Fig. 58
Now let us return to the volt meter itself. By referring to [Fig. 55], we see that it requires .024 ampere to move the needle of the volt meter clear across the scale, and we have found that one fluid cell was able to send enough current through the resistance of the armature to move the needle two thirds of the way across the scale. At this point we find [Fig. 1], which might be read "one-cell pressure." We prefer to commemorate the name of one of the workers in the field of electricity and call this pressure a "volt" after Alessandro Volta (1745–1827), born at Como, Italy. It is the electric pressure which is produced by one fluid cell of a certain kind. We say, then, that one volt pushes through the resistance of this armature .016 ampere. Half a volt would push through the resistance of the armature half as much current or .008 ampere. At this point we put .5. Thus each of the figures in the lower row ([Fig. 55]) shows what part of a volt is required to send enough current through this particular armature to move the needle to that point.
Fig. 59
We found out how much wire was wound upon the armature and put exactly the same amount in the outside resistance, R ([Fig. 59]). The needle now showed that one volt is able to push through twice the resistance of the armature only half as much current, and the needle stopped at .008 ampere. If this were to be the resistance in the volt meter circuit one volt should stand under .008 ampere and two under .016 and three under .024. It is evident then, that, if we know the internal resistance of a volt meter, we may make it capable of measuring greater electrical pressures by adding the proper amount of resistance. By putting at R, ([Fig. 59]) nine times the internal resistance of the instrument, thus multiplying the total resistance tenfold, the figures upon the scale of volts may be read as whole numbers from one to fifteen. In this case it will require fifteen cells to push the needle clear across the scale and ten cells to push it two thirds of the way across. If now we add enough external resistance to multiply the resistance of the armature a hundred fold it will require 150 volts to push .024 of an ampere through the armature and pull its needle clear across the scale. In this case the figures upon the scale of volts are multiplied by one hundred and read from ten to one hundred and fifty. Such a scale would adapt this volt meter for use with our 110-volt lighting circuit. Volt meters are made with a series of such external resistances, called "multipliers," attached so that they may be easily thrown into the circuit.
It is evident that we need some term so that we may speak of quantities of resistance. This need has given rise to a unit of resistance called an ohm, after George Simon Ohm (1789–1854) born at Erlanger in Bavaria. Two inches of No. 36 German silver wire, such as is wound upon the armature of this volt meter, gives one ohm of resistance. There are 125 inches of this wire upon the armature. Its resistance is, therefore, 62.5 ohms, and we may, therefore, say that one volt of electric pressure can push through 62.5 ohms of resistance .016 of an ampere of current. Ohm discovered this relationship in 1827, and formulated it as follows:
volts/ohms = amperes (not, however, using these words).
(1 volt)/(62.5 ohms) = .016 ampere.