Total inductance of two Parallel aiding inductors each with an inductor in series

what would be the total inductance of this circuit?

• 'Parallel aiding inductors' = transformer? – Chu Jan 17 at 9:19

As mentioned by Chu, $$\L_1\$$ and $$\L_4\$$ are coupled in your schematics. With defining the coupling as $$\M = M_{34}\$$, the resulting inductance becomes

$$\L = \frac{(L_1+L_2) (L_3+L_4) - M^2}{(L_1+L_2+L_3+L_4) - 2 M}\$$

In case they are not coupled (not forming a transformer) we set $$\M = 0\$$ and get the equation from Joe Mac.

Inductors add just like resistors. So combine the inductors in series to get just 2 inductors in parallel. Then add those two parallel inductors by adding together their reciprocals.

$$\\frac{(L_{1}+L_{2})(L_{3}+L{4})}{L_{1}+L_{2}+L_{3}+L_{4}}\$$

Formula for calculating inductors in parallel is the following:

$$\ \frac{1}{L_{total}} = \sum\limits_{n = 1}^{N}\,\frac{1}{L_{n}} = \frac{1}{L_{1}}+\frac{1}{L_{2}}\,...+\frac{1}{L_{N}}\$$

Formula for calculating inductors in series:

$$\ L_{total} = \sum\limits_{n = 1}^{N}{L_{n}} = L_{1}+L_{2}\,...+ \,{L_{N}}\$$

• Your equation should be1/ /(L1+L2)+/(L3+L4). You are taking the reciprocal of the reciprocals. Do you know Mathjax or LaTex or any markup language? – Sparky256 Jan 17 at 6:54
• You’re right. I forgot about the last reciprocal. – Joe Mac Jan 17 at 8:11
• L1 and L4 form a transformer. – Chu Jan 17 at 9:07