7.3 Mutually coupled inductors in parallel:

Consider inductors $A$ and $B$ are connected in parallel. The self inductance of coil $A$, ${L_A}$ is and that of coil $B$ is ${L_B}$. $M$ is the mutual inductance between them. There may be two conditions.

(i) The direction of magnetic flux produce by one is in same direction of other.

$${L_T} = \frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} - 2M}}$$

(ii) The direction of magnetic flux produce by one is in opposite direction of other.

$${L_T} = \frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} + 2M}}$$

Consider the first inductor branch ${L_A}$, (Inductor ${L_5}$ in parallel with inductors ${L_6}$ and ${L_7}$)

$${L_A} = \frac{{{L_5} \times ({L_6} + {L_7})}}{{{L_5} + {L_6} + {L_7}}} = \frac{{20 \times (20 + 20)}}{{20 + 20 + 20}} = 13.33mH$$

Consider the second inductor branch ${L_B}$, (Inductor ${L_3}$ in parallel with inductors ${L_4}$and ${L_A}$)

$${L_B} = \frac{{{L_3} \times ({L_4} + {L_A})}}{{{L_4} + {L_3} + {L_A}}} = \frac{{20 \times (20 + 13.33)}}{{20 + 20 + 13.33}} = 12.49mH$$

Consider the equivalent circuit inductance${L_{EQ}}$ (Inductor L1 in parallel with inductors ${L_2}$ and ${L_B}$)

$${L_B} = \frac{{{L_1} \times ({L_2} + {L_B})}}{{{L_1} + {L_2} + {L_B}}} = \frac{{20 \times (20 + 12.49)}}{{20 + 20 + 12.49}} = 12.37mH$$

Then the equivalent inductance for the above circuit was found to be: $12.37mH$.