194 ALTERNATING CURRENTS 



former is then reversed, the low side being short-circuited, and 220 volts 

 applied to the high side. Instruments having the proper ranges are con- 

 nected in circuit as shown in Fig. 180. The ammeter now reads 10.5 amp., 

 the wattmeter 410 watts and the voltmeter 220 volts. 



Find: (a) Transformer core loss. (6) Equivalent resistance referred to 

 high side, (c) Equivalent resistance referred to low side, (d) Equivalent 

 reactance referred to high side, (e) Equivalent reactance referred to low 

 side. (/) Regulation of transformer at 0.8 power-factor, lagging current. 

 (g) Efficiency of transformer at full and at half load, load being at 0.8 power- 

 factor, lagging current. 



(a) Core loss is indicated directly by the wattmeter and is equal to 148 

 watts. Ans. 



410 



(6) #01 = = 3.72 ohms. Ans. 



(c) R oz = 3.72 255 2 = 0.0372 ohm. Ans. 



99ft 



(d) Zoi = ^ = 21.0 ohms. 



lU.o 



Xoi = V(21.0) 2 - (3.72) 2 = A/427 = 20.7 ohms. Ans. 



(220 \ 2 

 f^J = 0.207 ohm. Ans. 



(/) Work on high side. 



The rated high-side current is 20,000/2,200 = 9.1 amps. Using 

 equation (49), page 183. 



Vi = \/(2,200 X 0.8 + 9.1 X 3.72) 2 + (2,200 X 0.6 + 9.1 X 20.7) 2 

 A/5,492,000 = 2,340 volts. 



Regulation = = 6 .36 per cent. Ans. 



ZjZOO 



The same result is obtained using the low-side constants. 

 Vi = V(220 X 0.8 + 91 X 0.0372) 2 + (220 X 0.6 + 91 X 0.207) 2 



= A/54,920 = 234 volts 



234. _ 220 

 Regulation = - - -- = 6.36 per cent. Ans. 



(g) Full-load eff. (using high-side constants) 



= _ 20,000 X 0.80 _ = 16,000 = 

 20,000 X 0.80 + 148 + (9. 1) 2 X 3.72 16,460 



97.2 per cent. Ans. 

 Half-load eff. 



= _ 10,000 X 0.80 __ = 8,000 = 

 10,000 X 0.80 + 148 + (4.55) 2 X 3.72 8,225 



97.3 per cent. Ans. 



The same values of efficiency are obtained if the -low-side 

 current and resistance are used. 



This method of determining the efficiency is much more accu- 

 rate than an actual measurement of the output and input, 



