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Dec 9 at 16:39 history edited JYelton CC BY-SA 4.0
Removed superfluous stuff in title
Dec 8 at 2:53 comment added The Photon If the goal of the exercise is to demonstrate that you know mesh analysis, then you need to start over and do mesh analysis from the beginning. If I were instructor I would not accept an answer that used part of the answer from the first method to "jump start" the mesh analysis --- because that wouldn't be "doing mesh analysis". If the goal is just to solve the circuit, then you already solved it by your first method so why do you need a second method?
Dec 7 at 19:16 comment added G36 You must understand that mesh analysis is based on KVL. So in mesh analysis, we sum the voltage drops around the loop.
Dec 7 at 18:42 comment added RiFF RAFF @vir But let's say those two parallel resistors were one so we do only have two loops. We still know I1 is 5mA. This is just where I don't get why the equation is not working. The answer I still get is 10mA. And is that the missing part, 10mA is explicitly current I3 at the source, but it's value changes during the split. But that's what I thought the 10(I3-I1) accounts for?
Dec 7 at 18:29 comment added vir You're doing analysis by inspection which is fine but you can't use it to "shortcut" your loop analysis when one of the "known" loop currents is part of two loops. This is the same problem you are running into with your current source.
Dec 7 at 18:23 comment added RiFF RAFF @vir Okay, please bear with me, I think I'm close to understanding with that answer. The only part that is still tripping me up is I understand that the 5mA splits at where it meets the parallel resistors. But if we just add the resistors together in parallel, I will still get 5mA traveling through that loop. So It still feels like I should be able to get the same answer from the mesh equations whether I did them simultaneously or not?
Dec 7 at 18:17 answer added G36 timeline score: 0
Dec 7 at 18:13 comment added vir You don't know how much current goes into each branch as it leaves the current source. Therefore you don't know how much current goes into the center loop. All three must be solved simultaneously.
Dec 7 at 18:11 history edited RiFF RAFF CC BY-SA 4.0
added 651 characters in body
Dec 7 at 18:08 comment added RiFF RAFF @vir Perhaps I just didn't ask directly enough. I am just asking, knowing I already know the value of I1, can I not just use mesh analysis on the loop I drew for I3, get the equation, and then plug in the value of I1 to get my value of I3? If no, why not? If yes, what am I doing wrong?
Dec 7 at 18:00 comment added vir Relabel the loops as Ia, Ib, and Ic. I1 is part of two loops.
Dec 7 at 17:50 comment added RiFF RAFF @ThePhoton Either method should produce the same answer, no? I already know I1, so using that information I have, I can see I am not doing something correctly in my mesh analysis. It should be as simple as getting the correct I3 mesh analysis equation and then I can plug in I1 to get I3, since I already know I1. I am not able to do so, so I am doing something incorrect and am trying to understand where my error is?
Dec 7 at 17:46 comment added The Photon Because "using the results from the current divider method" is not part of mesh analysis. Mesh analysis is a specific algorithm. If you do some other method first and use that as part of the solution, then you're not doing mesh analysis.
Dec 7 at 17:45 comment added RiFF RAFF @ThePhoton I'm not sure that's necessary. The only current that matters in the case of me trying to use mesh to get I3 is I1. The only thing I would add is I1 would be a clockwise loop in the direction already shown. This is the solution provided. Given I know from the solution I1 is 5mA, can I not just use that information using mesh analysis to get I3?
Dec 7 at 17:33 comment added The Photon Your diagram indicates branch currents I1 and I2. But mesh analysis is done with mesh or loop currents, not branch currents. Please modify your drawing to show which loops your currents are defined on.
Dec 7 at 17:22 history asked RiFF RAFF CC BY-SA 4.0