478 



Prof. T. E. Lyle on 



Similarly the energy leaving the transformer on the 

 secondary side in the same element of time dt is equal to 



dF 

 dt 



— n 9 C a ~dt; 



hence in the time dt the transformer absorbs energy to the 

 amount 71? 



{p^i + n 2 lu % )-^dt, 



so that in one cycle, of duration T, the energy absorbed 

 is equal to f*t+T 7 ™ 



1 ( w iCi + w 2 C 2 ) -jt dt = area A. 



7. It is easy to show that when n 1 C 1 + n 2 C 2 and F are 

 expressed in the forms 



n 1 Ci + WgC2=wiisin (wt—^) -f »? 3 sih 3(g>£— /a 3 ) + ra 3 sin 5(g>£ — /z 5 ) -f &c, 

 F=/ 1 sin {a)t — 4>i) 4-/ 3 sin3(W— 3 )-f/ 5 sin5(o>£ — </> 5 )-f&c, 



the integral or area A and therefore the total core loss, in 

 ergs, per cycle is equal to 



t{wi/i sin (0i— /*i) + 3m 3 / 3 sin 3($ 3 — yu 3 ) + 5m 5 / 5 sin 5 (0 5 — fi b ) -f &c.}, 



which when divided by 10 7 T, where T is the period, gives 

 the total power lost in the core in watts. 



This form of expression has the advantage of giving 

 separately the power absorbed by the iron by means of the 

 harmonics of different orders, and from the analytic expres- 

 sions for E.M.F.s and currents we can also obtain for the 

 different orders of harmonics the input, output, and copper 

 losses. Hence we can draw up a debit and credit account 

 for the individual harmonics which will afford a good test of 

 the accuracy of the wave-tracer, as the account for each order 

 should balance. 



This has been done for the first and third harmonics of the 

 three series given in this paper and the results shown in the 

 following tables. The quantities for the fifth harmonics are 

 negligible. The figures represent watts. 



Table I. — No Load. 





1st Harmonic. 



3rd Harmonic. 



Total. 







 29-93 



•06 . 





 29-99 

 29-95 





 -•28 



•002 





 -•28 

 -•30 







29 65 



•06 







29-71 



2965 





Copper loss, I. ... 

 Copper loss, II.... 







