unting the distance only one way, in feet,
and doubling the resistance constant, 10.6, which, for convenience is
taken as 22, instead of 21.2.

_Examples of Transmission Lines_

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 x 22 x 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 x 22 x amperes
--------------------------------- = Number of volts lost
circular mills

Substituting values, we have, in this case, 600 x 2