INDUCTANCES AND RESISTANCES 



happens we need to know: (1) the turns ratio n; (2) the coupling factor ^; 

 (3) the primary inductance L^,; and (4) the winding resistances. Armed with 

 this information we construct the equivalent circuit (Figure 4.34). If the 

 instantaneous currents after closing the switch are as shown 



also 

 Further 



K + Rl 



'AB 



'AB 



n" 



Lp ^^ ih - '2) 



d/i _ . Rs + Rl , d^2 

 dt '2 «2^„ "^ d? 



h{R. + r) ^^^{\ - k^)L,-^ vj^j, 



Close at 



frO 



Rp*r 





Figure 4.34 



Combining the equations and differentiating gives 



d^/g 

 df2 + 



Rr, + r + 



«' 



+ 



«' 



(1 -k^)Lj 



)d/2 , 

 dt ^ 



Wil - a 







The general solution for this equation is extremely cumbersome, and we 

 shall not concern ourselves with it. There are three particular cases which 

 are of interest, and these are : 



R I p 

 (1) — ^ — ^ — ^> Rp + r which we might call the 'lightly loaded case'; 



(2) 



(3) 



Rs + Rl 

 Rs + Rl 



«2 



Rjy-\- r which is approximately the matched condition ; 



<^Rp-\- r the 'heavily loaded case', where the secondary 



winding is practically short-circuited. 



The solutions are in the form of values of i^- More useful is y^_B which is 

 the input voltage to the ideal transformer and is /.,(i?s + Rz)ln^. The actual 

 voltage delivered to the load is then n times this. 



Case (1) (lightly loaded) v^_s = E .e ^^ 



Case (2) (matched) 



Va-b -2V 





2.{Rj,+r)l 



_ g (l-k^)L 



'•) 



67 



