400 Mr. E. C. Kimington on an Air- Core Transformer 

 If the coils have the same time-constants, or a 1 = « 2 = a, 



1 



a = 



(15 6) 



V3-/3 2 

 and I t~' gT 2 



Ii V 1+ 4^F~ s/^p*' 

 if also /8=1, 



«=— -5, and y" = v 1'3=1'155. 



Obviously, if a x and /3 are known the value of a 2 that 

 I 

 Ii 



makes f- a maximum is by symmetry, 



*= £ or « 2 (2-/3 2 )- a " 



or in the case of no magnetic leakage, . ^ a = — • 



If a v and « 2 are both variables, we have the two equations 

 to be satisfied, viz.: — 



1— a, 2 , 1— « 2 



2 



a 2= — To W\ an( ^ ^ 1 = 



«,« 2 (2- / 8 2 ) = l-a, 2 = l-« 2 2 , 



a 1 = a 2 =a say, 

 and 1— a 2 1 



So that to get r a maximum, the primary and secondary 



should have the same value of a, each equal to , or 



v o — p 



1 . " . I 



m the case of no magnetic leakage = —ja ; in which case j- 



will be 1*155, or a 15^ per cent, increase in impedance 

 caused by short-circuiting the secondary, and this is the 

 greatest that can be obtained. 



Consider, then, the case of a transformer having coils with 

 equal time-constants, and suppose there is no magnetic 

 leakage. 



For values of a below \/2 the impedance is increased, and 

 putting /3=1 in equation (14) gives 



- Vl + 4^ 2 

 Ii~" l+« 2 



