Kirchhoff's Current Law resources. - Page 1
Forum section: electronics / beginners

author: Mint.
author_rank: Frequent Contributor
timestamp: March 10, 2012, 07:43:35 am
content: I need some material to help me learn KCL, I understand most simple circuits, but when we get up to things such as the photo attached, I get very lost. This is for MITx 6.002x as you can see. I understand it a bit, but I need more practice and maybe to watch more videos. Can anybody suggest something please?

author: bfritz
author_rank: Regular Contributor
timestamp: March 10, 2012, 08:56:13 am
content: My guess is that you understand the theory, but that you are having problems with solveing these, because you are getting a little confused, as you don't know how to organize your thoughts. So, try making it easy to see what you know. Doing these on paper will make them easier, as you can put marks on paper easily.I will lead you through how I solved this.So the rule is that the sum of currents into the node, equals the currents out.orThe sum of currents meeting at a point is zero.So, if you mark each known by circling it, it makes it easier to recognize what you know. So, circle i1, i2, and i3, as they are known.Always look for the easiest case to solve. Every node in that circuit has 3 currents associated with it, so if you know two of them, you can write the equation for the third. Look for a node associated with the current in question, that has two known currents. In other words, look for a node associated with the current in question, that has two circled currents.So solving the first, you are trying to determine i4. You see that you know two of the currents into the rightmost node of i4, so you can write that equation directly. i4 + i3 -i2 = 0 Then you only need to solve for i4, as the others are known. i4 = -i3 + i2 = i2 - i3So, now you also know i4, so circle it.Now you want i5. The leftmost node has two currents of the three know, so you can solve that. So, i5 = -i4 - i1, and since i4 = i2 - i3 substitute in for i4 => i5 = - (i2 - i3) -i1 = -i2 + i3 -i1 = i3 - i2 - i1.Now you know i5, so circle it.Now you want i6. The top node has two of the three currents know, so you can solve it. So, i6 = i5 - i3, and i5 is know above so substitute in for i5 => i6 = (i3 - i2 - i1) - i3 = i3 - i2 - i1 - i3.So, try and make it easier to see what you know, and as you know more, update the drawing to show what you know.Hope that helps.

author: Bored@Work
author_rank: Super Contributor
timestamp: March 10, 2012, 10:16:36 am
content: This exercise is rigged, they are leading you by the nose, or, if you prefer they are taking you by the hand.The questions are asked in exactly the way that you need the solution of the previous answer to solve the current one. This provides a sense of a particular order you have to observe, and everything will be fine. But this is not the case in reality. There is no meaning in the order of the solutions, there is no magic in the order, other than that the exercise is rigged so you can do it exactly in the provided order without having to think too hard.There are systematic ways to solve such circuits. Constructing a system of linear equations and solving the system. There are also some receipts to muddle along. I assume they will teach you a systematic way later. If not, the course is shit. For now all they want to do is to hammer in your head thatthe sum of all currents (observing direction) in a node is zero.

author: Mint.
author_rank: Frequent Contributor
timestamp: March 11, 2012, 12:30:52 am
content: Thank you guys. This course is bloody horrible for beginners, absolute junk.EDIT: But I'm still gonna do it, I want that certificate real bad!

author: harnon
author_rank: Regular Contributor
timestamp: March 11, 2012, 01:06:52 am
content: Hey mint,Yeah it does assume a fair bit of knowledge although in fairness they did say that upon signing up. A fair chunk of the stuff so far is based on high school physics.@bored - this is an example problem from a lecture sequence, so does quite deliberately drag you through it step by step.

author: bfritz
author_rank: Regular Contributor
timestamp: March 11, 2012, 02:34:48 am
content: Quote from: Mint. on March 11, 2012, 12:30:52 amNote that the course is not a Freshman year College course. Most students would have likely been all the way through their Calculus classes before taking this course, and have had over a year off Physics as well. It looks pretty much what I remember my first College Circuits class to be. It is a serious "weed out" class in most colleges. It will seperate those who want to put in the many hours, from those expecting to be spoon fed.So I would agree that it is not for beginners. It will expect really good math and physics background. That is taught in other courses. But this exercise is pretty simple... It requires application of one and only one concept, and they lead you through the solution, by making you solve the intermediate steps.Looks like good solid content. I think this exercise is to make sure you understand how to apply KCL by itself. I'm sure they will soon go through loop and mesh equations to solve all the currents and resistances using a multiple equation approach. That is pretty classic. It was the way I learned it in college in the early 80's.

author: slateraptor
author_rank: Frequent Contributor
timestamp: March 11, 2012, 02:56:21 am
content: Quote from: bfritz on March 11, 2012, 02:34:48 amYou've failed to consider the caliber of MIT freshmen; university-level calculus and physics is high school academics for these kids, making 6.002 a de facto freshman course. It's practically the norm to gain acceptance as a declared engineering major into a reputable university these days.

author: MrPlacid
author_rank: Frequent Contributor
timestamp: March 11, 2012, 05:16:41 am
content: I am lost. I thought the current entering I5would equal I1and I4. But why are the current going in opposite direction?Simplifying the circuit. I get----R(1||4)---it<----------->i5---R---Is this possible?

author: slateraptor
author_rank: Frequent Contributor
timestamp: March 11, 2012, 05:22:18 am
content: Do a google search for example problems using mesh-current analysis. Read up on passive sign convention too.Actually, study up in this order:1) passive sign convention2) what is a node; difference between essential and non-essential node3) mesh-current analysis

author: MrPlacid
author_rank: Frequent Contributor
timestamp: March 11, 2012, 05:52:57 am
content: Quote from: slateraptor on March 11, 2012, 05:22:18 amThanks for the heads up. It seemed like studying electronics never end.

author: Zad
author_rank: Super Contributor
timestamp: March 11, 2012, 05:59:57 am
content: The content is exactly what I would expect at this level. It is broadly the same as I did. Unfortunately, for some reason, my lecturer was better at getting ideas across. Perhaps because he assumed a lower mathematical ability, so he didn't cut so many corners. Then again, it isn't rocket science, but it is really bringing home how rusty my maths skills are.One thing it isnotis a beginner's course. No serious Degree level engineering modules are.

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2012, 06:16:18 am
content: Quote from: Mint. on March 10, 2012, 07:43:35 amThey give you the answers here. They are giving you a massive amount of help by removing any need to think.For example:Q1: We can write down without thinking that,0 = i1 + i4 + i5andi4 + i3 = i2We want an answer in terms of known currents, so obviously we must use,i4 + i3 = i2=> i4 = i2 - i3Q2: As with Q1 we can write down without thinking that,0 = i1 + i4 + i5andi5 = i3 + i6but immediately we see that we know i4 from Q1, so,0 = i1 + i4 + i5=> 0 = i1 + i2 - i3 + i5=> i5 = i3 - i1 - i2Q3: This is a trick question. We can write down without thinking that,i5 = i3 + i6andi6 + i2 + i1 = 0However, now we should think and observe that we already know i1 and i2 because they were measured. So,i6 + i2 + i1 = 0=> i6 = - (i1 + i2)It would help to understand where you are getting lost? This exercise is not about Kirchhoff's current law, it is about simple logic. Probably you are looking at the diagram and getting confused because it looks complicated? It is important to take a deep breath, pause for a moment, and think about what you are being asked. For example, question 1 asks about i4. Now i4 only appears in two places in the diagram, where it enters the resistor (or whatever it is) and where it leaves the resistor. So you can forget about everything else on the diagram. It is all irrelevant. The only two things that count are the nodes on each side of the resistor.

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2012, 06:19:51 am
content: Quote from: Mint. on March 11, 2012, 12:30:52 amWhat is your current level of educational attainment?Unless you are an advanced learner, I think you would find a course like this a bit difficult unless you were already able to take a few AP level courses in American high school (or A-level mathematics/physics in the UK).

author: MrPlacid
author_rank: Frequent Contributor
timestamp: March 11, 2012, 06:44:30 am
content: Quote from: IanB on March 11, 2012, 06:16:18 amNot Op, but I got confused looking at the arrows of the current. At one node, I only saw currents leaving. The bottom node has three currents entering. I am just wondering where one would place the voltage(s) to get something like this.

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2012, 06:52:07 am
content: Quote from: MrPlacid on March 11, 2012, 06:44:30 amThe rectangles in the diagram are "circuit elements". One or more of them could be voltage sources like batteries. They are all drawn as anonymous rectangles because this is a thought exercise and it shouldn't distract you from the problem at hand.As to the arrows, these do not denote the direction of the currents, they give only the sign convention. Positive in the direction of the arrow, negative against the arrow. So i5 could actually be a current entering the node and leaving as i4 and i1. In such a case i5 would have a negative sign in the solution, telling you it is flowing in the opposite direction to the arrow.

author: MrPlacid
author_rank: Frequent Contributor
timestamp: March 11, 2012, 07:01:54 am
content: Oh, i thought it was current direction. Thanks for clearing that up.Anyway, before your reply, I thought it was a sub circuit that I had to connect an external power supply to then the whole circuit would make sense.Edit: I thought the rectangular shapes were resistors

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2012, 07:30:48 am
content: Quote from: MrPlacid on March 11, 2012, 07:01:54 amYeah, it's a bit confusing that rectangles can be used as resistor symbols. That's why I wrote resistor in my post just from force of habit.If you connected an external power supply you would invalidate the exercise, as you would then have to consider the current through that power supply in your node balances.

author: Bored@Work
author_rank: Super Contributor
timestamp: March 11, 2012, 09:52:49 am
content: Quote from: harnon on March 11, 2012, 01:06:52 amAnd I was surprised they dumbed it down to that level, being the MIT and all that. With the inadvertent consequence that people look for some magic way to establish that order in advance. But there is no order, just that the sum of all currents (observing direction) in a node is zero.

author: bfritz
author_rank: Regular Contributor
timestamp: March 11, 2012, 08:05:57 pm
content: Quote from: MrPlacid on March 11, 2012, 06:44:30 amIt is important to realize that the direction of the arrows does not denote the solution for direction of current flow. When doing a problem like this, you just pick a direction for current, and then write the equations. Then you solve the equations. You may find that some currents are negative... which means the current flows opposite the direction you chose the arrow.Since that node shows all currents flowing out of the node, you can simply write 0=i5+i4+i1. If the arrow for i5 was rotated 180 degrees, since the rule is that all currents flowing into (or out) of a node is zero, you would need to change the sign of the i5 term... in other words 0=-i5+i4+i1.So, the direction of current is blindly chosen, and the math ends up showing you the actual direction in the circuit. Does that help?

author: slateraptor
author_rank: Frequent Contributor
timestamp: March 12, 2012, 02:57:43 am
content: Quote from: bfritz on March 11, 2012, 08:05:57 pmTo add to bfritz's comment above, current directions in mesh-current analysis are typically chosen to simplify arithmetic in the same way as a ground node is chosen for node-voltage analysis; the choice is arbitrary, but making the "right" choice can help to reduce the amount of actual arithmetic performed.

author: MrPlacid
author_rank: Frequent Contributor
timestamp: March 12, 2012, 04:58:05 am
content: Thanks for all the help guys.I can see the importance of mesh analysis in simplifying things. It looks like one of those neat electronic tips. You know I have been in electronics for so long and yesterday was the first time I've heard of mesh current analysis.

author: IanB
author_rank: Super Contributor
timestamp: March 12, 2012, 05:35:11 am
content: Quote from: MrPlacid on March 12, 2012, 04:58:05 amI think the importance is more relevant to computer analysis than to human analysis. As a human you would tend to take simplifying steps such as applying Thévenin's theorem to the problem. In the general case when you end up with a bunch of simultaneous equations you wouldn't try to solve the equations by hand. The theory is useful to the theoretician to understand what's going on, but not so useful to the practitioner when a quick answer is needed.