POLYPHASE SYSTEMS 91 



vectorially ($> a = Ebo + Eoa)> The current Ibo is given. The 

 anide between #& a and /&<> is 30 0. Therefore, the reading 

 of this wattmeter is 



TTi = E ba I bo cos (30 - 0) 



= Ei ine In n e cos (30 - 0) 



Likewise, the wattmeter TF 2 reads the product of E ca , I eo 

 and the cosine of the angle between them. From the vector 

 diagram, Fig. 89 (6), E ca is found by adding vectorially E co and 

 Eoa (#ra = ECO + #). The current I w is given. The angle 

 between E ca and I co is 30 + 0. 



Therefore the reading of this wattmeter is 

 W 2 = E C J CO cos (30 + 8) 



= Eline I line COS (30 + 0) 



Summarizing 



Wi = El cos (30 - 0) 

 TF 2 = El cos (30 + 0) 



where E and I are the line voltage and line current, respectively, 

 the system being balanced. 



Wi and W z will read alike when = and = 180. Both 

 conditions correspond to unity power-factor. When equals 

 180, however, the power has reversed. The two instruments 

 also read alike at zero power-factor (0 = 90), although this 

 condition is seldom realized. 



When = 60, corresponding to a power-factor of 0.5, W* 

 reads zero, as cos (30 + 60) = cos 90 = 0. In this case, the 

 readiim of H", tfives the total power. For angles greater than 

 60, corresponding to power-factors less than ().">, cos (30 + 0) 

 becomes in^ative. ll' : reads negative and the total power becomes 



r = \\\ - n 



Therefore, di>cretinn must be used when two single instni- 

 II 1 1 .loved. M the total power may be either the sumor 



the difference of the readings. 



It may also be shown that 



(34) 



.. 



where n i^ th coil jx.wer-faetor an^le. Then-fore it is possible to 

 obtain the pnwer-fartnr in ;( balanced three-phase system by 

 mi-an> of the wattmeter n-adings alone. 



