Farmer Jones says this is more money than he cares to spend for transmission. As a matter of fact, he says, he never uses his motor except in the daytime, when his lights are not burning; so the maximum load on his line at any one time would be 26 amperes, not 45. What size wire would he use in this instance?
Substituting 26 for 45 in the equation, the result is 61,300 circular mills, which corresponds to No. 2 wire. It would cost $57.00.
Now, if Farmer Jones, in an emergency, wished to use his motor at the same time he was using all his lights and his wife was ironing and making toast—in other words, if he wanted to use the 45 amperes capacity of his dynamo, how many volts would he lose? To get this answer, we change the formula about, until it reads as follows:
Substituting values, we have, in this case,
nearly, less than 10 per cent. This is a very efficient line, under the circumstances. Now if he is willing to lose 10 per cent on half-load, instead of full load, he can save still more money in line wire. In that case (as you can find by applying the formula again), he could use No. 5 wire, at a cost of $28.50. He would lose 11 volts pressure drawing 26 amperes; and he would lose 18 volts pressure drawing 45 amperes, if by any chance he wished to use full load.
In actual practice, this dynamo would be regulated, by means of the field resistance, to register 110 plus 11 volts, or 121 volts at the switchboard to make up for the loss at half-load. At full load, his voltage at the end of the line would be 121 minus 18, or 103 volts; his motor would run a shade slower, at this voltage, and his lights would be slightly dimmer. He would probably not notice the difference. If he did, he could walk over to his generating station, and raise the voltage a further 7 volts by turning the rheostat handle another notch.
