Mesh analysis with Kirchoff law, VERY likely done wrong - Page 1
Forum section: electronics / beginners

author: Turrican3
author_rank: Regular Contributor
timestamp: July 04, 2020, 02:22:38 pm
content: Hey there, right when I thought I was making progresses with (basic) electronics, here's one example that unfortunately brought back confusion.Given:E1= 12VR1= 1 ΩR2= 3 ΩR3= 4 ΩR5= 2 Ωi2= 2 AI need to find the value of R4This is what I did (no subscripts anymore for the sake of making the post easier to write)1. Identified three meshes: left one with R5, R1, E1 / center one with E1, R1, R2 / right one with R2, R3, R42. at first, I ignored the right mesh and just went on with calculationsAssumed clockwise current flow, wrote the following equations, with IL/ IC/ IRbeing the currents of each mesh (Left / Center / Right respectively)-E1 - R5 * IL - R1 * IL = 0E1 - R1 * IC - R2 * IC = 03. I chose to temporarily ignore the righmost mesh and went on with calculations, ended up with IL = -4 A and IC = 3 A; since the first number is negative I understand it just means the actual current flow is the opposite of the assumed one, but the number is supposed to be right... right?And here's the issue: by reproducing the circuit with the assumed solution (R4 = 2 Ω)in this circuit simulatorI get different results, namely 3A instead of 4 at R5, and 2A instead of 3 at R2. Now, I understand R2 might be wrong due to having to take into account the other resistors, but shouldn't R5 be right anyway? Or perhaps the whole approach is flawed and hence led to wrong results in the first place?===============================To cut a long story short: I think I don't get this law as well as I thought.And that is especially true since I've seen the dedicated, excellent EEVBlog video (#820) about nesh/nodal analysis.Problem is, I don't think I've ever seen what he did with the middle resistor (which is shared between the left and right mesh), i.e. take into account the current from the right mesh when writing the left equation and viceversa. And that, albeit extremely clear in the video, definitely confused me as a beginner.So the question is: do I have to apply that kind of calculation in this example, too? And what would be the equation of the rightmost mesh, i.e. the one with R2, R3, R4 ?Would it be something like - R2 * IR - R3*IR - R4 * IR = 0 or does this not even make sense?

author: Benta
author_rank: Super Contributor
timestamp: July 04, 2020, 05:51:56 pm
content: "at first, I ignored the right mesh and just went on with calculations"This is where you crashed.For three meshes you need three equations. You can't just ignore things.

author: Turrican3
author_rank: Regular Contributor
timestamp: July 04, 2020, 07:11:23 pm
content: Hmm ok, so is the third one actually right (end of post) or is that wrong too?

author: SparkyFX
author_rank: Frequent Contributor
timestamp: July 04, 2020, 07:27:41 pm
content: Just go in sequence... the voltage across all three meshes will equal the voltage source (and R1), so start with what you got, i2 and R2. Vab = R2 * i2 = 3Ω * 2A = 6VThese 6V are shared between R3 and R4, but to get to the Resistance of R4 you got to know its current.You also know E1 = 12V, therefore 6V need to drop at R1 (1Ω), which means there need to be I = (E1-Vab)/R1 = 6V/1Ω = 6Amps for this to happen. This means the supply delivers 6A in total, 2Amps through R2, 3 Amps through R5 (6V/2Ω = 3A), which leaves 1A through R3 and R4.6V/1A = 6Ω, you know that R3 is 4Ω, therefore R4 must be the remaining 2Ω.

author: Turrican3
author_rank: Regular Contributor
timestamp: July 04, 2020, 07:39:56 pm
content: Thanks, I will write that down so that it (hopefully) better sticks in my head.

author: SparkyFX
author_rank: Frequent Contributor
timestamp: July 04, 2020, 10:20:39 pm
content: Well, its not strictly KVL or KCL, its a mix of both (which can get you in trouble should someone insist on using only one). But you might always use this approach as a sanity check for findings with KVL/KCL.Also consider to redraw the schematic in your head with source on one side and loads on the other, but keep the nodes in their proper place (don't change the function of the schematic itself, just rearrange it to better visualize the flow of energy). Sometimes this makes it easier to understand.

author: Turrican3
author_rank: Regular Contributor
timestamp: July 05, 2020, 05:15:39 am
content: Thanks again.I guess both are factors that are making things more confusing (to me) than they should: i.e. the fact that apparently you often have more then one way to find the exact same solution. And since the exercise was in the Kirchoff section I thought I had to stick with just one of them so I kept forcing myself to use KVL (although I tried at first to simplify the right part of the circuit, I assume that would be correct too: I calculated an equivalent resistor consisting of R3 and R4 in series, in parallel with R2... but this is Thevenin right? so I thought I wasn't solving it the intended way)And for the record, the provided solution from the author of the exercise did exactly what you are suggesting now, he redrew the circuit and moved the source to the far left side. :-)