52 ALTERNATING CURRENTS 



finding the power in a three-phase circuit, a method which applies 

 to every possible case, and which only involves the use of two watt- 

 meters. For this reason it is known as the two-wattmeter method. 



A large number of methods of measurement and calculation in 

 connection with three-phase systems is based on various more or less 

 arbitrary and frequently unjustifiable assumptions, such as sine waves 

 of p.d. and current, perfect balance of the system, accessibility of the 

 neutral point, or phase differences of exactly 120 between the three 

 p.d.'s or the three currents. Now, when the load is not a balanced 

 one, not only are the three p.d.'s in general unequal, but their phase 

 differences are also unequal, and different from 120. 



The two- wattmeter method of power measurement is entirely free 

 from any such arbitrary assumptions, as will be evident from a study 

 of its theory. In Fig. 36, A, B, and C are the three line wires con- 

 veying power to the three terminals A', B', C', to which the load is 

 connected. The load may be connected either delta- or star-fashion. 

 In either case, we may always arrange a star-connected load, as shown 



in Fig. 36, having a neutral 

 point O, which is equivalent in 

 every respect to the actual load. 

 Let the instantaneous values of 

 the currents along the three wires 

 (reckoned positive when flowing 

 from left to right) be i\, i 2 , and i 3 , 

 and let v it v 2 , and v 3 be the instan- 

 taneous p.d.'s between the neu- 

 tral point and the line wires 

 (reckoned positive if directed to- 



FIG. 36. Power Measurement in Three- wards O). The wattmeter COn- 



phase Circuit. nections are arranged as shown. 



It is to be noted that the series 



coils of the wattmeters may be placed in any two of the line wires, 

 provided the free ends of their shunt coils are in connection with 

 the remaining line wire. If w = total instantaneous power, we 

 evidently have 



w ss V&! + vftia + v a i a (1) 



Now, i\ and * 3 are the currents traversing the series coils of the 

 wattmeters, while * 2 does not flow through any wattmeter coil. It 

 will, therefore, be convenient to eliminate i% from the expression for the 

 instantaneous power. This is easily done by means of the equation 



*i + *a + *3 = 



which must obviously be true in every possible case. Putting 

 i 2 = ii i s in (1), and rearranging the terms, we find 



w = (vi - 0ft) ^ + (v a - 0ft) i 3 . . . ., . (2) 



