Inductive Coupling in Series

       When two inductors having self inductances L₁ and L₂ are coupled in series, mutual inductance M exists between them. Two kinds of series connection are possible as follows :

1.1 Series Aiding

       In this connection, two coils are connected in series such that their induced fluxes or voltages are additive in nature.

Fig.  1

       Here currents i₁ and i₂ is nothing but current i which is entering dots for both the coils.

      Self induced voltage in coil 1 = v₁ = -L₁ (di/dt)

       Self induced voltage in coil 2 = v₂  = -L₂ (di/dt)

       Mutually induced voltage in coil 1 due to change in current in coil 2 = v₁' = -M(di/dt)

       Mutually induced voltage in coil 2 due to change in current in coil 1 = v₂' = -M(di/dt)

       Total induced voltage = v₁ + v₂  + v₁' + v₂'

                                         = (L₁ (di/dt) + L₂ (di/dt) + M(di/dt) + M(di/dt))

                                         = - (L₁ + L₂ + 2M)(di/dt)

       If L is equivalent inductance across terminals a-b then total induced voltage in single inductance would be equal to -L_eff  (di/dt). Comparing two voltages,

                         L_eff  = L₁ + L₂ + 2M

1.1 Series Opposing

       In this connection, two coils are connected in such a way that, the induced fluxes or voltages are of opposite polarities.

Fig.  2

       Here i₁ and i₂ is same series current ''i'' which is entering dot for coil L₁ and leaving dot for coil L₂.
       Self induced voltage in coil 1 = -L₁ (di/dt)
       Self induced voltage in coil 2 = -L₂ (di/dt)
       Mutually induced voltage in coil 1 due to change in current in coil 2 =  v₁'+ M (di/dt)
       Also mutually induced voltage in coil 2 due to change in current in coil 1 = v₂' +M(di/dt)
       Therefore total induced voltage = v₁+ v₂ + v₁' + v₂'
                                                        = -L₁ (di/dt) - L₂  (di/dt) + M(di/dt) + (di/dt)
                                                        = -(L₁ + L₂ - 2M)(di/dt)

       If L is equivalent inductance across terminals a and b then total induced voltage in single inductance would be equal to -L_eff (di/dt)). Comparing two voltages,

                                           L_eff  = L₁ + L₂ - 2M

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