VOLTS, AMPERES, OHMS AND WATTS
UNDERSTANDING TERMS.--We must now try to ascertain the meaning of some
of the terms so frequently used in connection with electricity. If you
intended to sell or measure produce or goods of any kind, it would be
essential to know how many pints or quarts are contained in a gallon, or
in a bushel, or how many inches there are in a yard, and you also ought
to know just what the quantity term _bushel_ or the measurement _yard_
means.
INTENSITY AND QUANTITY.--Electricity, while it has no weight, is capable
of being measured by means of its intensity, or by its quantity. Light
may be measured or tested by its brilliancy. If one light is of less
intensity than another and both of them receive their impulses from the
same source, there must be something which interferes with that light
which shows the least brilliancy. Electricity can also be interfered
with, and this interference is called _resistance_.
VOLTAGE.--Water may be made to flow with greater or less force, or
velocity, through a pipe, the degree of same depending upon the height
of the water which supplies the pipe. So with electricity. It may pass
over a wire with greater or less force under one condition than another.
This force is called voltage. If we have a large pipe, a much greater
quantity of water will flow through it than will pass through a small
pipe, providing the pressure in each case is alike. This quantity in
electricity is called _amperage_.
In the case of water, a column 1" × 1", 28 inches in height, weighs 1
pound; so that if a pipe 1 inch square draws water from the bottom it
flows with a pressure of 1 pound. If the pipe has a measurement of 2
square inches, double the quantity of water will flow therefrom, at the
same pressure.
AMPERAGE.--If, on the other hand, we have a pipe 1 inch square, and
there is a depth of 56 inches of water in the reservoir, we shall get as
much water from the reservoir as though we had a pipe of 2 square inches
drawing water from a reservoir which is 28 inches deep.
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:
1. 28 inches = height of the water in the reservoir.
2 square inches = size of the pipe.
Multiply 28 × 2 = 56.
2. 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
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.
CALCULATING THE VOLTAGE.--In order to arrive at the voltage we must use
a conductor, which, with a resistance of 1 ohm, will produce 1 ampere.
It must be remembered that the volt is the practical unit of
electro-motive force.
While it would be difficult for the boy to conduct these experiments in
the absence of suitable apparatus, still, it is well to understand
thoroughly how and why these standards are made and used.