MEASUREMENT OF POWER IN TRI-PHASE CIRCUITS. 195 



This formula can be easily proved as follows : Let ABO 

 (Fig. 83) be the voltage triangle. Then if be the centre of 



gravity of masses -p, -, and - placed at A, B, and C respec- 



jAj V T 



tively, OA, OB, and 00 will be the three P.D.s to the centre of 

 the star-box. 



Also by II. (5)- 



R r 



= 3 . 



where G is the centre of gravity of the triangle ABC ; 



04 E 



Now, if the arms had been equal GA would be the P. D. across 



GA 

 the volt-coil. Hence the required multiplying factor is 3-x-r, and 



T 



this we have shown equals 2 4- # 



MISCELLANEOUS EXAMPLES. 



1. If a coil of resistance r and self-induction L is subjected to an alternating 

 P.D. e of varying frequency, and if n lt n 2 , and i lt a' 2 , are corresponding values of 

 frequency and current, show that 



e / n 2/ 2 n 2 7 ^ 



a) - = ^-x/ 2 1 2 !" 



(2) 



2 



2. If an inductance coil is subjected to an alternating P.D. of varying fre- 

 quency, and if T lt T 2 , T &9 and t lt i 2 , i a are corresponding values of periodic time 

 and current, show that 



- - 



3. An induction motor takes 75 amperes at full load when its power factor is 

 0*83, what is (1) the energy current and (2) the wattless current ? 



Answers. Energy current = 62-25 amperes. 



Wattless current = 41 '86 amperes. 



4. What is the capacity of a condenser which allows a current of 1 ampere 

 to pass, when a P.D. of 1000 volts, having a frequency of 100 alternations per 

 second, is applied between its terminals ? 



Answer. 1'59 microfarad. 



