Problems for beginners - Page 1
Forum section: electronics / beginners

author: Picuino
author_rank: Super Contributor
timestamp: April 08, 2025, 04:22:16 pm
content: A problem, not so simple, for those learning circuit theory. Already solved.https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxABZCQkLsaBTAWjDACgAnEPSEbQvCDSD+g3mEiduvUdKEVe4yVx7zewtVGRSNKBUPx8BWiVNwU55g2O1cwGaqquqlU+9Q3u5ruw+t9ICw0ff1krYNt-PRlA70ivVVVok2VwPw0nY1cAdz4UI0FMsTZc7HyygIsK9lLYiudFNhQwanry8qyoWHYgA

author: MrAl
author_rank: Super Contributor
timestamp: April 09, 2025, 03:25:24 pm
content: Quote from: Picuino on April 08, 2025, 04:22:16 pmHi,That's interesting, but not sure I could recommend that technique except with the simplest of circuits. That's because recognition of symmetries is not always easy and subject to error.

author: golden_labels
author_rank: Super Contributor
timestamp: April 09, 2025, 06:40:07 pm
content: I believe this is not about offering a universal method for solving circuits. It’s an exercise in spotting patterns and simplifying circuits to a structure easier to grasp.Education is not about infusing a learner with encyclopedic knowledge or making them proficient in using a tool. The first can and should be checked in sources, the second comes naturally with tool use. An indispensable element of teaching and learning is exercising the brain and building a scaffolding for the actual knowledge. This remains true both in professional or academic education, and in primary schools.

author: BeBuLamar
author_rank: Super Contributor
timestamp: April 09, 2025, 06:53:58 pm
content: When I started I didn't know how to solve it so I wired up the circuit and use an ohm meter. Got close though.

author: Picuino
author_rank: Super Contributor
timestamp: April 09, 2025, 06:58:20 pm
content: This is a problem I call the “happy idea” problem. Very difficult, close to impossible, if you don't manage to simplify the initial problem with a method that, once you are told about it, is simple.The level is first university course in circuit theory. Some teachers teach the basic tools (including mesh solving and symmetries) and then ask these types of problems to check if you have internalized the circuit solving tools.

author: Grandchuck
author_rank: Super Contributor
timestamp: April 09, 2025, 07:38:26 pm
content: I remember using delta - wye transforms for this

author: golden_labels
author_rank: Super Contributor
timestamp: April 09, 2025, 09:51:13 pm
content: Without simplification, finding the solution is tedious and it gives nothing but a number. No understanding, of how the circuit behaves. That I may agree with.But certainly it’s not “close to impossible” or even difficult. KCL gives a very neat, nicely behaving system of six equations with six unknowns. With resistances being equal, the equations can ignore the resistances altogether, leaving us with an even simpler problem with a few small integers. This is high school level maths.A bit tedious, naïve, not elegant, but certainly far from impossible.

author: TimFox
author_rank: Super Contributor
timestamp: April 09, 2025, 09:55:19 pm
content: In physics, we learned to look for symmetries to solve problems.Past that, we looked for broken symmetries.

author: golden_labels
author_rank: Super Contributor
timestamp: April 09, 2025, 11:06:12 pm
content: Earlier I skimmed over the video, only paying attention to the final result. I rewatched it now again, and also checked the channel.I see the didactic value in the “resistor cube” problem, as drawn by Picuino. Not so much for the video. The animation shows the solution, withou ever explaining why it works. Worse, the audience may get a wrong idea of why it works and then incorrectly solve other circuits. And then some minor stuff, like the maths suggesting the author has little intuition about using fractions.Since this hasn’t been explained, I’ll permit myself to deliver the details. Hopefully that helps somebody.“Symmetry” is mentioned, but it is never shown where that symmetry really lies. I’ll take the cube, and label the nodes like that:Now, let’s turn it into a pancake. The colors should make the transformation clearer. The “Y” node is now the entire circle:We can now see, that this pizza can be divided into three identical parts.We may rotate it by 1/3 of the circle, and end up with exactly the same thing. This is the symmetry we seek. This ability to rotate things and the circuit remaining the same is why we may say some nodes are the same and collapse them into one. Observe this approach fails, ifcorrespondingresistors were of different values. This is the part the video missed, and which could easily build a wrong intuition.

author: Nominal Animal
author_rank: Super Contributor
timestamp: April 10, 2025, 04:10:03 am
content: If you want a more general net, for solving the resistance between any pair of vertices instead of only the diagonally opposing vertices, you can also flatten the net thus:(Click to embiggen)

author: Picuino
author_rank: Super Contributor
timestamp: April 10, 2025, 06:40:05 am
content: Yet another problem.Calculate the current through the middle resistor without solving the problem by the traditional method of using 3 linear equations, nor simplifying with the star-delta method.https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgrCAMB0l3AWAnC1b0DYQEYYCY48B2IgZkgA48MUjxJwQFTwBTAWm2wCgA3EdggohSBAUJBIsDBggYQZUaGG4AnEBgSSsmkAWENccNRq37TIsYYKRuAd3HDRDXc6j2Lb7EjyWZHwWEpC2DbB11g719QkyjtHB89SigceFjE8zi3QzT1OPNdcxzjAA8BJGwEpAEaJN8tbGEwADVuIA

author: Picuino
author_rank: Super Contributor
timestamp: April 10, 2025, 07:15:00 am
content: SPOILER ALERTThe problem can be solved simply by Thevenin's theorem, simplifying the two vertical branches by two generators and two resistors, leaving three resistors in series with two generators.

author: Zero999
author_rank: Super Contributor
timestamp: April 10, 2025, 07:52:11 am
content: Quote from: Picuino on April 10, 2025, 07:15:00 amThat's how I would do it, because I don't remember the other methods, which I used to pass by exams, then forgot.Here's an example.https://www.eevblog.com/forum/beginners/dumb-question-about-resistors-in-seriesparallel/msg1511881/#msg1511881

author: Picuino
author_rank: Super Contributor
timestamp: April 10, 2025, 01:04:23 pm
content: Yet another one, easy. Classic voltage divider R-2R.What is the voltage across each of the resistors? It is not necessary to apply the node method or the mesh method.https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3EFwkXG7mF5hiKKOJpUh4qjARsAToOGjwYMSLFT4itRtWsx-cWjhsA7sr6QBh6wMgW99kCkkvHSt1NXePrnSU7TVd3EO0zSz9jbEI8D11Y+JCk-1NPUJ8jOPBVCMdLVOMaYl5jDJKVMUr-MEC+HJCa8OQzAA9uGl4UbHomFDEe3ltbQgA1NiA

author: Picuino
author_rank: Super Contributor
timestamp: April 10, 2025, 01:21:05 pm
content: Yet another one. Source superposition.What is the voltage across the final resistor R4?It is not necessary to apply the node method or the mesh method.https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3EQmkXXnuGIooomlTCFRVGAjYAnEChS8wwvghFqRE+Ao1b12TX0i9dcNgHclK072Mj+UNgA8+vFNnoI8S7FKqqoQAam5Cnt5KrP6B4Lw0YVzKDma2quoytFTm0tByXGA0qUEZOmJUflmy4ayR9GDYEF5xRSB4YYopIMRSjj1SFpD63UUOJmOiYHo2k9rxpS6zxQvpq8M2gvOCvUvcvLvduxurzt3OJ4LOkxfWBgP3kyf9NxOZ4dgYVF702HjkLVo8RAACUaGwgA

author: Picuino
author_rank: Super Contributor
timestamp: April 10, 2025, 04:05:02 pm
content: Quote from: golden_labels on April 09, 2025, 09:51:13 pmSolving circuits according to brute force attack will always work. That's what simulators, such as falstad's, do. They set up the meshes and then solve huge systems of linear equations in a short time.It's different for people. For us, solving all these problems with the mesh method is a waste of time and energy that we don't need to waste in that way. In addition, the mesh method (brute force), although it will always work, does not give us an understanding of what is really going on in the circuit and how to look at it in a more intelligent way.The circuit solving tools that can be used in these problems (simplification by symmetries, Thevenin's theorem, series-parallel simplification, source superposition) and many others available, allow us to approach circuit problems in a smarter and more fun way.

author: Picuino
author_rank: Super Contributor
timestamp: April 10, 2025, 04:09:47 pm
content: Quote from: Picuino on April 10, 2025, 01:04:23 pmSOLUTION:This circuit is solved by simplifying resistors in series and parallel from right to left. By making the series of the two rightmost resistors, we find a 2K resistor, which can be simplified by making a parallel with the other 2K vertical resistor. The resulting parallel returns to a 1K resistor and we are back to square one. So until we get to the stack, with a divisor by 2. We can see how each stage divides by 2 the previous voltage.

author: Picuino
author_rank: Super Contributor
timestamp: April 10, 2025, 04:14:51 pm
content: Quote from: Picuino on April 10, 2025, 01:21:05 pmSOLUTION:This circuit is solved by shorting to ground all but one battery (for example the 4V battery). Then we calculate the voltage on the resistor R4 by making the parallel between 3 resistors of 1K and the voltage divider with the 1K resistor of the battery. The result is always a divisor by 4. Therefore the 4V battery produces a voltage at R4 of 1 volt.If we repeat the procedure for each of the two remaining batteries, we have 2 volts produced by 8 volts battery and 4 volts more by 16 volts battery.Now we just have to add up all the calculated voltages, resulting in 7 volts.

author: Zero999
author_rank: Super Contributor
timestamp: April 10, 2025, 09:52:45 pm
content: Quote from: Picuino on April 10, 2025, 04:14:51 pmThat circuit also has some practical use. The resistor values are all the same, so if the resistor on the output is eliminated, or in reality is as higher impedance as possible, say the input of an op-amp, then the output is simply the average of all the voltage sources. Setting the voltage gain of the amplifier to equal the number of resistors/voltage sources, results in a non-inverting summing amplifier.

author: MrAl
author_rank: Super Contributor
timestamp: April 11, 2025, 01:49:38 pm
content: Quote from: golden_labels on April 09, 2025, 06:40:07 pmHi,Yes, I realize that, and I am arguing the cons of trying to spot patterns until some proficiency is attained by the practitioner. That's because a prospective symmetry may look like a genuine symmetry but not be, and so the results would be in error.If you saw this cube for the first time would you be able to spot that symmetry. On the other hand, math software is so commonplace now that doing a full Nodal Analysis is not that out of the question even for 8 nodes.It is interesting though how much this particular problem was able to be simplified, and this is actually an example of how that can help.So next time we have a cube of resistors on hand powered from two opposite corners, we will know the answer to the total resistance right away (ha ha):it is R/1.2 exactlyThere are a lot of other interesting problems like this too. Put a lot of those cubes together to form a 3d grid and solve for the total resistance, then let each resistor tend to zero while the number of resistors on one edge tends to infinity, calculate the resistance.Even the flat 2d grid is interesting when we do that because it ends up having a well-known field pattern related to the Laplace Equation.I love the infinite network problems too as there are different ways to solve those. This would be similar to the telegrapher's equation.

author: Picuino
author_rank: Super Contributor
timestamp: April 11, 2025, 04:16:18 pm
content: Quote from: MrAl on April 11, 2025, 01:49:38 pmThat looks like problems that are solved with coupled quadripoles. It's a bit more advanced than basic circuit theory (in some first courses they don't get to see it), but it would also be interesting to see a problem like that. The R-2R resistor ladder problem can be turned into one of those.Quote from: MrAl on April 11, 2025, 01:49:38 pmThat is one of the objectives of the thread. Many teachers look for problems that can only be solved on the exam (with little time) by applying simplifications. If you already know them, they can't catch you off guard.

author: Picuino
author_rank: Super Contributor
timestamp: April 11, 2025, 04:36:35 pm
content: Solve the voltage at RL without mesh analysis.https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxGwBYQlLsaBTAWjDACgA3cFPCyasD3CVqkKCEpjB4sTARsATt15gRykIVEgUbAB7hshbZGLgE5FNIEpuANUXaU1TY+q4tlByica3h3+I6Stj+LiFG7uKewaHUIrwuUg7xASmRYtjJlAnUCIRiieAOeQW5+XwebADuNOWRafxQ1RLZFRT+6c3hbd5ujZDNvW2CvJ01Qy4jAQP6KP5gGDYoCBALtNwgAEo6+qoQxAJ5IGQSG5vss4sgeEg85Hi81luZ+th49HhuxLx45E+b8lerQW9EM9AWRn+nn0lGISE+NDAYl+p2kmwwegkhFWOAkGCMENRNk2hEx5gEVwQeFMayJWwAMmwgA

author: TimFox
author_rank: Super Contributor
timestamp: April 11, 2025, 04:41:46 pm
content: Hint: work the problem backwards from RL.

author: Picuino
author_rank: Super Contributor
timestamp: April 11, 2025, 04:56:08 pm
content: Another approach is to convert the networks Pi (4-2-4 Ohms) and T (1-3-1 Ohms) into transmission matrices (ABCD-parameters) that are multiplied.I am trying to solve it, but I am not good with transmission matrices and do not get the correct result.https://en.wikipedia.org/wiki/Two-port_network

author: Picuino
author_rank: Super Contributor
timestamp: April 11, 2025, 05:40:00 pm
content: The network in PI is formed by R2=4, R3=2 and R4=4ABCD Parameters (https://en.wikipedia.org/wiki/Two-port_network#ABCD-parameters) are:A = 1.5B = 2C = 0.625D = 1.5Matrix = [ 1.5 2.0 ][0.625 1.5 ]The network T is formed by R5=1, R6=3, R7=1ABCD Parameters (https://en.wikipedia.org/wiki/Two-port_network#ABCD-parameters) are:A = 1.333B = 1.333C = 0.333D = 1.333Matrix = [1.333 1.333][0.333 1.333 ]Definitely, this particular problem is not easily solved by quadripoles (or I am not fluent enough).

author: MrAl
author_rank: Super Contributor
timestamp: April 12, 2025, 04:29:32 pm
content: Quote from: TimFox on April 11, 2025, 04:41:46 pmHello there,That technique I like to call "collapsing the network" which just means that components can be combined using their impedances (and possibly more types of components). The question then comes up, can every network be collapsible.I tried this with a grid network of lots of resistors (not infinite though) and had not gotten to a reasonable solution yet. Perhaps you have seen one.The networks I am presenting next are all "collapsible" and so there comes and interesting solution that is not difficult. Next post.

author: MrAl
author_rank: Super Contributor
timestamp: April 12, 2025, 04:35:01 pm
content: Quote from: Picuino on April 11, 2025, 04:16:18 pmHi there,Perhaps you can briefly show your proposed solution to the resistor network. And yes I understand that sometimes we just one a single solution to a problem without worry if we will be able to solve every network that way. We get some pretty interesting stuff that way too.Here are several infinite network problems that are not too difficult (attachment).These are all about calculating the input impedance which is often a concern.If you or anyone else wants to show your solution or just how to solve these we can compare notes.I also posted a solution on Wikipedia a long time ago to the transmission line which breaks it down into functional blocks. That's a different view on how it works.

author: temperance
author_rank: Super Contributor
timestamp: April 12, 2025, 06:59:57 pm
content: You have an RC network, 10R with 1 nF connected to a signal generator. The signal generator output is a 10 V 50% duty cycle square wave, 10 KHz.Case 1: the signal generator rise and fall time are both 0. (just assume something like that is possible)Case 2: the signal generator rise and fall time are both half the RC time constant.For both cases, calculate the power loss in the resistor.Have fun.

author: TimFox
author_rank: Super Contributor
timestamp: April 12, 2025, 07:31:51 pm
content: Quote from: MrAl on April 12, 2025, 04:29:32 pmRecently, I needed to evaluate a simple non-linear network comprising several resistors and diodes. I started from the voltage across the diodes, computed the diode current with the diode equation, then worked back to the input from that voltage and current.I graphed the result with the output voltage on the y-axis and input voltage on the x-axis, even though the calculation had the input as the dependent variable.

author: golden_labels
author_rank: Super Contributor
timestamp: April 14, 2025, 12:37:35 am
content: Let’s see some unexpectedly practical example.We have a grid of 1 Ω resistors, n columns, n+1 rows in size. What is the equivalent resistance? How does it change if we change n? What if n goes to infinity?Don’t miss that power is connected across the entire width of the square!Falstad’s circuit simulation

author: MrAl
author_rank: Super Contributor
timestamp: April 14, 2025, 12:49:03 pm
content: Quote from: golden_labels on April 14, 2025, 12:37:35 amHi,I would think that particular one would be simpler because there are a lot of equipotential nodes along the horizontal. If there is no voltage drop across any resistor then there is no current flow, and therefore that resistor would not be able to contribute to the total resistance.Connect the power source in a different way though, and everything can change drastically.There is another interesting way to solve these too and that is to use methods taken from the numerical solutions to partial differential equations. The voltages at every node can then be calculated even with a very, very large grid, even with different resistor values and power connections. Solutions can take a while to complete though on a home computer.

author: golden_labels
author_rank: Super Contributor
timestamp: April 14, 2025, 08:39:26 pm
content: Quote from: MrAl on April 14, 2025, 12:49:03 pmThe value in this exercise is not calculating the equivalent resistance. This is pretty easy indeed. The value is in observing what happens whennchanges and, more importantly, recognizing the general rule. The very same principle explains a counter-intuitive behavior in some practical situations.

author: MrAl
author_rank: Super Contributor
timestamp: April 14, 2025, 11:20:48 pm
content: Quote from: golden_labels on April 14, 2025, 08:39:26 pmThat goes without saying because you already explained that, and I have no problem with that.Here's the quote from your original post:START_QUOTEWe have a grid of 1 Ω resistors, n columns, n+1 rows in size. What is the equivalent resistance? How does it change if we change n? What if n goes to infinity?END_QUOTETake note of the question, "What is the equivalent resistance?"Even if it is not SOLELY about a single calculation of the equivalent resistance, it still involves the equivalent resistance and that has to be either calculated or functionalized, your choice. I submit that you can't come up with a general rule if you can't calculate at least one instance of the equivalent resistance. That would be a prerequisite.It's also a little related and interesting to calculate the total resistance when we combine resistors of equal value in parallel and series and obtain the same resistance as the original resistor. For example, two 1 Ohm resistors in parallel makes a 1/2 Ohm resistor, then two of those sets in series makes 1 Ohm again. Now that we have 1 Ohm again, we can combine four of those sets in the same way to create yet another 1 Ohm resistor, and then do that again and again.Of course we could start with two 1 Ohm in series to create 2 Ohms, then place two of those sets in parallel. We get 1 Ohm again. With multiple sets of those we can also create a 1 Ohm resistor, then another, then another, by keep adding sets of the previous set of 1 Ohm resistors. (Maybe just a little bit comical)

author: golden_labels
author_rank: Super Contributor
timestamp: April 15, 2025, 12:48:00 am
content: Quote from: MrAl on April 14, 2025, 11:20:48 pmCalculating equivalent resistance is needed to continue into the latter part of the exercise. Without a doubt it’s easier than all the examples earlier, but it was never my point to make that part of the exercise hard. The value is in the other part. I hope this is clear now, and I’m not going to discuss it further.Quote from: MrAl on April 14, 2025, 11:20:48 pmYes, it works not only for 1 Ω, but for any resistance. In fact this is the takeaway from this exercise: the equivalent resistance is always the same as the “unit resistance.” It also works for networks with other cell shapes. So the magic isn’t caused by the horizontal resistors not carrying any current.(1)The principle here is the same as the one behind the “ohms-per-square” unit. Which may go against intuition. It basically says that the area of a flat conductor (e.g. copper on a PCB) is irrelevant for the edge-to-edge resistance. A 1 mm square has the same resistance as a square that covers the entire solar system. Resistance is instead proportional to the aspect ratio of the rectangle.(1)I used square cells, because they are trivial to calculate, it is easy to tell what “unit” is, and they work down to n=1. Other lattices may require calculation with n ≫ 1. That may be not doable by hand and provides no educational value.

author: MrAl
author_rank: Super Contributor
timestamp: April 15, 2025, 01:38:25 pm
content: Hello again,For some reason you seem to think that a reply must follow exactly, to the letter, what you were saying. That's not how conversation works.If you talk about a truck with four tires and I talk about a car with four tires, so what? It's regular human conversation.I still appreciate your input in these matters however.Quote from: golden_labelsAgain, you already made that clear not sure why you are repeating it again.I was talking about the equivalent resistance; you were talking about the equivalent resistance plus how it changes. I get that but do you as you seem to be strongly implying that there was some kind of mistake made somewhere.Quote from: golden_labelsIt does not have to be caused by a zero horizontal resistance, it is a entirely different circuit.If I have a circuit with four series munchkins holding hands, it does not have to have any zero horizontal resistance it's a different circuitAs I said, I still appreciate your interest in these matters and appreciate input from anyone who loves electrical/electronic circuits. Thanks for that.