ELECTROMOTIVE FORCE AND CURRENT. 



31 



of this line circles of radius equal respectively to the two 

 other voltages. By joining the intersection of the circles 

 to each end of the horizontal line complete the triangle. 

 This will then be a triangle of electromotive forces for the 

 circuit. (See Fig. 14.) 



The electromotive force measured at the terminals of 

 the non-inductive resistance will be in phase with the current 

 in the resistance, since there is no self-induction. The electro- 

 motive force V 2 is, therefore, always equal to C x R, the 

 product of the current, and the constant resistance. 



Consequently the phase difference between this electro- 

 motive force and the electromotive force of the remaining 

 part of the circuit will be the same as the phase difference 

 between the current and the voltage of the remainder of the 

 circuit. Thus in the triangle (Fig. 14) the angle <P t will be 

 the angle of lag in the portion r of the circuit. 



SCALE OF VOLTS 



10 20 30 40 50 60 70 80 90 100 110 

 FIG. 14. TRIANGLE OF E.M.F. 



The angle <P between F 2 and V 3 will be the angle of phase 

 difference between the current and total voltage of the 

 circuit including the added non-inductive resistance. 



The readings entered on the table are those taken from 

 an actual experiment, the corresponding vector diagram for 

 the first set of readings being drawn in Fig. 14. 



Example. (1) An alternator giving a voltage of 200 

 supplies an alternating circuit in which is inserted a non- 

 inductive resistance of - 2 ohm. The current is found to be 32 

 amperes, and the voltage of the part of the circuit not 

 including the -2 ohm resistance is 198 volts. Draw a 

 diagram of the various voltages in the circuit, and ascertain 



