By Edward Brackett Raymond
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At the load. This is caused by the presence of the inductance in the To line. quired, calculate the line, first AN, or the total volts re- separate the voltage of the load into its two components, one, AC, in phase with the current, AD, and the other, AH, out of phase with the current. f. f. which from the triangle ACG : in reactance, = E" gives CG = since the square of the long side of a right-angled sum of squares of other two sides. triangle = The applied voltage -h all AN = E "'? + (2 f equals TrnLI the values of which are known.
F. of all the coils of the armature between the brushes; that is, the averThis is shown to be ex- age value of the sine curve. pressed as follows volts : =- where g < = magnetic number of through the armature, jV turns in series between brushes, n cycles per second. This relation follows, since four times in each revflux passing = olution the coils are rilled since volts now and emptied = rate of change of flux -*- io of flnx, and From this 8 . f. f. tends to tricity; that is, TT 2 = 2 IO produce a flow of tends to produce current.
Max) x = amp. square 2 ""^ =- ; V2 mean square root of as read on an ammeter, hence, E square root of mean square = 2 irnLI, where / is the square root of mean square current and 2 TrnL is a value in ohms, as can be seen, since from Ohm's law E Thus the IR, hence back R must equal 2 irnL. f. f. is a maximum 2 irnLI. f. of self-induction lags behind the current 90. f. f. drop of resistance first, IR, and second, 2 irnLI. f. of self-induction = = This can be shown in two ways: first, as in Fig. 9; and second, as in Fig.