0
\$\begingroup\$

what would be the total inductance of this circuit? https://i.stack.imgur.com/gZHEB.png

\$\endgroup\$
  • 1
    \$\begingroup\$ 'Parallel aiding inductors' = transformer? \$\endgroup\$ – Chu Jan 17 at 9:19
0
\$\begingroup\$

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.

\$\endgroup\$
0
\$\begingroup\$

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.

Edit: Then take the reciprocal of that answer. I forgot about this last part. Very important.

\$\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}}\$

\$\endgroup\$
  • 1
    \$\begingroup\$ 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? \$\endgroup\$ – Sparky256 Jan 17 at 6:54
  • \$\begingroup\$ You’re right. I forgot about the last reciprocal. \$\endgroup\$ – Joe Mac Jan 17 at 8:11
  • 1
    \$\begingroup\$ L1 and L4 form a transformer. \$\endgroup\$ – Chu Jan 17 at 9:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.