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Find the equivalent inductance looking into the terminals a and b of the circuit below.

enter image description here

( Answer : L(eq)= 7H )

Now I try to transform it to be the following.


simulate this circuit – Schematic created using CircuitLab

Then, I get the following formula :

$$ L\left(eq\right)=4+\left(\frac{1}{9+3}+\frac{1}{\left(10^{-1}+0^{-1}\right)^{-1}+\left(12^{-1}+6^{-1}\right)^{-1}}\right)^{-1} $$

However, $$0^{-1} = Error$$

Therefore, how to deal with this question?

Thank you for your help.

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By inspection, L4 is shorted out. Re-draw circuit by replacing L4 with short. – Marla Jun 21 '14 at 14:16
Alternatively, mathematically we can see that \$0^{-1}\to\infty\$ and thus \$(10^{-1}+0^{-1})^{-1}\to0\$ and we can evaluate: $$L_{\text{eq}}=4+\left(\frac{1}{9+3}+\frac{1}{(12^{-1}+6^{-1})^{-1}}\right)^{-1‌​}=7$$ – Shaktal Jun 21 '14 at 17:06
up vote 3 down vote accepted

L4 = 10 H is shorted. So it can be removed from the circuit.

You have to calculate parallel inductance of

  • 9 H + 3 H (L2 and L3)
  • 12 H (L5)
  • 6 H (L6)

which is 3 H. Then you add the series 4 H (L1) and you get 7 H equivalent inductance.

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Since there is a wire across the terminals of L4, no current will pass through L4. Therefore, we can omit the effect of L4 and remove it out? – Casper LI Jun 21 '14 at 14:34
@CasperLi The wire has a inductance that is very small when compared to the L4 inductor. In this particular situation you can omit it. – Cornelius Jun 21 '14 at 14:35
It is helpful. Thank you so much. – Casper LI Jun 21 '14 at 14:36

Wires actually have a bit of inductance and you could probably get the right answer with your formula by using 0.1uH for the wire, but that's so tiny in comparison to the 10H you might as well ignore the 10H. And it's in series with a few H as well, so another tiny bit won't make much difference.

Hopefully this is enough of a hint.

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