202 



POLYPHASE CURRENTS. 



[Exp. 



From a 3-wire 2-phase supply, connect two resistances as load, 

 one on each phase. Measure the currents, 7 A and 7 B , in each 

 resistance and the total current 7 in the common conductor. If 

 the two currents 7 A and 7 B differ in phase by 90, we will have 

 1= V-^A 2 + 7 B 2 . This will be true for an inductive as well as 

 A for a non-inductive load, provided the 



load on each phase has the same 

 power factor, i. e., A B . 



If EA and EE are not at right 

 angles, or OA and 0B are not equal, the 

 currents 7 A and 7 B will no longer be at 

 right angles; the branch currents will 

 still, however, add as vectors to give 

 the total current, as in Fig. 7. 



14. Line drop. To illustrate line 



FIG. 7. Addition of currents. 



drop, with the same circuits and re- 

 sistances just used, insert a small additional non-inductive resist- 

 ance in the supply wires to represent resistance in a long supply 

 line. 



Construct a triangle OAB for the supply voltage and O'A'B' 

 for the delivered voltage for the following three cases: 



O IB B' B 



FIG. 8. Resistance in 

 lines A and B. 



B' B 



FIG. 9. Resistance in FIG. 10. Resistance in 

 common conductor. all three lines. 



With resistances in lines A and B only, Fig. 8; 

 With a resistance in the common conductor O only, Fig. 9; 

 With resistances in all three lines, Fig. 10; in this third case 

 measurements of voltages O'A and O'B are also to be taken. 

 For the first case (Fig. 8), the supply voltages, OA and OB, 



