This is the circuit:

enter image description here

I saw it in a book. It said that this is a 7th order circuit.

From theory we know that the order of linear passive circuits is determined by the sum of capacitors and inductors, minus the number of Inductor-nodes (a node where only inductors meet), minus the number of capacitor-meshes (a mesh that has only capacitors on it's branches).

This is my attempt to verify the order.

Capacitors + Inductors = 11

That component to the leftmost part of the schematic, judging from the way the book is illustrated previously, is just a complex electric impedance, so I believe that we can't do anything with that component. It doesn't count neither as capacitor, or inductor. It may contain 10 inductors inside but that's irrelevant, in order to find the order of the circuit (I'm not sure about that though).

Far to the right (my rightmost arrow) the inductor and capacitor can change their place without the circuit being affected in any way, so we have an inductor node.

Another inductor node exactly to the left of the previous one.

And one capacitor mesh, that I have marked to the left.

Therefore the order is 11 - 2 inductor nodes - 1 capacitor mesh = 8, not 7. So I'm clearly missing something here and I have no idea what it is. The way this problem was presented in the book seemed more like a trick question.

I will appreciate any help on this! Thanks in advance!

  • \$\begingroup\$ I wasn't familiar with the rule as you describe it, but why have you left the topmost and the rightmost capacitors out of your equation? \$\endgroup\$ – jippie Sep 2 '15 at 16:45
  • \$\begingroup\$ When did i say i left them out? They are here. I counted them all, 6 capacitors + 5 inductors equals 11. However i don't know whether they have a further role to play here. My guess is they don't. But who am i to say. What role do you think they play on the order? \$\endgroup\$ – Nikos Sep 2 '15 at 16:49
  • \$\begingroup\$ where from are you taking the output? \$\endgroup\$ – MaMba Sep 2 '15 at 16:54
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    \$\begingroup\$ @jippie actually you were right about that mesh i think. Sorry, i got confused and thought you meant something else. The answer was right in front of me but i couldn't picture it. I was stuck. Thanks. \$\endgroup\$ – Nikos Sep 2 '15 at 21:32
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    \$\begingroup\$ Since I seem to understand how it works, I guess I've learned something too. Making note of this nice trick in my notebook. \$\endgroup\$ – jippie Sep 3 '15 at 6:39

You are missing the mesh/loop which includes the left-most capacitor, the upper capacitor and the right-most capacitor. 11 reactive components - 2 inductor nodes - 2 cap meshes = 7.

If you're feeling up to the challenge you can also attempt to find the transfer function of the circuit (s-domain). The circuit order is whatever is higher: the order of the numerator polynomial or the order of the denominator polynomial.

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    \$\begingroup\$ @RestlessC0bra Not sure if it isn't a simple mesh as you can simply follow the outline of the diagram. In this case you would draw the vector to visualize current more conveniently at the outside of the circuit rather than on the inside of the mesh. \$\endgroup\$ – jippie Sep 3 '15 at 6:42

A capacitor in series with a capacitor, or an inductor in parallel with an inductor, doesn't count. You have 2 instances of capacitor feeding capacitor, which drops the count to 7.

  • \$\begingroup\$ I'm sorry. I don't get it. Where are the two series capacitors? And i actually didn't know this one rule you said, but i did the math on paper and i will guess why this rule is true. Is it because there is no \$s^2\$ term arising, in series capacitors, or shunt inductors? (when we do the analysis on the s-domain that is) \$\endgroup\$ – Nikos Sep 2 '15 at 16:47

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