Alternate Current Transformer. 11 



where M, A, and D have the values given in the last paragraph. 

 14. When we multiply both sides of the first equation in § 7 

 by ^Cj, we get 



AEiCiOosa=:i/'iCi''(l + -^(Sin8 +^'CosY)\ 



which expresses the power P^ supplied to the transformer as the 

 sum of three terms, of which the first 



is equal to the copper loss in the primary coil ; the second term 



*^iC;^Il^^=H,,say, 

 A " 



is equal to the total iron loss in the transformer ; and the third 



terra 



/r,CosY _ 



is equal to the power passed down to, and developed in the 

 secondary coil : — 



For, neglecting rC^ losses, the energy entering the transformer 



-dF 

 on the primary side in any element of time dt is nfiy—j:dt, and 



the energy leaving the transformer on the secondary side in the 



dF 

 same element of time dt \& — ti,f^^--j-dt, hence in the time di the 



transformer absorbs energy to the amount 



^F 



//jCi + ^A-^^A 



of which a part ^M goes to increase the magnetic energy of the 



iron, while the remainder ^W is dissipated as heat by hysteresis 



and eddy currents. 



But amp. WjCi + ^aCa = F/o- SO that we may write «iCi + «.3C.2 = 



F - 



-^intvt in which case F = FSin(7t/^— 8) where 8 is the angle of 



magnetic lag: — 



F- . 

 hence <f M + dW = w— ^\nzvtCo?,{wt — 8)d/. 



Integrating over a complete period T, M returns to its original 

 value and we get the core loss per cycle 



