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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 = v = -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 + v + 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 -Leff  (di/dt). Comparing two voltages,
                         Leff  = 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) - L (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 -Leff (di/dt)). Comparing two voltages,
                                           Leff  = L1 + L2 - 2M

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hamada i'm hamada rageh electrical power engineer my talent to write articles about electrical engineering and i depend on google books site to write my articles

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