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The voltage across the two parallel aiding inductors above must be equal since they are in parallel so the two currents, i1 and i2 must vary so that the voltage across them stays the same. Then the total inductance, \$L_T\$ for two parallel aiding inductors is given as: $$L_T=\frac{L_1 L_2-M^2}{L_1+L_2-2M}$$

But, if the two inductances are equal and the magnetic coupling is perfect (\$M=k\sqrt{L_1 L_2}\$, where \$k=1\$ and \$L_1=L_2=L\$), using the formula above we get:

$$L_T=\frac{L^2-L^2}{2L-2L}$$

and my workbook says we should get \$L_T=L\$.
What am I missing?

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  • \$\begingroup\$ you are missing the fact that the singular case (where the division by zero occurs) should be defined too. \$\endgroup\$ – Eugene Sh. Jun 20 '17 at 15:48
  • \$\begingroup\$ Is there a way to derive the singular case, the same way the formula above is derived? \$\endgroup\$ – A6EE Jun 20 '17 at 15:51
  • \$\begingroup\$ Try taking the limit of M->L. My calculus is rusty. \$\endgroup\$ – Eugene Sh. Jun 20 '17 at 15:53
  • \$\begingroup\$ Oh, of course, limits.. I'll check it and get back to you. \$\endgroup\$ – A6EE Jun 20 '17 at 15:54
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    \$\begingroup\$ It's all much simpler than that. Two coils, same phase, perfect coupling is just like winding one single coil running two parallel wires instead. \$\endgroup\$ – carloc Jun 20 '17 at 16:52
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Intuition tells us the answer is \$\small L\$, but it's always wise to keep an eye on intuition.

Thus, we have \$\frac{0}{0}\$, which is an indeterminate form.

One way out of the dilemma is L'hopital's rule:

$$ \lim_{k\to 1} \small \left (\frac{L^2-k^2L^2}{2L-2kL}\right )=\normalsize \lim_{k\to 1}\: \small \left (\frac{\large\frac{d}{dk}\small (L^2-k^2L^2)}{\large \frac{d}{dk}\small (2L-2kL)}\right)=\normalsize\lim_{k\to 1} \small \frac{-2k\:L^2}{-2L}=\normalsize L$$

No surprises; intuition checks out.

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