Inductive Coupling in Parallel

       When two inductors having self inductance L₁ and L₂ are coupled in parallel, we have two kinds of connections as follows.

1.1 Parallel Aiding 

      Consider parallel coupling of two inductors as shown in Fig.1.

Fig. 1

       Applying KVL to both loops, we get,

                   - jω L₁ i₁ - jω M i₂ + v = 0

                   - jω L₂ i₂ - jωM i₁ + v = 0

i.e.                 v = jωL₁ i₁ + jωM i₂               ...............(1)

                      v = jωL₂ i₂ + jωM i₁                 ..............(2)

We have,        jωL₁ . i₁ + jωM .  i₂ = jωL₂ .  i₂ + jω M . i₁

But                  i = i₁ +  i₂

i.e.                    i₂ = i -  i₁

       Putting value of i₂ in above equation, we get

.^..                      jωL₁ i₁ + jω M(i - i₁) = jω L₂ ( i - i₁) + jω M i₁

.^..                       jω i₁ (L₁ + L₂ - 2M) = jω i(L₂ -M)

.^..                       i₁= {(L₂ - M)/(L₁ + L₂ -2M)} i

Similarly,            i₂ = {(L₁ -M)/(L₁ + L₂ - 2M)} i

       Putting values of i₁and i₂ in equation (1), we get,

       If L is effective inductance of parallel combination then,

                          v = jωL_eff . i                              ............(4)

       Comparing equations (3) and (4) we have

                          L_eff  = (L₁ L₂ - M²)/(L₁ + L₂ - 2M)

1.2 Parallel Opposing

      Consider two inductors connected in parallel as shown in Fig.2.

Fig. 2

       Applying KVL to both loops, we get,

                - jω L₁ i₁ + jω M i₂ + v = 0

                - jω L₂ i_{2 +} jωM i₁ + v = 0

i.e.             jωL₁ i₁ - jωM i₂ = v                      ...............(5)

                  jωL₂ i₂ - jωM i₁ = v                      ..............(6)

We have,    jωL₁ i₁ - jωM  i₂ = jωL₂  i₂ - jω M  i₁

But             i = i₁ +  i₂

.^..                i₂ = i -  i₁

Substituting value of i₂ in above equation, we get

                  jωL₁ i₁ + jω M(i - i₁) = jω L₂ ( i - i₁) - jω M i₁

.^..                jω i₁ (L₁ + L₂ + 2M) = jω i(L₂ + M)

.^..                i₁= {(L₂ + M)/(L₁ + L₂ + 2M)} i

Similarly,     i₂ = {(L₁ + M)/(L₁ + L₂ + 2M)} i

       Putting values of and in equation (1) we get,

       If L is effective inductance of parallel combination then,

                    v = jωL_eff . i                                 ............(8)

       Comparing equations (7) and (8) we have

                    L_eff  = (L₁ L₂ - M²)/(L₁ + L₂ + 2M)

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