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I vaguely know there are basically the loop/mesh analysis and the nodal analysis corresponding to Kirchhoff’s voltage law and Kirchhoff’s current law, respectively. And one can do Fourier transform of a periodic circuit network and then solve a system of linear equations. However, in the following example, it seems that the two analyses look quite different. enter image description here

In the periodic network shown, the building blocks are three different capacitors C0, C1, C2 on the hexagonal edges. For simplicity, I only show a few of them. In nodal analysis, the independent nodes are just the red and blue ones, hence two unknown voltages. On the other hand, in loop/mesh analysis, it appears to me that only one unknown, the loop charge/current, would be enough. This means that, after Fourier transform, the numbers of linear equations are different, 2 and 1, respectively.

This looks rather mysterious to me. I am not very sure of the loop analysis. Please correct me if anything wrong.

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I have no idea of how your analysis proceeds from here, however the number of variables is the same.

Keep in mind that voltage is a relative variable while current is an absolute one.

When you say the “voltage of the node” you have implicitly added a variable to your system. Your reference “ground” potential, which is not defined for your problem setup and would only show up in the infinite set of initial conditions for the system.

Instead if you say “the voltage between two nodes” you have removed the need for that arbitrary reference (and it directly relates to the associated current).

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  • \$\begingroup\$ Thank you for your answer! Please allow me to confirm if I understand you correctly. Let's say we make each node simply grounded or via something like inductor. Then we have 2 independent node voltages in nodal analysis. And in loop analysis, in addition to a hexagonal loop charge/current, we also need to introduce another charge/current variable from red node to blue node via the ground. Therefore, the number of variables is always 2. Is this correct? \$\endgroup\$ – xiaohuamao Nov 19 '18 at 1:09
  • \$\begingroup\$ @xiaohuamao basically yes. But it is much simpler than that. Think of a single isolated capacitor (or any other element for that matter) its equation relates the potential across it to the current through it. The potential to ground has absolutely no influence on it. \$\endgroup\$ – Edgar Brown Nov 19 '18 at 1:25
  • \$\begingroup\$ Yes, it’s simply the reference issue. It’s a stupid mistake. Is the equal number of variables in two analyses always true (at least in simple RLC circuits)? \$\endgroup\$ – xiaohuamao Nov 19 '18 at 4:12
  • \$\begingroup\$ @xiaohuamao I haven’t seen a mathematical proof of it, but I don’t see how it can be otherwise. All elements have a one-to-one relationship of one variable to another. If that was not the case, one circuit analysis technique would produce more unknown variables than another and “schematic algebra” would be impossible. \$\endgroup\$ – Edgar Brown Nov 19 '18 at 4:24

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