in a current of (50,000kW)/(80kV)=600 A, a much more manageable value. The ratio of currents between 80 kV and 120 V is 670 -- the energy lost in the transmission lines would be that ratio squared, approximately one-half million, since the power dissipated by the transmission lines is proportional to the current squared.
If a device exists that can raise or lower voltages while maintaining power,
with very little loss of energy, then power can be transferred over greater
distances with less loss. Electric power stations can thus be situated further
from residential areas. Such a voltage conversion is difficult with dc;
however, the conversion from one ac voltage to another is straightforward
when the electrical component known as a transformer
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Last updated April 6, 2000.
© George Watson, Univ. of Delaware, 2000.