As an example, let us say that Farmer Jones has installed a water-power electric plant on his brook, _200 yards distant_ from his house. The generator is a 5 kilowatt machine, capable of producing _45 amperes_ at _110 volts pressure_. He has a 3 horsepower motor, drawing 26 amperes at full load; he has 20 lights of varying capacities, requiring 1,200 watts, or 10 amperes when all on; and his wife uses irons, toasters, etc., which amount to another 9 or 10 amperes--say 45 altogether. The chances are that he will never use all of the apparatus at one time; but for flexibility, and his own satisfaction in not having to stop to think if he is overloading his wires, he would like to be able to draw the full _45 amperes_ if he wishes to. He is willing to allow _5 per cent loss_ in transmission. _What size wires will be necessary, and what will they cost?_ Substituting these values in the above formula, the result is: Answer: 600 × 22 × 45 ------------- = 108,000 circular mills. 5.5 [Illustration: Transmission wire on glass insulator] Referring to the table, No. 0 wire is 105,534 circular mills, and is near enough; so this wire would be used. It would require 1,200 feet, which would weigh, by the second table, 435.6 pounds. At 19 cents a pound, it would cost $82.76. 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: Distance in feet × 22 × amperes --------------------------------- = Number of volts lost circular mills Substituting values, we have, in this case, 600 × 22 × 45/66,373 (No. 2) = 9 volts, 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.