How to solve this resistors question? - Page 1
Forum section: electronics / beginners

author: TG91
author_rank: Newbie
timestamp: October 24, 2021, 04:57:47 pm
content: I am currently practicing what should be quite easy Kirchhoff's Laws and Ohms Laws questions but this question has stumped me!Apologies for the poorly quality hand drawing but it all I have access to today. I need to find the missing currents which I have done and then calculate the resistor values, I am struggling with the latter, can someone help and tell me how you calculate each resistance please?

author: harerod
author_rank: Frequent Contributor
timestamp: October 24, 2021, 05:14:20 pm
content: First of all, let's assume that your calculated currents are correct.Then have a look at the right vertical branch: 10V have to drop over those two resistors. Their currents are supposed to be 1:4, therefore their voltages are also 1:4. The other Kirchhoff-law...After that you do the same with the left vertical branch.Last you look at the voltage difference over the horizontal resistor and its current and feed that into Ohm's law.Check?Recommendation:keep using those funky arrows to reference current direction and potential difference. They represent a powerful tool that goes by the beautiful name of "Zählpfeil" ("counting" - "arrow") in Kirchhoff's and Ohm's mother tongue.Edit:you could use a simulation tool to verify your eventual results. I work with the free LTspice professionally, but there are a lot of other tools around.Edit:Wow, I didn't intend to troll the TO. I crossed out the part, where I went wrong. Kudos to Picuino for pointing out that the original question is underdefined. With that insight one can start and have fun with that network.

author: Picuino
author_rank: Super Contributor
timestamp: October 24, 2021, 05:56:24 pm
content: Solve this:10V = I2*R2 + I4*R410V = I1*R1 + I5*R510V = I1*R1 + I3*R3 + I4*R4Edit:You need set 2 resistors to solve the problem

author: tooki
author_rank: Super Contributor
timestamp: October 24, 2021, 06:06:32 pm
content: This came as a very opportune question for me since I have an exam on exactly this topic on Tuesday, and I needed to practice anyway!I plugged my results into a simulator to verify them, and I’ve saved it here:simulation

author: Picuino
author_rank: Super Contributor
timestamp: October 24, 2021, 06:11:50 pm
content: You can get help with the delta-star transformation:https://en.wikipedia.org/wiki/Y-%CE%94_transform

author: Picuino
author_rank: Super Contributor
timestamp: October 24, 2021, 06:56:46 pm
content: There are infinite solutions. 3 equations and 5 variables.If you set R2 = 1 kOhm and R1 = 1 kOhm thenR4 = 181.8 OhmR5 = 6 kOhmR3 = 1333.3 Ohm

author: Picuino
author_rank: Super Contributor
timestamp: October 24, 2021, 07:07:01 pm
content: Formulas

author: Picuino
author_rank: Super Contributor
timestamp: October 24, 2021, 07:21:04 pm
content: If you choose other values:R2 = 500 kOhmR1 = 500 kOhmThe solution is:R4 = 545'5 kOhmR5 = 8000 kOhmR3 = 666'7 kOhm

author: tooki
author_rank: Super Contributor
timestamp: October 24, 2021, 08:15:25 pm
content: Quote from: Picuino on October 24, 2021, 07:07:01 pmFYI, the OP had μA, not mA.

author: Picuino
author_rank: Super Contributor
timestamp: October 24, 2021, 08:27:02 pm
content: Then the solutions will be the same multiplying all the resistance values ​​by one thousand.Edit:First solutionSecond solution

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 25, 2021, 09:48:36 am
content: Quote from: Picuino on October 24, 2021, 07:21:04 pmSuppose you set R1 and R5; what do you get for a solution?

author: Picuino
author_rank: Super Contributor
timestamp: October 25, 2021, 02:11:47 pm
content: If you set R1, you set the voltage of R1. So, you know voltage and current of R5. Then R5 is defined by R1R1 and R5 are not independient.10V = I1 * R1 + I5 * R5R5 = (10 - I1 * R1) / I5

author: Picuino
author_rank: Super Contributor
timestamp: October 25, 2021, 02:17:12 pm
content: Other restrictions to take into account so that the 10V voltage is not exceeded:R1 < 10V / I1 --> R1 < 2.5 MOhmR2 < 10 / I2 --> R2 < 1.25 MOhmI1 * R1 < I2 * R2 --> R1 < 2 * R2

author: Picuino
author_rank: Super Contributor
timestamp: October 25, 2021, 02:50:28 pm
content: The best way to see it is with the knot method.Node 1 voltage is fixed at 10 volts.The voltage of node 2 (where R2, R3 and R4 meet) can range from 0 to 10V.The voltage of node 3 (where R1, R3 and R5 meet) can range from 0V to 10V.V_node_3 > V_node_2 according to I3 direction.You can set V_node_2 and V_node_3 and solve the problem.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 25, 2021, 03:23:13 pm
content: Quote from: Picuino on October 25, 2021, 02:11:47 pmSo apparently it's not always possible to set any two resistors and then solve the problem.Is it always possible to set any 3 resistors and then solve the problem? For example is it enough to set R1, R3 and R5?

author: harerod
author_rank: Frequent Contributor
timestamp: October 25, 2021, 03:32:46 pm
content: With Picuino's finding we have to accept that this task may have a philosophical side to it. One could ask how many resistors could actually be preset and while still satisfying the given voltage and currents. Since neither R4 nor R5 are given, the potentials over R3 may float around between, but are limited by the supply rails.One could demand R2==R4. With i3=0µA, this would yield 526kR.Mmmh, i3==3µA over 526kR is only 1,58V. So why not set R2==R4==R3?From here we only have to solve for R1 (659kR) and R5 (7MR36).I attached my scribble to illustrate the voltage/current arrows once more.I also attached an LTspice .asc-file, in case somebody, quite rightfully, wouldn't trust me after that blunder in the first post.As a final remark: in the real world the impedances of sources and sinks will almost certainly be known or specified. So if this were a measurement bridge application, u0 (supply) and R3 (load) would most likely be a given. Or if this was a sensor, the bridge resistances and the permissible current would be given. From there the rest of the application would then be developed. So the original question is pretty much backwards.Edit: added scribble, which was dropped when first sending the post.

author: Picuino
author_rank: Super Contributor
timestamp: October 25, 2021, 03:58:30 pm
content: Quote from: The Electrician on October 25, 2021, 03:23:13 pmThe problem have 5 variables and 3 equations so you need to set only 2 variables (resistors)Other equations are (solving problem by nodes):10 - V2 = I2 * R2V2 = I4 * R4V3 - V2 = I3 * R310 - V3 = I1 * R1V3 = I5 * R55 equations and 7 variables. You need to set only 2 variables, in this case V2 and V3 from 0 to 10V (rails) and you can solve for other 5 variables (resistors)

author: antenna
author_rank: Frequent Contributor
timestamp: October 25, 2021, 11:02:14 pm
content: Although useless here, the refresher on kirchoffs current law led me to delta-star transformations (among other transformations) which are really cool! If we knew the resistor values, we could use the transformation of one of the delta loops into a star and be able to simplify it down to a single resistor

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 26, 2021, 07:37:09 am
content: Quote from: Picuino on October 24, 2021, 05:56:24 pmYou have not fully characterized the solution. You should say more than "set 2 resistors to solve the problem". The reader might think that all that is needed is to setanytwo resistors.You began to derive the rest of the full characterization in reply #11, but you didn't carry it through to a conclusion.The problem cannot be solved by setting the pair R1/R5 or the pair R2/R4 to arbitrary values, but any other pairs will do.It's possible to find 5 equations for the 5 variables. There are 4 paths from the top node to the bottom node, and 1 path around the whole thing. The first 4 paths yield equations set equal to 10 volts, and the 5th path yields an equation set to zero volts. See this image; the paths are numbered 1 through 5:Here are the 5 equations:These equations can be put in matrix form:The reader is wondering at this point how we can possibly have 5 equations for 5 unknowns when earlier it was pointed that we need to set two resistors to arbitrary values, then we can solve for the other 3 resistors. The choice of values for the initial two resistors is not completely arbitrary--some choices can lead to negative values for the remaining 3.Had we followed this procedure from the beginning, the first thing to do is to determine if enough of our 5 equations are independent to give a unique solution to the problem. We can use a linear algebra concept for this. We simply compute the rank of the 5x5 coefficient matrix, which would be 5 if we had 5 independent equations, but this matrix only has a rank of 3. This means there are an infinite number of solutions, and two of the variables are free variables and can be chosen arbitrarily (with the aforementioned limitation).Ordinarily, this would mean that we could choose any pair of resistor values arbitrarily and then our equations would give a solution for the rest. But this problem is peculiar in that the pairs R1 R5 and R2 R4 cannot be chosen arbitrarily because when you choose one of the pair the other one is immediately determined. The values of R1 R5 and R2 R4 are not independent as Picuino mentioned.The linear algebra procedure that can show this is to set a pair of variables in the matrix to a constant and see if the rank of the matrix increases to 5. This happens for all pairs of variables except the R1 R5 pair and the R2 R4 pair.I used a math program to find all the solutions to the system and here's the result. The two resistors on the left of each line are the two arbitrarily chosen, and the rest of the line shows the calculation for the other resistors.Notice that all pairs are present except the R1 R5 and R2 R4 pairs:

author: harerod
author_rank: Frequent Contributor
timestamp: October 26, 2021, 08:42:41 am
content: Quote from: antenna on October 25, 2021, 11:02:14 pmSadly students are usually under so much pressure, that they don't have the time to really enjoy playing with a circuit. If I weren't on vacation, I wouldn't take the time either.The circuit may also serve as a demonstrator for Thévenin's Theoreme, which allows the reduction of complex networks to a single source and impedance. The wiki has links to additional tools for network analysis:https://en.wikipedia.org/wiki/Th%C3%A9venin's_theoremHere is yet another scribble, which shows the Thévenin-reduced left and right branches of the bridge. Not to lose information, I also noted the restrictions regarding currents in R1,R2,R4,R5.The whole playing around with network conversions (and superposition, if there were several sources), shows how far one can get without actually resorting to the even more powerful tool of matrices.

author: The Electrician
author_rank: Frequent Contributor
timestamp: October 26, 2021, 11:32:05 am
content: Quote from: antenna on October 25, 2021, 11:02:14 pmSomebody did all the calculations in 1947:https://docs.google.com/viewer?a=v&pid=sites&srcid=Y2xldmVyNGhpcmUuY29tfGNsZXZlcjRoaXJlLWNvbXxneDozMDk2OWRjNGZmNmZlZDlh