Alternate Current Transformer. 31 



,ri = .00129 .T,= -.00024 

 .Ti + .r2=. 00105. 



30. If Kj and K.^ be the permissible copper losses per second 

 per cm.'' at full load, Ki for the primary being decided on as say 

 15.10^ ergs, and Kg differing from Kj by a small amount which 

 will depend on how the copper losses are to be divided between 

 the two coils ; then 



where c\ and c^ are the amplitudes of the current densities, and 

 p the specific resistance of copper at the expected working- 

 temperature. 



Let us take /5=1800 abs. 

 then fi = 12.91 abs. 



Let j-j, ^2) ^^e the sectional areas, /j, 4 the mean turns, and 

 ^1, ^2 tlie space factors of the two coils ; so that ^iXj, j^.^^a, are 

 their total copper sections, and ^iS-^/i, ^^^.J^i their total copper 

 volumes; then (.see § 6) since /j^/i, 



= ^1 + K say, 



^2h T2 



where k is a small quantity to be determined, depending on the 

 current densities in the two coils at full load. 



This equation, together with Si + s.^ — i^'^, give us the copper 

 sections, 



^252=2Q/;^{i-^k} 



where Q= — -— ^the harmonic mean of a, and Oo. 



Let us assume for the copper space factors 

 ^i=.5, ^2=.7 

 then Q=.583 

 and ^i5i=l. 166(1 +.42k)^^ 

 ^^j^= 1.1 66(1 -.58k)^^ 



31. If we arrange so that the current densities in the two 

 coils shall be equal at full load, then 



