INDUCTANCES AND RESISTANCES 



Self inductances in parallel, but possessing some mutual coupling 



If two coils, which considered separately have self inductances L^ and Lg, 

 are connected in parallel, their effective inductance is not {L^L^j^L-^ + L^ if 

 there is mutual coupUng between them. It is not difficult to show that the 

 effective inductance is now 



M 2 - L1L2 



^eff 



2M - (Li + L2) 



Generally speaking the only case of coupled inductances in parallel which 

 is likely to occur is when L^ = L^. Putting M equal to k{L-^L^^l'", we get 



L^Uk^ - 1) 

 -t-eff 



and if L^ = L^ 



L{\ + k) 



Thus if A: = 0, the effective inductance of the parallel combination is L/2, 

 as anticipated. If A: = 1, the effective inductance is merely L — a result which 



M=kL 



Effective \nductar\ce A-B =L('\*^<) Effective inductance /4-8 = A.f1 -/<J 



2 2 



Figure 4.17 



may at first sight seem surprising. This assumes that the sense of connections 

 is such that the m.m.f.'s of the two coils act in the same direction. If they 

 oppose one another the effective inductance for two equal coupled inductances 

 becomes 



L{\ - k) 



Leu = 2 



When A; = 0, the effective inductance is L/2 as before. When k = \, the 

 effective inductance is zero (Figure 4.17) 



Tightly coupled mutual inductance connected to a constant direct voltage 

 generator 



If a tightly coupled mutual inductance be connected to a generator of 

 e.m.f. E {Figure 4.18), then the current rises in L^, as we have seen, according 

 to the equation 



d//d? = L/Li 



This current will induce in Lg a constant e.m.f. E2 = M dz/d/ = (MfL-^Ei 

 but M/Li == N^lN^, therefore E^ = (NjNi)Ei. 



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