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Consider this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

(this part is preceeded and followed by other components).

$$I_1 + I_2 = \displaystyle \frac{V_1}{sL_{12}} - \frac{V_3}{sL_{12}} \left( \displaystyle \frac{L_1}{L_1 + L_2} \right)$$

with

$$L_{12} = L_1 || L_2 = \displaystyle \frac{L_1 L_2}{L_1 + L_2}$$

How can this relation be obtained? Any hint?

The ratio \$ \displaystyle \frac{L_1}{L_1 + L_2} \$ seems a voltage divider; but I can't see anything similar. Moreover, \$ L_1 \$ and \$ L_2 \$ are not parallel.

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    \$\begingroup\$ Just realize that \$\frac 1 {s L_{12}} (\frac {L_1} {L_1+L_2})=\frac 1 {sL_2}\$ \$\endgroup\$
    – Roger C.
    Jan 29, 2015 at 18:49
  • \$\begingroup\$ Thank you! It is a substitution. If you give it as an answer, I will chose it. \$\endgroup\$
    – BowPark
    Jan 29, 2015 at 18:53

2 Answers 2

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Just realize that \$\frac 1 {sL_{12}} (\frac {L_1} {L_1+L_2})=\frac 1 {sL_2}\$

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Maybe it would be more direct to see that $$I_1 + I_2 = \displaystyle \frac{V_1}{sL_{1}} + \frac{V_1 - V_3}{sL_{2}}$$ and that $$L_{1} = \displaystyle L_{12} \left(\displaystyle \frac{L_1+L_2}{L_2} \right) $$ From there it's just algebra to see that it's equivalent to what you've got.

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