Conceptual paradox with resistance in loop/node analysis

Question

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?

Answer 1

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.

Answer 2

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.