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There are loop/mesh analysis and node analysis corresponding to Kirchhoff’s voltage law (KVL) and Kirchhoff’s current law (KCL), respectively.

Let's consider three elements, R,L,C, connected serially in the simplest way to some external voltage source. With KVL and a charge variable \$q\$, we have \$(-\omega^2L+\frac{1}{C}+j\omega R)q(\omega)=v_\mathrm{ext}\$.
Let's consider three elements, R,L,C, connected parallelly in the simplest way to some external current source. With KCL and a flux variable \$\phi\$, we have \$(-\omega^2C+\frac{1}{L}+\frac{j\omega}{R})\phi(\omega)=i_\mathrm{ext}\$.

We see that such a resistor in the KVL of an AC simple RLC circuit can lead to a term proportional to \$j\omega R\$, which becomes a term proportional to \$\frac{j\omega}{R}\$ in KCL. And all other terms are real.
(Surely, if you divide or multiply these equations by \$j\omega\$, you get the impedances like \$j\omega L,\frac{1}{j\omega C},R\$.)

We know resistance entails dissipation, which is naturally related to imaginary terms. So I don't think these equation forms are meaningless. I hope to understand.
However, the imaginary \$\frac{j\omega}{R}\$ term in node analysis KCL seems to be smaller and smaller as the resistance increases. It appears somewhat counterintuitive. How to understand this?

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    \$\begingroup\$ Please post an example schematic and your workings. \$\endgroup\$ – Andy aka Nov 19 '18 at 7:53
  • \$\begingroup\$ Capacitors present as a lower impedance path to higher frequency signals (and an open circuit to DC signals). Inductors present as a higher impedance path to higher frequency signals (and a short circuit to DC signals). Are you asking for physical intuition for these rules of thumb? \$\endgroup\$ – vicatcu Nov 19 '18 at 13:40
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    \$\begingroup\$ Resistors have no frequency dependence on their impedance ideally... only capacitors and inductors \$\endgroup\$ – vicatcu Nov 19 '18 at 13:43
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    \$\begingroup\$ ......"will lead to a term proportional to jwR...." ??? Please, demonstrate this statement with an example. \$\endgroup\$ – LvW Nov 19 '18 at 14:18
  • \$\begingroup\$ @LvW Updated. Please kindly let me know your comment. \$\endgroup\$ – xiaohuamao Nov 19 '18 at 18:31
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Although those are imaginary terms, and you are expressing your equations in terms of flux and charge, which makes things a bit confusing.

Note that in one equation you are calculating a voltage. In a voltage equation resistance implies dissipation.

In the other equation you are calculating a current. In a current equation conductance implies dissipation.

You always have to keep track of your units.

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  • \$\begingroup\$ It's great that someone finally understand what I mean! Thank you, again. Let me think about your answer. \$\endgroup\$ – xiaohuamao Nov 19 '18 at 19:05
  • \$\begingroup\$ Your answer is very helpful. It's also basically related to the two expressions of dissipation power, \$i^2R\$ and \$v^2/R\$. But I still feel somewhat confused that the dissipation or imaginary term goes very large when we have a more and more perfect conducting wire (\$R\$ goes down). Naively, a perfect conducting wire is dissipationless. \$\endgroup\$ – xiaohuamao Nov 19 '18 at 19:37
  • \$\begingroup\$ No. If R goes to zero the conductance goes to infinity. Infinite conductance means no dissipation. Unless you somehow manage to put a voltage across it, of course. That an abstract imaginary term grows means nothing unless you take in consideration the units that that imaginary term represents. \$\endgroup\$ – Edgar Brown Nov 19 '18 at 19:42
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I hope I'm understanding your concerns...

As @vicatu wrote: Resistors have no frequency dependence on their impedance ideally... only capacitors and inductors.

Think of a capacitor as two metal plates, suspended one above the other, separated by air. Connect a battery's positive terminal to one of the plates, and its negative terminal to the other. Is there current flow? No. This is the limit case where frequency goes to zero for a capacitor - no current.

Now think about a length of wire, connecting the same two battery terminals - you'd get near infinite current flow (it's a dead short). Now wrap that wire around a wooden pole. That's a simple inductor now. Would you expect the (infinite) current from the battery to change? No. That's the limit case for an inductor as the frequency goes to zero.

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  • \$\begingroup\$ Please see my update. I understand these basics. \$\endgroup\$ – xiaohuamao Nov 19 '18 at 18:58

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