In this case the value of the steady current as calculated by Ohm's law is 10 amperes, but Helmholtz's law shows us that with the great self-induction which we have assumed to be present, the current, even at the end of 30 seconds, has only risen up to within 5 percent. of its final value; and only at the end of two minutes has practically attained full strength. These values are set out in the highest curve in Fig. 54, in which, however, the further supposition is made that the number of spirals, S, in the coils of the electromagnet is 100, so that when the current attains its full value of 10 amperes, the full magnetizing power will be Si = 1000. It will be noticed that the curve rises from zero at first steeply and nearly in a straight line, then bends over, and then becomes nearly straight again, as it gradually rises to its limiting value. The first part of the curve--that relating to the strength of the current after _very small_ interval of time--is the period within which the strength of the current is governed by inertia (i.e., the self-induction) rather than by resistance. In reality the current is not governed either by the self-induction or by the resistance alone, but by the ratio of the two. This ratio is sometimes called the "time constant" of the circuit, for it represents _the time_ which the current takes in that circuit to rise to a definite fraction of its final value. E = 10 r = 1 R = 100 L = 10 Si 1000 + _..------------------------------- | . _ _--------- | . .---- | . .- 2 IN SERIES | . .- | - | .: - : | .: . : 500 | . : __- -:--------------------------- | . : _.- - : 2 IN PARALLEL | . :. - : | . / : - : | . / - : |. / - : : |./. : : |/_____:_____________:____________________________ t 10 20 40 60 80 100 120 FIG. 54.--CURVES OF RISE OF CURRENTS. This definite fraction is the fraction (e - 1)/e; or in decimals, 0.634. All curves of rise of current are alike in general shape, they differ only in scale, that is to say, they differ only in the height to which they will ultimately rise, and in the time they will take to attain this fraction of their final value. _Example (1)._--Suppose E = 10; R = 200 ohms; L = 8. The final value of the current will be 0.025 amp. or 25 milliamperes. Then the time constant will be 8 ÷ 400 = 1-50th sec. _Example (2)._--The P.O. Standard "A" relay has R = 400 ohms; L = 3.25. It works with 0.5 milliampere current, and therefore will work with 5 Daniell cells through a line of 9,600 ohms. Under these circumstances the time constant of the instrument on short circuit is 0.0081 sec.