Suppose, for instance, I needed to supply fifty amperes at one hundred-volt pressure ten miles distant from the generator, and had a conductor the size of a trolley wire to bring the current. The resistance of the trolley wire is one ohm for every two miles, or five ohms. The drop in voltage is found by multiplying the amperes of current by the ohms of resistance. Ten miles from the central station, therefore, the drop on fifty amperes would be 50 × 5 = 250 volts. It would, therefore, be necessary to maintain a pressure of 350 volts at the generator to deliver the fifty amperes at 100 volts. The energy supplied by the generator is 350 volts × 50 amperes = 17,500 watts = 17.5 K. W. The energy delivered to the consumer is 100 volts × 50 amperes = 5000 watts = 5 K. W. In order to deliver fifty cents' worth of electricity per hour to the consumer it would, in this case, be necessary to generate $1.75 worth of electricity at the central station. That is, about seventy per cent. of the energy generated would be wasted in transmission. If now we decide to generate this electrical energy at ten times as high voltage it will be necessary to transmit only one tenth as many amperes, or five. In this case the drop in voltage would be 5 amperes × 5 ohms = 25 volts. It would be necessary to maintain 1025 volts of pressure at the generator to deliver to the consumer the five amperes at 1000 volts = 5000 watts. That is, to deliver 5000 watts in this case we must generate 1025 volts × 5 amperes = 5125 watts, and less than 2-1/2 per cent. of the energy generated would be lost in transmission. If now the consumer must have his energy delivered at 100 volts, we must introduce a step-down transformer at his end of the line which may give him 50 amperes at 100 volts = 5000 watts. This transformer, being small, will cause a loss of 15 or 20 per cent., but if there were a very large amount to transform it might be done with a loss of only 4 per cent. [Illustration: Fig. 131] [Illustration: Fig. 132]