166 TRANSFORMERS. [Exp. 



Thus, let IV = 4I.6 watts. The total losses are found by adding 

 this constant core loss to the copper losses for each load, as in 

 Fig. 8. The efficiency* should be computed for ^, , J, J, i 

 and ij load. 



The computations for full load and half load are as follows: 

 At full load. 



Core loss = 41.6 



Copper loss = 51.4 



Total loss = 93.0 



Output = 2,000.0 



Input =2,093.0 



Per cent, loss = 100 X t0tal ! - = 100 X -^- = 4. 



input ^ 2,093 



Efficiency^ 100- 100 X 

 At half load. 



input 

 = 100 4.44 = 95.56. 



Core loss = 41.6 

 Copper loss = 12.85 

 Total loss = 54.45 

 Output =1,000. 

 Input =1,054.45 



Per cent. loss=iooX 54 ' 45 =5.16. 

 IP5445 



Efficiency = 100 5.16 =94.84. 



* (3ia). The efficiency will be different for different frequencies and 

 for different rating of voltage and current ; see 48. See 57 for a more 

 exact method of determining Wo for full-load voltage. 



Referring to Fig. 5, the core loss at 100 volts is 41.6 for the frequency 

 (66.2 cycles) used in the test. Corrected for 60 cycles, Wo = 43.3, it 

 being possible to thus determine the efficiency for a frequency not used in 

 the test. This is useful in comparing guarantees. 



t (3ib). This formula will be found much better for making computa- 

 tions than the equivalent and more usual form, 



Efficiency = Output -=- Input. 



