Circuit Analysis Question Has Me Stumped - Page 1
Forum section: electronics / beginners

author: sshoptaugh1991
author_rank: Contributor
timestamp: August 01, 2017, 08:03:38 pm
content: I did not do so well when I took circuit analysis the first time, so I am independently studying the material to understand what I didn't do well with. So far going through the material is fine, but there is one question that has me stumped in the book. This is early on, in the second chapter about resistive networks, so this is starting to be discouraging.The goal is to find the voltage of R5 and the voltage of R3My first thought was to combine R4 and R5, but trying to do KCL on that loop has me stumped.My next thought was to redraw the circuit so that R1 and R5 were in parallel, and combine them to yield 3K. But then there is the issue of doing KCL on the branch containing R2 and R3.Am I thinking too hard about this? This is chapter 2 in my book, so the idea of nodal loop analysis has not been brought up.Any help in pointing me in the right direction would be great, but I do not want the total solution.

author: rstofer
author_rank: Super Contributor
timestamp: August 01, 2017, 08:19:50 pm
content: Actually, R1, R4 and R5 are all in parallel and total 2k so redraw the circuit with the 2k in series with the 10k as a substitute for R2 and blow off R1, R4 and R5. The current source still goes between the new 12k R2 and the 4k R3.Now combine the new 12k R2 in parallel with the 4k R3 and the source as before. But, wait, the new 12k is in parallel with 4k which is now a 3k equivalent. So, what we really have is a 20 mA source and a 3k equivalent resistor.20 mA * 3k = 60VI sure hope I did that right... If I didn't, I'm sure I will hear about it.

author: sshoptaugh1991
author_rank: Contributor
timestamp: August 01, 2017, 08:28:55 pm
content: So the two branches that are just lines are negligible in this case, then? It could be seen as the branch below R4 doesn't exist, and R5 is just connected in parallel, yeah?

author: dmills
author_rank: Super Contributor
timestamp: August 01, 2017, 08:33:34 pm
content: Combine R1 and R5 making 3K, then notice that this 3K is in parallel with R4 making 2K.That 2K is now in series with R2 meaning the current splits as ratio of 4/16 and 12/16, so current in R2 is 5mA and R3 is 15mA…. Voltages I leave as an exercise in multiplication.It often helps to redraw these things, especially to make common loops and parallelism obvious (I suspect you missed the parallel R4).Regards, Dan.

author: rstofer
author_rank: Super Contributor
timestamp: August 01, 2017, 08:59:02 pm
content: Quote from: sshoptaugh1991 on August 01, 2017, 08:28:55 pmThey try to make a node at the down side of R4 by connecting the current source to the downrunning wire. But it's still a wire and you can draw a separate wire between the current source and ground. Once you do that, it is obvious that R1 R4 and R5 are all in parallel.But then the equivalent 2k is in series with R2 totalling 12k. But then the 12k is in parallel with the 4k giving 3k and the voltage just falls right out.

author: Ratch
author_rank: Regular Contributor
timestamp: August 02, 2017, 12:34:04 am
content: Quote from: rstofer on August 01, 2017, 08:19:50 pmThe current source has to output -60 volts. The upper bus will be -10 volts.Ratch

author: rstofer
author_rank: Super Contributor
timestamp: August 02, 2017, 12:57:55 am
content: Quote from: Ratch on August 02, 2017, 12:34:04 amI had some muddy thinking about the current source. According to LTspice, that node voltage at R3 is -60V and the voltage at the top of R5 is -10V.It's fun to play with these problems but it is always worth verifying the results with LTspice.

author: sshoptaugh1991
author_rank: Contributor
timestamp: August 02, 2017, 04:25:29 am
content: I think I found the logic to the latter part of the problem, to find the voltage of R5.Since R5 is in parallel with R1, the voltage across the resistor will be the same.Using the same logic, the sum of voltages across R2 and R3 would be the same as R1 and R5. So, with -60V as the voltage across R3, using Ohm's Law to calculate the current through R3 yields 15mA. Using KCL to find the current through R2 yields 5mA (5mA coming into R2, -15mA coming into R3, totalling the 20mA source coming out of the node), and the voltage of R2 is then 50V. The voltages of the branch total of -10V using KVL.Though, I am confused on the signs of the numbers. If 20mA is coming out of the node between R2 and R3, why is the current at R3 negative if it is entering the node? This is something that is confusing me after all the math and circuit logic is finished.

author: John B
author_rank: Frequent Contributor
timestamp: August 02, 2017, 05:08:24 am
content: I got VR5=10V with the negative side on the top of R5. I have a page and a bit of handwritten working if you want to see it, but the general process was:1. Define polarities of resistors (even if they're wrong, they just need to be logically consistent), ie R2 and R3 have their negative side on the node that connects to the left of the current source, and R5 R4 R3 and R1 have their positive side on the node that connects to the right side of the current source. (You can determine that by the direction of the current source)2. By using KVL and defining your voltage loops, you can reduce the unknown voltages across resistors to 3. Ie Vr5=Vr4=Vr1, Vr5-Vr3+Vr2=03. Define your 3 KCL equations/nodes. There's one to the left of the current source, one to the right, then there's the node at the top.4. Express each current quantity as a V/R quantity (ohms law). eg I5=Vr5/12000, I4=Vr5/6000 (remember Vr5=Vr4=Vr1)5. Substitute the V/R values into your KCL equations.6. Then by simplifying and substituting through the 3 KCL equations you can find the value of VR5. Compare that value against the assumed polarities of the resistors, eg VR5=10V, but remembering that the negative side is at the top.

author: BravoV
author_rank: Super Contributor
timestamp: August 02, 2017, 05:33:23 am
content: Just sharing, I'm a self taught noob, when it times like this, usually I love to move around in my mind, as I have mentally better to understand it if I view them this way.Equipped with basic and simple resistor's formulas, then I can solve them easily. CMIIW.

author: sshoptaugh1991
author_rank: Contributor
timestamp: August 02, 2017, 05:42:10 am
content: BravoV,Thank you very much for sharing this different view. This is yet another skill I need to develop, seeing circuits in different ways. This helps a lot with determining the sign of the values.I have quite a while before I take the course again, so hopefully studying this stuff again on my own will fill in the gaps that I have from class. I am quite certain when I hit RC and LC step responses, I will be posting frequently.Thanks again for the help!

author: John B
author_rank: Frequent Contributor
timestamp: August 02, 2017, 05:45:29 am
content: Quote from: BravoV on August 02, 2017, 05:33:23 amI like that solution! I guess mine is a no-no, I think it will count as a nodal analysis.

author: xavier60
author_rank: Super Contributor
timestamp: August 02, 2017, 06:45:15 am
content: I had it all wrong.I now get 10 volts across R5 also.

author: Ratch
author_rank: Regular Contributor
timestamp: August 02, 2017, 12:36:55 pm
content: Quote from: sshoptaugh1991 on August 01, 2017, 08:03:38 pmSimple decision. Four equations for KVL or two equations for KCL. Use the node equation method as shown below. Easily solved for v1 =-60 and v2 = -10Ratch

author: rstofer
author_rank: Super Contributor
timestamp: August 02, 2017, 01:53:44 pm
content: KCL and KVL are the proper way to solve the problem but...QuoteIt would seem to me that we should therefore restrict the solution to combining resistors and using just Ohm's Law.On that basis, I suspect LTspice is out of the question as well but it would be one of the first tools I added to my toolbox. Second would be wxMaxima - in fact, I can't imagine taking the EE program without making extensive use of this tool.I have attached wxMaxima.png which shows the solution given by Ratch's equations. The first 2 lines are housekeeping, eq1 and eq2 are the equations. Res is the actual solution and the following lines just clean things up and print the results in sorted order. This is not very important for two equations but it is most helpful when there are several voltages and currents.One cool thing about tools like wxMaxima, you can copy and paste from previous problems. About the only original work in the attached sheet was defining the equations and setting up the arguments to solve().