Ohm's Law/Kirchhoff's Law Question - Page 1
Forum section: electronics / beginners

author: Ron Reeland
author_rank: Newbie
timestamp: January 29, 2022, 08:31:16 pm
content: Hi:I ran across this Popular Electronics Magazine (Dec. 1970) quiz while looking for another article. Electronic quizzes were a fairly regular feature in the magazine.I have enlarged the paragraph above the circuits, so it is legible. Can anyone explain how the formula works? Ohm's Law and ratios?It mentions that Ohm's Law and Kirchoff's Law applies but does not explain how the formula is derived from them.My experience with Ohm's law is that if two quantities are given, the third can be determined.The quiz uses voltages only. It does some squaring, addition and subtraction, but neither resistance nor current is specified.Answers, too, in case you want to try your hand at manipulating the basic algebraic relationship for the various circuits.(Not sure if my two images will appear in my post. I cannot view them in the preview option. If they do not appear, what must I do to have them in my post?)Thanks,Ron Reeland

author: Slh
author_rank: Regular Contributor
timestamp: January 29, 2022, 09:56:00 pm
content: You can get to it from complex numbers. In these puzzles they're only concerned with the magnitude, not the phase.V = IZZ(R) = R, Z(L) = jwL, Z(C) = -1/(jwC)Z(RLC) = R +j(wL - 1/(wC)V(RLC) = I(R + j(wL - 1/(wC)))The magnitude issqrt(V) = sqrt((IR)^2 + (I(wL-1/(wC)))^2)so if you square both sides and notice that the magnitude of V(L) = IwL and the magnitude of V(C) = -I/(wC) then you should have your answer.

author: Ron Reeland
author_rank: Newbie
timestamp: February 01, 2022, 07:27:17 pm
content: Thank you for the reply. Complex and imaginary "j" numbers are beyond my basic mathematics knowledge. And at age 78 I just do not retain what I study in textbooks related to higher math. I will just have to be content with the basic ohm's law functions.Thanks Again,Ron Reeland

author: bob91343
author_rank: Super Contributor
timestamp: February 01, 2022, 11:38:36 pm
content: Ron, it's not that tough. The most complex thing about complex numbers is the name. They aren't complex at all. I am 89 years old and am studying Laplace transforms and particle physics because I find them interesting and useful.

author: rstofer
author_rank: Super Contributor
timestamp: February 01, 2022, 11:54:13 pm
content: Age is NOT a factor, I'm 76 and still working with math nearly every day. Preferably in MATLAB.I do know how to do the vector math for the resulting complex numbers but I would rather have a machine do the crunching. Octave will do the same thing and it's free.Here's Khan Academy's Electrical Engineering course section on Circuit Analysis:https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topicThis is a really good program starting from the very beginning. There is also the Math courses at:https://www.khanacademy.org/parentThis stuff is all learnable, even by more experienced (older) people. Just take it one step at a time.There also Real Analog at Digilent:https://digilent.com/reference/learn/courses/real-analog/startIf you find you need to draw graphs of functions:www.desmos.comIf you need help solving equations and stuff:www.symbolab.comI contribute to both of these sites.If you want what I consider to be the best math instruction on the Internet:https://calcworkshop.com/This site costs money (monthly or annual subscription) and is worth every dime. Every single dime!If I had a kid in high school or college, they would absolutely have a subscription. From their first day in Algebra I...Khan Academy is also very good but I don't think it is as structured as CalcWorkshop.I contribute to Khan Academy as well.3Blue1Brown.com is also very good but for more advanced stuff.EE is ALL math and math is fun! Really!Try the EE program at Khan Academy and see how it goes. It beats playing TicToc... Well, I don't really know that, I'm just guessing. I have never used TicToc.

author: rstofer
author_rank: Super Contributor
timestamp: February 02, 2022, 12:08:10 am
content: Quote from: bob91343 on February 01, 2022, 11:38:36 pmI'm 76 and still playing with math, almost every day!We should start a Geezer subforum... You can be President!The problem isn't with the math, it's with the way it is/was taught such that everybody hates it. They don't do well, math kicks their ass and they tend to avoid it with a passion. It really doesn't need to be that way. The Internet is the great 'leveler' in terms of the math playing field. All of human knowledge is accessible through Google. If one explanation doesn't work, well, try the next link down.All it takes is desire and an Internet connection.Maybe this is a decent introduction to Complex Numbers:https://www.mathsisfun.com/numbers/complex-numbers.html

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 02, 2022, 05:35:02 am
content: I always think the maths for the item at hand is often overkill, especially for junctions of resistors, makes it seem harder than it actually is.Ordinary weighted sums is how I teach engineering.

author: ledtester
author_rank: Super Contributor
timestamp: February 02, 2022, 06:46:48 am
content: Quote from: Capernicus on February 02, 2022, 05:35:02 amHow would you solve the first problem using ordinary weighted sums?

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 02, 2022, 07:06:32 am
content: questions 7->10 are the circuits with junctions in them. (and it puts them later because they are the tough ones right?) the current will divide from the battery and come back together to the battery, as a weighted sum of current.The other questions there's not even a junction in them, so its not even required.If u swap the inductors for caps (cause I havent actually studied inductors enough yet.) then the capacitor will act as a resistor to the ac. farads vs frequency (probably use t=rc, then u could compute the max volts it gets to.)Then I'm not sure what the metric they want for the answer, but I could probably answer it if I needed to.

author: Ron Reeland
author_rank: Newbie
timestamp: February 02, 2022, 12:09:18 pm
content: Hi:I certainly did not mean to imply that all seniors are incapable of learning or continuing to contribute to society or pursuing math for personal satisfaction.In my case, I am the custodian at Peru Public Library. Six days per week part time. I have been working there for 22 years and counting. The ladies find many challenges both mental and physical for me besides basic cleaning and maintenance chores. The job satisfaction is why I continue working. Nothing like a "pat on the back" to reward me for my efforts. I support libraries and feel I do my part to promote their use and value to society.I also fool around and restore antique radios, vacuum cleaners, and anything that needs repair. I get involved with Tesla Coils and such. And there is nothing like a Hamfest of Amateur Radio Operators to find a treasure that needs loving care. I enjoy combing mental and hands-on activities that result in tangible accomplishments.I have studied vectors and phase and have been introduced to imaginary numbers there. But I just do not have a practical use for these concepts. I appreciate the links to the math sites. There is so much on the internet for almost any interest.I really appreciate Forums of EEV nature as they are populated my super knowledgeable folks that are friendly and eager to assist.Thanks,Ron Reeland

author: LaryPant
author_rank: Contributor
timestamp: February 03, 2022, 11:00:50 pm
content: Quote from: Capernicus on February 02, 2022, 07:06:32 amBy a "weighted sum of current" I assume that you mean some variant of the typical current divider formula for finding the current through each branch of a parallel network, but this does not apply to these questions because the impedance is unknown.This is the beauty of the (VT)^2=(VL-VC)^2+(VR)^2 relationship presented in the article. It allows one to find the voltages or currents without knowing the impedance. The questions with junctions are not the hardest due to the junctions, but because the provided formula was for voltages, but one needs to use the very similar current formula (IT)^2=(IL-IC)^2+(IR)^2, which was not given directly in the article.Quote from: Capernicus on February 02, 2022, 07:06:32 amThis makes no sense and should be corrected:Yes, both inductors and capacitors will oppose the current flow in an AC circuit, but the phase is different and neither should be referred to as "a resistor" under AC conditions, the correct term is impedance.I have no idea of what you mean by "farads vs frequency"(?) but the time constant (tau=rc) has to do with the transient response of a RC circuit and not with steady state AC. Even if the formula was relevant, neither R nor C is known. The time constant is also irrelevant for finding the max voltage in a RC circuit as it is equal to the supply voltage.Quote from: Capernicus on February 02, 2022, 07:06:32 amI highly encourage you to try and answer the questions and derive the provided formula, see the first reply for inspiration.

author: tooki
author_rank: Super Contributor
timestamp: February 03, 2022, 11:15:19 pm
content: Quote from: Capernicus on February 02, 2022, 05:35:02 amYou "teaching" is a positively terrifying thought. Someone with as poor an understanding of things should be learning, not teaching.Quote from: Capernicus on February 02, 2022, 07:06:32 am"Junctions" isn't the term for these. They're parallel circuits. 1-6 are series circuits. Series vs. parallel is yet another electricity fundamental you apparently do not understand.I know I'm hard on you, but it's that you REALLY need to stop trying to teach people, answer questions, etc. when in reality you are just not there yet (and as I've said before, filling in the gaps with your own invented explanations isn't equivalent to education). You need to erase your homemade "knowledge" and re-learn electronics from the beginning, properly.

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 04, 2022, 05:36:24 am
content: Understanding series vs parallel is important, but knowing junctions in general is even more powerful if u want to plan out your own discrete machines.if the hz of the ac power isnt high enough, and the amount of farads isnt high enough on the capacitor you will start to lose current, via that impedance thing you're talking about.I'm no good for computing exact values, but I just know the general concepts, I find its served me well making lots of things, like motor drivers, oscillators, led screens, microphones, data registers. (even analogue data registers). computing the exact values of things isnt anywhere near as important as knowing simply what they are doing in general.

author: LaryPant
author_rank: Contributor
timestamp: February 04, 2022, 09:49:38 pm
content: Quote from: Capernicus on February 04, 2022, 05:36:24 amUnderstanding series vs parallel is not important, it is a necessity! If you understand series and parallel networks you understand Kirchhoff's Laws which are electronics 101. And if you know these laws then you can analyze circuits. (I am once again unsure of what you actually mean by "knowing junctions".)Quote from: Capernicus on February 04, 2022, 05:36:24 amI do not want to argue the merit of what you have build and designed, but this has to be corrected once again: Current cannot belost. The current flowing into a circuit/junction/node/banana/whatever must flow out of it again - Kirchhoff's first law.Any impedance (think AC resistance, for now) will reduce the current (and change the phase, but let's ignore that for know) as given by Ohm's LawI=V/ZIt is not something thatstartsonce a certain threshold has been reached, it is a linear relationship.Electronics is certainly an exciting world, which we all enjoy, and one can quickly start building fascinating things, but I would highly recommend to study/learn electronics more in depth and properly as it will open the doors to an even larger world.I really hope that this helps,Lary

author: Canis Dirus Leidy
author_rank: Regular Contributor
timestamp: February 05, 2022, 01:40:43 am
content: By the way. There is a graphical method for solving such problems that does not require(*) the use of complex numbers:phasor diagrams.(*)Mainly because it becomes inapplicable as soon as it comes to something more than a simple series and parallel RLC circuits. But it would be too cruel to immediately beat first year students with the full course of circuit analysis theory.

author: basinstreetdesign
author_rank: Frequent Contributor
timestamp: February 05, 2022, 06:43:06 am
content: Quote from: Canis Dirus Leidy on February 05, 2022, 01:40:43 amRon, I believe the easiest way to reconcile the formulas given in the articles you reproduced with the problems given is to remember that the relationship between current and voltage of both inductors and capacitors is that of a right triangle. When you draw the voltages of the RLC components around a series loop with a source voltage as vectors connected head-to-tail and the common current through them as a vector then the voltages of the C and L are at right angles with the current while the voltage across the R is in line with it. Also the voltage across the L is 180 deg apart from the voltage across the C so subtracts directly.Thus the formulas reduce to the application of the Pythagorean equation. Once drawn on paper it all becomes very simple and you will kick yourself for not realizing it sooner.

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 05, 2022, 08:32:53 am
content: I bet there heaps of ways to look at this problem, and they are all great.Whenever I see complex number methods the first thing that crosses my mind is the fact there probably is another way to model the situation without them and avoid the complexity.Especially when I see the square root of -1, it is absolutely not necessary to do anything with that, theres always an equivilent solution without it.

author: tooki
author_rank: Super Contributor
timestamp: February 05, 2022, 11:43:46 am
content: Quote from: Capernicus on February 05, 2022, 08:32:53 amSome ways might be great, but certainly not “all”. Some ways will be incomplete (e.g. applicable only in some situations), others will be outright wrong.Quote from: Capernicus on February 05, 2022, 08:32:53 amWell, the next semester of school starts in 2 weeks, and one teacher has promised to introduce complex numbers because of how itsimplifiesexactly these problems. We shall see!

author: tooki
author_rank: Super Contributor
timestamp: February 05, 2022, 11:58:48 am
content: Quote from: Capernicus on February 04, 2022, 05:36:24 amWhat is this junction theory you have in mind? Because it’s not some broadly known electronics fundamental. Or it is, but you’re using your own private vocabulary to describe it, which is useless. Can you kindly explain it in detail?Quote from: Capernicus on February 04, 2022, 05:36:24 amYou don’t say…Quote from: Capernicus on February 04, 2022, 05:36:24 amNo, you don’t. Youthinkyou do, but you don’t.As I’ve said before: you’ve learned bits and pieces and then filled in the rest with your own theories and mental models which aren’t correct. Yet you hold those DIY mental models to be qualitatively equivalent to the actual proven principles and models developed empirically by actual experts.Quote from: Capernicus on February 04, 2022, 05:36:24 amUnderstanding the general principles is absolutely essential, but a) I don’t think you understand them as well as you think, and b) they don’t help you figure out why something isn’t working when you think it should, nor c) how to optimize for specific situations, d) calculate limits, etc.Knowing how to calculate things can save a lot of time compared to (potentially destructive!) experimentation. For example, rather than testing dozens of resistor values to find what I need in a circuit, if I can just calculate it, then I know exactly what to use.

author: TimFox
author_rank: Super Contributor
timestamp: February 05, 2022, 02:41:00 pm
content: One might accomplish certain things without understanding mathematics by trial and error.This is a forum for beginners, and I would never recommend willful ignorance as a design method.To be useful, it is better to suggest mathematical methods that are useful and recommend sources for further information.Specifically, one serious problem for a literal beginner is knowing what terms are used relevant to a new (to him) problem.For example, when contemplating connecting a coax cable between two objects, it may not be obvious that the answers to his questions are filed under the term "transmission line" in the literature.

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 05, 2022, 04:48:22 pm
content: Quote from: tooki on February 05, 2022, 11:58:48 amWhen electricity meets an intersection, if its fanning apart, the current divides (fractions), if its fanning together, the current adds.So theres 2 mathematical functions/calculations able to be used straight away just with having the wires and electricity itself.Quote from: tooki on February 05, 2022, 11:58:48 amIts hardly important at all to me! =) Alot of the time when I put caps in my circuits the actual farads dont even matter at all! And when I'm deciding resistances as long as the machine still outputs acceptably, a range of values is acceptable. I've got my next oscillator on the way, I don't know what the frequency will be exactly but as long as its oscillating that's the main part right?Quote from: tooki on February 05, 2022, 11:58:48 amYou obviously dont believe in me one little bit! thats ok, I don't expect ppl to and I'd probably rather people didn't cause it gets awkward as hell, best sneak around without the spotlight its alot less embarressing.Quote from: tooki on February 05, 2022, 11:58:48 amI don't see that as anything but a plus and I recommend everyone do it!Quote from: tooki on February 05, 2022, 11:58:48 amI actually am quite foggy on the whole thing! I do things approximately, electronics is tough to work out whats actualy going on for real, but I know enough to get what I do done, I understand my own circuits, I'm not one to fix pro equipment when it fails, nor do I do calculus much, but I understand the surface of it.Quote from: tooki on February 05, 2022, 11:58:48 amI guess so, but do no experimentation at all and u'll get absolutely no-where for sure.

author: TimFox
author_rank: Super Contributor
timestamp: February 05, 2022, 05:28:58 pm
content: "When electricity meets an intersection, if its fanning apart, the current divides (fractions), if its fanning together, the current adds.So theres 2 mathematical functions/calculations able to be used straight away just with having the wires and electricity itself."Congratulations. You have just renamed Kirchhoff's Current Law, abbreviated "KCL". Other beginners, looking for information on this topic, won't find it under "junction theory", but can find it under "KCL" or "Kirchhoff" (note the double "h" in the proper name). Seehttps://www.allaboutcircuits.com/textbook/direct-current/chpt-6/kirchhoffs-current-law-kcl/, one of many results for Googling the standard term.

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 05, 2022, 05:44:37 pm
content: Another thing to add to it.If you divide, then divide again, youll have get two consequtive! divides happening, so it actually computes log as well.Its how DACS work.U can also do multiply with a voltage divider, (if its between 0 -> 1) As long as its a fixed quantity. How about that for an optimized analogue multiply.

author: TimFox
author_rank: Super Contributor
timestamp: February 05, 2022, 06:31:41 pm
content: Profound, if confusing. How does that produce a logarithm?

author: IanB
author_rank: Super Contributor
timestamp: February 05, 2022, 06:55:55 pm
content: Quote from: Capernicus on February 05, 2022, 08:32:53 amComplex numbers are simply an ordered pair of two numbers, combined with a set of rules on how to do arithmetic with them. A cool thing about the complex numbers is that once you have them, it becomes possible to define the square root of negative numbers (like -1), something that cannot be done with the real numbers. You could say that the complex numbers come first, the square root of -1 comes second. It is all a matter of perspective.Another thing about complex numbers is that because they are an ordered pair, it is possible to look at them as (x, y) coordinates on the plane. Since the x and y directions are at right angles to each other (i.e. 90° apart), and since inductors and capacitors have voltage and current 90° apart, the complex numbers can become a convenient shorthand for working with AC circuit calculations. The complex numbers can conveniently keep track of the 0° and ±90° components of voltage and current, permitting simpler formulas and more straightforward calculations.

author: rstofer
author_rank: Super Contributor
timestamp: February 05, 2022, 06:59:13 pm
content: Quote from: TimFox on February 05, 2022, 05:28:58 pmThey might also find it under "Nodal Analysis".

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 05, 2022, 07:11:57 pm
content: Quote from: TimFox on February 05, 2022, 06:31:41 pmJust as consequtive multiplies are a power, consequtive divides could be a logarythm, as consequtive adds are a multiply, to consequtive subtracts are a divide. log-> Its the count of consequtive divides it takes to get to 1. it would be cool to see that done with wire dividers!IanB - Ive got done work with the FFT before, and it has amplitude and phase as x+y as well, I never understood it but I got it working and used it. I found all of digital and analogue filtering virtually impossible to understand, so I always just nabbed code instead to get my projects finished. I have an idea of how I'd go about attempting my own filter tho, if there was some way to increase the falloff speed of a blur, you could make a simple blur turn into a lowpass filter, as it is without increasing the falloff, its on orders of octaves not harmonics how slow it attenuates the sound. How a capacitor and resistor does it is quite dumbfounding to me just an ordinary bin can do it.

author: TimFox
author_rank: Super Contributor
timestamp: February 05, 2022, 07:28:55 pm
content: The base 10 logarithm of 2 is 0.30103...How do you compute that useful result with your "junctions"?If it only works for base 2, how do you compute log2(3)?

author: TimFox
author_rank: Super Contributor
timestamp: February 05, 2022, 07:31:52 pm
content: Quote from: rstofer on February 05, 2022, 06:59:13 pmYes, "Nodal analysis" is a useful algebraic method to solve KCL on a practical circuit.

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 05, 2022, 07:38:23 pm
content: the integral log2 is all you need to make an ADC. fractional results aren't as important.log2(3) should abort on the second divider. 1^2. (the first divider is 0^2) (its 2.)the strange thing about the integer log2, is it actually truncates upwards instead of downwards.

author: TimFox
author_rank: Super Contributor
timestamp: February 05, 2022, 07:57:31 pm
content: That's a far less useful result than your original statement: "If you divide, then divide again, youll have get two consequtive! divides happening, so it actually computes log as well."Maybe it works better with punctuation.

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 05, 2022, 08:10:16 pm
content: Got me drawing up another SCHEMMY!!!adjust that top line of resistors and itll change the sensitivity, but it has a problem still because when it starts activating it keeps activating all past the threshold. so has a problem still. I think if you make the resistors resist less ohms towards the left, it actually make the inverters outputs precedent over each other. im still learnin'!Because its just a binary output, it can be heaps convoluted charge and it doesnt matter, its only an ADC so that makes it easier.the wire coming in the side is an inverter, that activates when the charge has divided off enough for it to activate.But then the question is, can u get the rest of the bits?

author: Electro Fan
author_rank: Super Contributor
timestamp: February 06, 2022, 05:34:15 am
content: Quote from: Ron Reeland on January 29, 2022, 08:31:16 pmNot sure if you are looking for any more help with Ohm’s Law but if you are and you haven’t come across the circle of Ohm’s Law (and the associated triangles) you might find this useful:http://electrical-mcqs.blogspot.com/2017/12/ohms-law-interview-question.html

author: george.b
author_rank: Frequent Contributor
timestamp: February 06, 2022, 03:50:45 pm
content: Quote from: Capernicus on February 05, 2022, 04:48:22 pmIt's very important, because you're getting bogus results most of the time. I could be mistaken, but I think the truth goes more along the lines of you being scared of doing the math, because that requires a modicum of formal education, even if it's self-education - you'll have to sit down with a book or an article and read through it andactually be taught about it by somebody else, directly or indirectly, which is something I think you're too arrogant to bring yourself to. It seems that, in your mind, you're this unparalleled, misunderstood genius who can solve complicated problems intuitively in his head and with some trivial experimentation using nothing but capacitors and resistors. In reality, you're someone with a seemingly weak maths background, scared of complex numbers because you can't understand them, so you'd rather go around them than just accept and use them, just like you'd rather go around all of the well-established circuit theory and just throw stuff at the wall hoping something will stick (and not explode). You're free to do so, of course, but then don't be surprised when your ideas are met by others (or by reality) with resistance and/or ridicule.Being afraid of learning will only set you back, you know.

author: Ron Reeland
author_rank: Newbie
timestamp: February 09, 2022, 07:54:05 pm
content: Hi:basinstreetdesign posted a reply that the quiz formula is based on the famous Pythagorean Theorem. And he is right. A closer look at the quiz formula reveals that it is the usual A squared = B squared + C squared of the classic example. Thanks, as I did not comprehend that fact. The author pointed this out to a degree when he mentioned the values follow the 3,4,5 solution.So, I printed up some graph paper and worked out the quiz #1 and quiz #2. My scans of them are shown below. Are they correct?But I am stumped by quiz#3. I cannot deduce how to form a triangle since there is not an in phase or horizontal base for a resistor. I believe the capacitor's 6 volts is to be subtracted from the unknown inductor voltage to form one leg of a triangle. But I just am getting myself more confused.Please assist. I am not ready to proceed in the quiz until I am taught how to solve quiz #3. I wonder if I will stumble on the balance of the quizzes.Thanks,Ron Reeland

author: ledtester
author_rank: Super Contributor
timestamp: February 09, 2022, 08:04:36 pm
content: Well, the introduction of the problem set says this:QuoteIn the case of #3, VR = 0, so you just have to solve VT = VL - VC.