Meaning of Watts.—It is obvious, therefore, that if we multiply the height of the water in inches with the area of the pipe, we shall obtain a factor which will show how much water is flowing.
Here are two examples:
- [p. 62]28 inches = height of the water in the reservoir.
2 square inches = size of the pipe.
Multiply 28 × 2 = 56. - 56 = height of the water in the reservoir.
1 square inch = size of the pipe.
Multiply 56 × 1 = 56.
Thus the two problems are equal.
A Kilowatt.—Now, in electricity, remembering that the height of the water corresponds with voltage in electricity, and the size of the pipe with amperage, if we multiply volts by amperes, or amperes by volts, we get a result which is indicated by the term watts. One thousand of these watts make a kilowatt, and the latter is the standard of measurement by which a dynamo or motor is judged or rated.
Thus, if we have 5 amperes and 110 volts, the result of multiplying them would be 550 watts, or 5 volts and 110 amperes would produce 550 watts.
A Standard of Measurement.—But with all this we must have some standard. A bushel measure is of a certain size, and a foot has a definite length, so in electricity there is a recognized force and quantity which are determined as follows:
The Ampere Standard.—It is necessary, first, to determine what an ampere is. For this purpose a standard solution of nitrate of silver is[p. 63] used, and a current of electricity is passed through this solution. In doing so the current deposits silver at the rate of 0.001118 grains per second for each ampere.
The Voltage Standard.—In order to determine the voltage we must know something of resistance. Different metals do not transmit a current with equal ease. The size of a conductor, also, is an important factor in the passage of a current. A large conductor will transmit a current much better than a small conductor. We must therefore have a standard for the ohm, which is the measure of resistance.
The Ohm.—It is calculated in this way: There are several standards, but the one most generally employed is the International Ohm. To determine it, by this system, a column of pure mercury, 106.3 millimeters long and weighing 14.4521 grams, is used. This would make a square tube about 94 inches long, and a little over 1/25 of an inch in diameter. The resistance to a current flow in such a column would be equal to 1 ohm.