Basic circuit analysis - Page 1
Forum section: electronics / beginners

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 02:20:46 am
content: Hello, everyone!I have a question regarding the circuit attached below.Can I choose any point to place the ground (reference)?And also, is there a voltage difference between the current source's terminals? If there is not, then I would place the reference at the point I show in the image.I've also tried to use the theory of Supernode but there is a resistance in parallel with the voltage source and I don't know how to work it put.Thanks in advance!Edit:I think I solved it, I would really appreciate if you could check if what I've done is correct. Thank you very much!!!

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 02:53:53 am
content: Here are the images, I was not able to upload them on the first message.

author: IanB
author_rank: Super Contributor
timestamp: September 10, 2017, 03:06:45 am
content: Just remember that voltages are differences in potential between two points in a circuit. Therefore you can choose any point in the circuit as your reference and describe all voltages as differences from that point.You may choose any point in your circuit as the reference, and therefore you can choose the point that makes the calculations most convenient. Whichever point you choose your answers must be the same, or you have done something wrong. This could be a way of checking your calculations.

author: Ian.M
author_rank: Super Contributor
timestamp: September 10, 2017, 03:21:17 am
content: A resistance in parallel with an ideal voltage source or in series with an ideal current source has no effect on the rest of the circuit and can be eliminated. The current through it (parallel) or voltage across it (series) is only dependant on the resistance and the source.The problem therefore simplifies to two resistors and two sources, which is then very easy to solve.

author: Ratch
author_rank: Regular Contributor
timestamp: September 10, 2017, 03:51:15 am
content: Nexo,Some ground nodes make the calculations easier that others. I would suggest you use the bottom line node where the current feeds into the current source.Of course a current source has voltage across it. How could it force current without voltage?Your sheet of calculations are unreadable. The focus is bleary and there is hardly any contrast to the drawing.Why is that cap in the drawing if it just shorts itself at a node?What are you trying to calculate? You never did say.I calculated all the voltages and currents. Which one did you want to check.Ratch

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 04:20:54 am
content: Quote from: Ian.M on September 10, 2017, 03:21:17 amThanks for your answer. When you say "A resistance in parallel with an ideal voltage source or in series with an ideal current source has no effect on the rest of the circuit and can be eliminated." you mean that it won't affect the circuit it self and I can just take it out of the picture and make the calculations with the other elements? That's a great tip, thanks!Ratch,Thanks for your reply!I did use the bottom line and the calculation was pretty easy."Of course a current source has voltage across it. How could it force current without voltage? "True, my bad."Your sheet of calculations are unreadable. The focus is bleary and there is hardly any contrast to the drawing."Sorry, my phone is not a good one."Why is that cap in the drawing if it just shorts itself at a node?"The idea behind all these exercises is to rewrite the circuits following these rules: if there is a cap, it acts as an open circuit so no current will flow and if there is an inductance, it has no voltage difference between its terminals so we can redraw it as a line."What are you trying to calculate? You never did say."True, my bad again. We are asked to find the voltage and current in each element."I calculated all the voltages and currents. Which one did you want to check."Well, I want to check all the elements, here is a, hopefully, better picture.

author: Ian.M
author_rank: Super Contributor
timestamp: September 10, 2017, 04:28:00 am
content: Quote from: Nexo on September 10, 2017, 04:20:54 amYes. Remove the parallel resistance and short out the series resistance. That of course means that the voltage across the current source in the replacement circuit will be that across the series resistor + the source in the original circuit. and similarly for the current through the voltage source. However, once you've solved the rest of the circuit, you can reinstate the resistors you eliminated and complete the solution.

author: IanB
author_rank: Super Contributor
timestamp: September 10, 2017, 04:35:27 am
content: If the voltage sourceV= 15 V, then the voltage acrossR2must be 15 V, must it not?

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 04:41:46 am
content: You're totally right, I copied incorrectly. The voltage is 15V. What about the other elements?Thank you'll for being this patient!

author: Ratch
author_rank: Regular Contributor
timestamp: September 10, 2017, 04:51:06 am
content: Quote from: Ian.M on September 10, 2017, 03:21:17 amThe current load on the voltage source and the voltage of the current source will increase if resistors are placed in parallel and series respectively. This can cause different calculation values to occur if the voltage source feeds into a node or a current source is part of a loop. I can show you what I mean after the OP finds the correct answer.Ratch

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 04:59:22 am
content: I attached a picture that has the voltage and current for each element, could you compare it to your answers, please? Thanks in advance!

author: Ratch
author_rank: Regular Contributor
timestamp: September 10, 2017, 05:04:21 am
content: Quote from: Nexo on September 10, 2017, 04:59:22 amLooks OK to two significant digits. What is the current load of the voltage source?Ratch

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 05:10:57 am
content: Sorry, I don't understand the question

author: Ratch
author_rank: Regular Contributor
timestamp: September 10, 2017, 05:13:55 am
content: Quote from: Nexo on September 10, 2017, 05:10:57 amThe 15 volt source supplies a finite value of current. What is the current value?Ratch

author: IanB
author_rank: Super Contributor
timestamp: September 10, 2017, 05:17:35 am
content: Quote from: Nexo on September 10, 2017, 04:41:46 amWhen you record the voltages across the resistors you should take care to indicate which side of the resistor is more positive. (Equivalently, show the assumed direction of the current through each resistor with an arrow, and then note the voltage with respect to this flow direction. Is the voltage +27.5 V or -27.5 V?)

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 05:21:50 am
content: Oh! Is 4.3 A, the same amount that flows through the resistor R4.Quote from: IanB on September 10, 2017, 05:17:35 amGood tip. Is +27.5V, well, I got +27V.Do you have a simpler way to do it? Or you use supernodes and KCL, as well?Thanks!

author: IanB
author_rank: Super Contributor
timestamp: September 10, 2017, 05:27:39 am
content: Quote from: Nexo on September 10, 2017, 05:21:50 amBut what about the current flowing through R2. Doesn't that current also come from the voltage source?By the way, if you keep your intermediate results to full precision, 4.25 A instead of 4.3 A, it will help with checking your results at the end.QuoteIs it really 27 V? For instance, all the voltages around the loop consisting of V, R4 and R3 should sum to zero. But 15 + 43 + 27 = 85 V, which is not zero. So one or more voltages are presumably negative when going around that loop?

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 05:43:35 am
content: I was mistaken, when I corrected the value from 16V to 15V I didn't correct the calculations and that's why I was getting 27V when in fact I should get 27.5V.My question now is the next one: Looking at the picture, are the arrows placed correctly? I rounded in a light blue box the ones that seem odd. Hope you can help me!How would a KCL look at the node where R1, R2, R3 and V meet?Thanks a lot, again.

author: IanB
author_rank: Super Contributor
timestamp: September 10, 2017, 06:07:57 am
content: You want to have just one current arrow for each path in the circuit. It doesn't matter which way the arrow points as long as you obey the sign of the arrow when you do the calculations. Essentially you can draw one arrow next to each resistor and one arrow next to the voltage source. The current source already has an arrow.In particular, you cannot have two opposing arrows on R3 as you have shown--you have to designate a current direction and be consistent with it in all the current balances.When I solved the network I only needed one arrow for the unknown current through R4 and one for R3. I know the current through R1 is 7 A, and the current through R2 is 1.5 A. I also used Ian.M's trick of temporarily removing R1 and R2 from the circuit when solving the rest of it.

author: Ian.M
author_rank: Super Contributor
timestamp: September 10, 2017, 06:14:33 am
content: Are you sure the circuit you posted is the one you actually need to solve? I've redrawn it to make it easier to put current directions on without getting them muddled up and it seems a bit odd that the capacitor is shorted out.Unsolved LTspice sim attached - try to resist using the sim to solve it until you are sure you understand the manual solution.

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 06:49:49 am
content: Quote from: IanB on September 10, 2017, 06:07:57 amThat was exactly what I've been thinking, it makes no sense to have two opposites arrows in the same element within one loop, it can happen when we have 2 loops and the current from one loop might be opposed to the current for the same element in the other loop.Thank you!Quote from: Ian.M on September 10, 2017, 06:14:33 amPretty sure, what the teacher is doing is to give us these circuits to practice for the exam and he places caps and inductors and we have to redraw the circuit having in mind that a cap basically means an open circuit and no current (so we can just delete it, although we are asked to find the voltage in it) and that an inductor means basically a short with no voltage drop so we can just redraw it using a line instead of it.

author: sibeen
author_rank: Frequent Contributor
timestamp: September 10, 2017, 08:09:09 am
content: Using the diagram provided by Ian, this is a very basic circuit to analyze as it effectively only has one node.I have called the node at the top of R3 = N1. Node 2, at the top of R4 is a supernode with N1. N2 = N1+15.R1 can be discounted as there is 7A flowing into N1. R2 can be rubbed out and C1 can be rubbed out. The Nodal equation then becomes:(N1/R3) +7 + ((N1+15)/R4) = 0Using figures provided of Rx = 10 we then have:(N1/10) +7 + ((N1+15)/10) = 0N1 = -42.5, therefore N2 = -27.5

author: Assafl
author_rank: Frequent Contributor
timestamp: September 10, 2017, 09:42:23 am
content: Quote from: sibeen on September 10, 2017, 08:09:09 amYour equations look okay. But are the signs okay?The sum of all currents coming in to the node should be zero. Wouldn't the 7A leg be a negative current direction (since the current is flowing out - and not into the node)?So V(n1)/10 + (V(n1)+15)/10 + (-7) = 0?(Stated another way, a rough qualitative check is that if indeed V(n1) and V(n2) are both negative, that means the current at all three legs that meet at the reference point are flowing out - that obviously cannot be...)

author: sibeen
author_rank: Frequent Contributor
timestamp: September 10, 2017, 11:27:28 am
content: YepShould be (N1/10) - 7 + ((N1+15)/10) = 0N1 = 27.5.D'oh.

author: Assafl
author_rank: Frequent Contributor
timestamp: September 10, 2017, 01:51:04 pm
content: Quote from: sibeen on September 10, 2017, 11:27:28 amYup. Nexo got the right answer (he got 27V, but in his calculation he used 16v for the battery - not 15v).

author: Ian.M
author_rank: Super Contributor
timestamp: September 10, 2017, 02:23:10 pm
content: Solved LTspice sim attached, with node voltage and component current .op labels (run the sim to update them) + manually drawn current arrows. I had to reverse R2 from the previous version to get the current direction consistent. I wont post a graphic till Nexo confirms he's happy with his solution.

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 07:09:04 pm
content: Yep, I'm happy with the solution, numerally it all adds up. I just have one question:If I use KLC at that node after using a supernode, then I'll get the answer we've all gotten but if I use KLC before using the supernode, then I'll have to analyze the current flowing through R2, as well (the resistance in parallel with the voltage source). So should I always use KLC AFTER I used supernodes, right?

author: IanB
author_rank: Super Contributor
timestamp: September 10, 2017, 07:29:16 pm
content: There are many different ways you can solve such a problem. The art is in spotting the way that minimizes the calculations you have to do. I solved the problem using two simultaneous equations. But then sibeen came along and solved it with one equation.If you cross out R2 and solve the problem using a supernode without R2 you will get exactly the same answer for the voltages. Then you can put R2 back in and add its contribution to the current through the voltage source. It's up to you how you do it, as long as you keep in mind what you are doing as you go along.

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 07:40:36 pm
content: I see. So, if I want, I can use KCL before using the supernode and I will have to take into account the current that flows through the resistance in parallel with the voltage source as well as the current that will flow (coming from the current source) through the voltage source? If so, then the calculations would be more difficult so, based on your suggestion, is better to use KCL at that same node but AFTER I used the supernode (with this I mean, after I shorted the voltage source) because no current will flow through R2 (the one that is in parallel with the voltage source). Right?

author: IanB
author_rank: Super Contributor
timestamp: September 10, 2017, 07:50:26 pm
content: It's not that no current will flow through R2 (it will), but rather you can delete R1 and R2 from the circuit and solve the simpler circuit that remains. If you look at sibeen's post, that's what he did. Notice he does not have R1 or R2 in his equation. He effectively crossed them out on the circuit diagram. We know that the voltage at node R2/R4 is +15 V greater than the voltage at node R1/R2/R3, and adding or removing R2 doesn't change that. So we may as well remove R2 before we do any calculations.

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 08:34:11 pm
content: I see, thanks!I have another circuit, should I upload it here to ask some things or should I create another post?

author: IanB
author_rank: Super Contributor
timestamp: September 10, 2017, 08:44:58 pm
content: You may as well continue the thread if it is a continuation of the same general topic.

author: Ian.M
author_rank: Super Contributor
timestamp: September 10, 2017, 08:50:32 pm
content: Ok, Here's the walk-through of one approach to the manual solution, illustrated by LTspice. If you don't have a high-def monitor, you'll probably want to open the image full size in another tab.I'm not going to slavishly calculate every component current or node voltage by hand - I'll let LTspice do the heavy lifting, but at each stage, I have simplified it till the manual calculations would be trivial.Step #:Draw circuit, clarifying what nodes there are and assign the reference node. From left to right, we have node 1 (I1--R1), node 2 (where R1..R3 meet) and node 3 (where the useless capacitor is attached), and our reference node Gnd is at the bottom..Simplify it by eliminating resistors in series with current sources or in parallel with voltage sources. R1 and R2 drop out..Prepare to apply the Principle of Superposition: Reduce V1 to 0 and solve remaining circuit..Restore V1, reduce I1 to zero and solve remaining circuit..Restore I1, reconnect R1 and R2 to their sources, disconnect them from the rest of the circuit and short the I1-R1 series pair. Solve for V(node 1) and I(R2) which is also -I(V1).Superimpose node voltages and component currents from step #3 and step #4 - just add them up. Also at this stage add in from step #5, V(node 1), I(R2) and I(V1).Job done with noorand hopefully no.

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 09:32:16 pm
content: Wow! Thank you very much! I'll study it.Now I have the next circuit.Is it okay what I've done? I don't analyze the part within the red square because the voltage at the node where R2 and the inductor meet is the same as the negative terminal of V3 so it is basically the same node B and there is no potential difference between the terminals of R4, thus, no current flow through it. Am I right?I'll take a picture of what I've done and the answers I've gotten.

author: Assafl
author_rank: Frequent Contributor
timestamp: September 10, 2017, 09:42:15 pm
content: Yes. Theoretical inductors are short circuits for D.C. Analysis. So getting rid of the resistor and inductor and replacing them with a short will make it easier to compute.

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 09:47:46 pm
content: Thanks! And what about the voltage source V3? Can I get rid of it as well?

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 10:04:39 pm
content: QuoteBut for me is.I don't understand where those 35V and 7.5V come from in steps 3 and 4. Could you please explain it?

author: sibeen
author_rank: Frequent Contributor
timestamp: September 10, 2017, 10:11:02 pm
content: Yes, V3 is shorted out by the inductor, so get rid of it.

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 10:13:18 pm
content: Quote from: sibeen on September 10, 2017, 10:11:02 pmThanks!Is this correct?

author: sibeen
author_rank: Frequent Contributor
timestamp: September 10, 2017, 10:23:05 pm
content: Looks like you've made an error somewhere.As IanB has stated there are many ways you can solve each circuit and the trick is to try and find the easiest solution for each circuit topology. In this case we do have a supernode so nodal analysis seems the best option with node N1 being at your node A.Hint: There is only ONE node that needs to be calculated in this circuit.

author: Ian.M
author_rank: Super Contributor
timestamp: September 10, 2017, 10:30:08 pm
content: Quote from: Nexo on September 10, 2017, 10:04:39 pmIts the Principle of Superposition - it lets you remove (or reduce to zero) all except ONE source in a linear circuit, solve for that source, then repeat for all the other sources and simply add up the results.See:https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-dc-circuit-analysis/a/ee-superpositionor:https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/superposition-theorem/or if you refer a more hands-on approach:Hopefully one of them will 'click' for you.N.B. Power in DC circuits is NOT linearly superimposable - you *must* do all power calculations after combining all the voltage and current results

author: Nexo
author_rank: Regular Contributor
timestamp: September 10, 2017, 10:38:33 pm
content: Exactly! What I've done was to use a supernode at V3 as well and that was my question because it is shorted so I didn't know if i could just get rid of it because I was afraid it would affect the circuit itself because there is a current coming out of R2 (produced by V3) and there is a current coming out of V1, and these two current will meet at the node I call B and I have no idea what happens there with the current, will the electrons fight until the largest charge win or something like that? I'm lost in this part because how would I be able to find out the current flowing through the inductor, will it be the current produced by V1 or V3?Thanks in advance!

author: Ian.M
author_rank: Super Contributor
timestamp: September 10, 2017, 10:43:43 pm
content: V3 isn't shorted by the inductor but the series combo R2--V3 is, so that section can be deleted from the main circuit and treated as a completely separate problem.The rest of it is easy to solve by superposition. For both sources it reduces to a three resistor series/parallel problem which can be solved by inspection: simply replace the parallel resistors by a single one of R/2 (noting the current will need to be halved to get the individual currents) which gives you a potential divider with a 1/3:2/3 ratio. However for academic credit in a test or paper, its better to solve it formally.If you need to solve for the current through the inductor, superposition is your friend.

author: sibeen
author_rank: Frequent Contributor
timestamp: September 10, 2017, 11:06:31 pm
content: Nexo, redraw the circuit taking out anything that isn't required. See attached.

author: IanB
author_rank: Super Contributor
timestamp: September 11, 2017, 01:05:55 am
content: Quote from: Nexo on September 10, 2017, 09:47:46 pmHopefully you have already realized this, but no. The inductor may short out R4, but V3 and R2 are still there and form a separate island. When you want to know the total current through the inductor this part of the circuit is important. You can ignore R2 and V3 when you solve the part of the circuit on the left, but you must come back and add them in at the end to complete the solution.

author: sibeen
author_rank: Frequent Contributor
timestamp: September 11, 2017, 02:33:23 am
content: I may have added some confusionNexo, if you use the diagram I provided in my last post you can easily work out the voltage at the node N1 at the top of R3.(N1-10)/5 + N1/5 +(N1-15)/5 = 0N1 = 8.333VSo the node at the negative terminal of the 15V battery = 8.333 - 15 = -6.667VThe current through R5 is -6.667/5 = -1.333A.The current through R2 is simply V3/R2 as the inductor is a short, 17/5 = 3.4A.You now put the two results together. You have two currents through the inductor, 3.4A flowing top to bottom and 1.333A flowing bottom to top. The resultant current through the inductor is therefore 3.4 - 1.333 = 2.067A.

author: Nexo
author_rank: Regular Contributor
timestamp: September 11, 2017, 03:59:23 am
content: Quote from: sibeen on September 11, 2017, 02:33:23 amNow we are talking, that was exactly what I didn't know how to do. Thank you very much!!!Continuing with the basic circuit analysis, I now present to you the next one:I'm asked to prove (demostrar means prove in Spanish) the following:

author: Ian.M
author_rank: Super Contributor
timestamp: September 11, 2017, 04:27:30 am
content: Ouch! that's some heavy physics if you need to formally take it any deeper than stating Kirchhoff's Voltage Law (KVL).I'm having difficulty even conceptualizing a circuit that violates KVL (in the absence of a time-varying magnetic field, and without any flow of charge external to the circuit), but I *think* that if the equation you have been asked to prove (and by extension, KVL) wasn't true it would permit circuits that were perpetual motion machines of the first kind, continuously producing heat from the flow of current though resistive elements of the circuit, without any energy input to maintain the flow. That violates the First Law of Thermodynamics: The Law of Conservation of Energy.

author: Nexo
author_rank: Regular Contributor
timestamp: September 11, 2017, 05:49:54 am
content: I think I understand you.I asked the teacher and he said that I should use KCL at the node that joins R1 and Rx but even doing that I can't come close to that equation, so I'm blind here.

author: IanB
author_rank: Super Contributor
timestamp: September 11, 2017, 06:05:35 am
content: Quote from: Nexo on September 11, 2017, 05:49:54 amLet the current in the loop be I. Now express I in terms of Vc, R1 and Vx. Lastly eliminate I by noting that VR1 = I x R1. You should get what you need to find.

author: Assafl
author_rank: Frequent Contributor
timestamp: September 11, 2017, 06:32:33 am
content: Quote from: IanB on September 11, 2017, 06:05:35 amSince it is DC analysis that is the approach you beat me to.However, one would probably be safe showing how the equation work in the case of, say, a capacitor or an inductor. So how this also works when the R approaches infinity (charged capacitor) and when it approaches zero (inductor happy with its magnetic field) (or open / short circuit).Yet another case is what if X is a voltage/current source (with/without a resistor). Even then it should work (albeit Thevenin would have made it easier...).Perhaps a more general approach would be to let v(x)=f(I).

author: Nexo
author_rank: Regular Contributor
timestamp: September 11, 2017, 06:51:37 am
content: I did it! Thank you very much you all.Now, let me show you the last one, I was able to solve all of them and this one is giving me troubles:I did it by solving the equations I have from KCL (solving for each unknown variable; first method) and the second method is to use matrixes (Kramer rule) but I get one answer with the first method and completely different one with the second method, so my questions are these:1. Which answer is correct?a) Va = 1.11v, Vb = -9.77v, Vc = -20.65v (first method)b) Va = 6v, Vb = 0v, Vc = -6v (second method)2. Which method do you prefer and why?Thanks!

author: sibeen
author_rank: Frequent Contributor
timestamp: September 11, 2017, 07:53:04 am
content: Knowing the value of V1 and V2 would possible be of some assistanceAlso, why did you put the reference node (earth) where you did?

author: Nexo
author_rank: Regular Contributor
timestamp: September 11, 2017, 08:15:21 am
content: My bad, v1 = vb = 12v. The reference point was already given in the problem.

author: sibeen
author_rank: Frequent Contributor
timestamp: September 11, 2017, 09:29:17 am
content: Nexo, neither of your results are correct.For this problem I'd go and use KVL (mesh analysis). Work out what the current flows are and then use some simple maths.

author: Nexo
author_rank: Regular Contributor
timestamp: September 11, 2017, 11:47:51 am
content: We can only use KCL

author: sibeen
author_rank: Frequent Contributor
timestamp: September 11, 2017, 12:55:40 pm
content: Just simplify the circuit to a 24V supply, a 7.52k resistor (10K||30K) in series with a 560k. Work out the voltage across the 7.52k and you can then just use the voltage divider equations to work out the individual nodes. Then just add 12 volts to each result to give you the result as a reference to the battery mid point.

author: Ian.M
author_rank: Super Contributor
timestamp: September 11, 2017, 02:52:42 pm
content: That will work to check your results,(if you don't goof and come up with something ridiculous like 7.52 instead of 7.5, and I get Vc=11.68V to 2 d.p. so any solution that differs significantly or has Va or Vb outside the range Vc to +12V is bad*)but to do it purely with KCL (+ Ohm's Law), you have to solve it by nodal analysis, without first simplifying the circuit.As you have tried to solve the same set of equations by two methods and got conflicting results, neither of which match the check method you'll need to show your workings.*I know I haven't had a major brain fart and goofed up a simple potential divider calculation <paranoid> because LTspice .op analysis gives:V(c): 11.6828 voltage ,</paranoid>

author: Nexo
author_rank: Regular Contributor
timestamp: September 12, 2017, 06:52:47 pm
content: Give me an hour, I'll try to solve it again and post my results.EDIT:This are my answers:Va = 11.894Vb = 11.789Vc = 11.683At least Vc is pretty close to your answer.

author: Nexo
author_rank: Regular Contributor
timestamp: September 12, 2017, 10:21:08 pm
content: Quote from: IanB on September 11, 2017, 06:05:35 amI did this exercise but I didn't do it using KCL, any idea on how to do it using it?

author: sibeen
author_rank: Frequent Contributor
timestamp: September 12, 2017, 11:35:16 pm
content: Quote from: Nexo on September 12, 2017, 06:52:47 pmThey are correct

author: Nexo
author_rank: Regular Contributor
timestamp: October 05, 2017, 02:15:09 pm
content: Thank you all!