Figure illustrates two coils coupled in series aiding. Let L1 be the self-inductance of coil1, L2 be the self-inductance of coil 2 and M be the mutual inductance among the two coils.
Figure: Inductive Coupling in Series (Flux Aiding)
Let us consider coil 1 first.
Self-induced e.m.f e2 =- L1 (di/dt)
Mutually induced e.m.f e1′ =- M(di/dt) [due to change of current in coil 2]
Let us consider coil 2 now.
Self-induced e.m.f e2 =- L2 (di /dt)
Mutually induced e.m.f e2′ =- M ( di/ dt) [because of change of current in coil 1].
Thus the total induced e.m.f of the series coupled coils may be written as
e = e1 + e2 + e1′ + e2′
or, e =- L1 (di /dt) - L2 (di /dt) - 2 M(di /dt) =- (L1 + L2 + 2M ) (di /dt )
If L is the equivalent inductance of the coil, then we may write
e =- L (di /dt)
From Eqs. (32) and (33), we get
or, - L(di /dt) = - (L1 + L2 + 2M ) (di /dt)
or, L = L1 + L2 + 2M