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Just so you have my approach to consider, as well. (I know you've already selected an answer.) Here's the redrawn schematic which I prefer: simulate this circuit – Schematic created using Circu …

You have four relatively obvious current loops that could use KVL. That's all you need for mesh analysis. (You don't need to include any KCL statements for the nodes.) simulate this circuit – S …

simulate this circuit – Schematic created using CircuitLab It's the case that \$R_i=\frac{2R\left(R+R_{i-1}\right)}{3R+R_{i-1}}\$, where \$R_{i-1}\$ is the prior source impedance leading up to sect …

The Photon probably gave you a way too terse answer. But it's correct, just the same. I'll provide a way too long answer. (tl;dr) Simplistic Geometric Approach It's best to think of a magnitude (the l …

I suspect the problem may arrive from the requirement of an impossible domain for passive component values. Your characteristic equation breaks down (as I'm sure you already know) into: \$\left(s+1\ri …

It's not complicated. You have two unknown voltage nodes: simulate this circuit – Schematic created using CircuitLab Note that I've made the two legs independent of each other, because they ar …

I had mentioned that I'd probably take a different approach than user600016's approach, using KCL instead of KVL. Both work for me, but I'm usually more comfortable with KCL as I tend to make fewer er …

Yes, diodes are used in circuits to discharge energy stored on capacitors. It's an entirely different circumstance, though. Inductors will reverse their voltage when discharging energy and if all the …

Assuming \$f=60\:\textrm{Hz}\$ and using the included schematic editor: simulate this circuit – Schematic created using CircuitLab From this, I get: $$\begin{align*} V_B &= V_C + V_2\\\\ \fra …

Just to be complete: simulate this circuit – Schematic created using CircuitLab I gather that's the circuit. And you'd like an expression for \$V_{\left(t\right)}\$ and \$I_{L\left(t\right)}\$ …

My main question is how the overcharge from the one plate can create an overcharge on the other. I think I understand your question. The initial conditions are with the capacitor fully charged …

I'd use KCL (not KVL) with the bottom node "grounded" and the top node as \$V\$: $$\frac{V}{R}+\frac1{L}\int V\:\text{d}t=0\:\text{A}$$ Taking the derivative with respect to time: $$\begin{align*}\ …

Since everyone seems to have been variously confused at different times about your question, let me re-draw and re-phrase it: simulate this circuit – Schematic created using CircuitLab (The ab …

With Verbal Kint's contribution, I'll expand on thoughts behind my earlier comment(s). Notes from: "A Practical Method of Designing RC Active Filters" I want to start out by reflecting on the TR-50 pa …

You know the current in the capacitor must be the same as the current in the inductor. Ground the bottom node, for a simple reference point. Then the current into the capacitor from the top mode (whos …