To determine the size of the transmission wires will require knowledge of the strength of current (in amperes) to be carried, and the distance in feet. In transmission, the electric current is again analogous to water flowing in pipes. It is subject to resistance, which cuts down the amount of current (in watts) delivered. [Illustration: Bringing wires into the house or barn] The loss in transmission is primarily measured in volts; and since the capacity of an electric current for work equals the _volts_ multiplied by _amperes_, which gives _watts_, every volt lost reduces the working capacity of the current by so much. This loss is referred to by electrical engineers as the "C^2R loss," which is another way of saying that the loss is equal to the _square of the current in amperes_, multiplied by _ohms_ resistance. Thus, if the amperes carried is 10, and the ohms resistance of the line is 5, then the loss in watts to convey that current would be (10 × 10) × 5, or 500 watts, nearly a horsepower. The pressure of _one volt_ (as we have seen in another chapter) is sufficient to force _one ampere_, through a resistance of _one ohm_. Such a current would have no capacity for work, since its pressure would be consumed in the mere act of transmission. If, however, the pressure were _110 volts_, and the current _one ampere_, and the resistance _one ohm_, the effective pressure after transmission would be 110-1, or 109 volts. To force a 110-volt current of _50 amperes_ through the resistance of _one ohm_, would require the expenditure of _50 volts_ pressure. Its capacity for work, after transmission, would be 110-50, or _60 volts, × 50 amperes_, or 3,000 watts. As this current consisted of _110 × 50_, or 5,500 watts at the point of starting, the loss would be 2,500 watts, or about 45 per cent. It is bad engineering to allow more than 10 per cent loss in transmission. There are two ways of keeping this loss down. One is by increasing the size of the transmission wires, thus cutting down the resistance in ohms; the other way is by raising the voltage, thus cutting down the per cent loss. For instance, suppose the pressure was 1,100 volts, instead of 110 volts. Five amperes at 1,100 volts pressure, gives the same number of watts, power, as 50 amperes, at 110 volts pressure. Therefore it would be necessary to carry only 5 amperes, at this rate. The loss would be 5 volts, or less than 1/2 of 1 per cent, as compared with 45 per cent with 110 volts. [Illustration: Splicing transmission wire] In large generating stations, where individual dynamos frequently generate as much as 20,000 horsepower, and the current must be transmitted over several hundred miles of territory, the voltage is frequently as high as 150,000, with the amperes reduced in proportion. Then the voltage is lowered to a suitable rate, and the amperage raised in proportion, by special machinery, at the point of use.