Help with Kirchoff's laws - Page 1
Forum section: electronics / beginners

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 04:40:27 am
content: Ok everyone, I'm hoping you can shed some light on this subject! The following 2 problems are part of my Physics 202 lab. I'm not looking for anyone to do the work and answer it for me, but I am hoping someone can guide me and my lab partners a bit more! Unfortunately our book only gives one example of a much simpler circuit consisting of 2 loops, with 2 batteries and 3 resistors. The 3rd resistor splits the 2 batteries and is shared between the 2 loops. Somewhat similar to this:http://ccphysics.us/henriques/p202lab/KirchhoffLoops.jpgexcept the resistors are all on the vertical legs, and the batteries on the bottom horizontal legs. (I realize electrically speaking, it's the exact same thing)Now we're to do two problems, as shown on the lab pages I scanned in. The resistor values are the same for both of the circuits and given on the table on the 2nd page. The + signs indicate the positive voltage side of the resistor, which we used that indication for hooking up the meter to measure the amps. The 3rd page is some stuff we worked out by hand, but likely with errors on it. After seeming to not make much sense of the first one, we decided to try the extension circuit. I'll include a picture of what we worked out on that problem on the whiteboard, but we really didn't seem to make a whole lot of progress on that one either.We're able to work the simple ones like the books examples show, and in class our demonstration circuits were kept simple to a maximum of 2 resistors per loop and 3 resistors for the whole system. But these extra resistors seem to really be kicking us in the rear.Now on the extension, we tried working it two ways, seeing that R1 and R2 are in essence a series pair, we tried to combine them to be I(12). This will be evident on the picture I'll post.I've watched some youtube videos and those make sense, but trying to work these two circuits seem rather convoluted and are quite honestly make me and my lab partner rather mad at the amount of time they're taking.Can anyone PLEASE offer some advice? Our book leaves much to be desired in this department. Our thought on the extension was perhaps that we should go after finding the center resistors current first since it's a shared part of both the left and the right loops. Is that reasonable logic? Or is there no logic? Or what should the logic be?In general the steps appear to be:1. Write the junction equations2. Write the loop equations3. Mathematical voodo magicA huge thank-you to anyone who can provide some guidance. Fortunately we've started on this earlier and it's not due until Friday afternoon (~ 39 hours from now depending one where and what day it is where you're reading it). However we've got over 3 hours into this and have made essentially zero progress so far.Collectively this seems like one of the larger and more helpful forums I've come across, so I'm hopeful for some direction. And I must say I'm hooked on Dave's videos!Thanks! Picture soon to follow (have to boot my laptop up for the SD reader)Chris

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 04:59:55 am
content: You recognize that circuit 7.1a does not appear to be the same as circuit 7.1b?

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 05:01:52 am
content: Here's a picture we took of the whiteboard for the extension problem. The sectioned off right half is where we tried to solve it with R1 and R2 combined. We could manage to work both until we ended up with 2 variables in the equations, but then we were sort of stuck there.http://airplanesathome.com/images//img_0759.jpgThanks!

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 05:06:37 am
content: Wow, you are getting into a bit of a pickle there. The extension problem is actually easier than the first problem, but it takes a certain amount of insight to see the easy solution...

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 05:10:58 am
content: You understand that you can analyze a circuit usingeitherKirchhoff's current laworKirchhoff's voltage law, and you can pick whichever is simpler to apply? Also that you can always use Ohm's law to express currents in terms of voltage differences, or voltage differences in terms of currents?For example, if you assume the battery terminal voltage is known then Figure 7.1b has two unknown voltages at points C and D. So you can write two simultaneous equations involving those voltages and solve them. Once you have the voltages the currents through each resistor fall out easily.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 05:13:23 am
content: While I would say that 7.1a and 7.1b are both rather non-standard styles of circuit schematic design, they are the same I think. The extra wires used in 7.1a are just because the lab board doesn't have a diamond layout shape that we could utilize. In both instances however, resistor pairs R1-R2 and R3-R4 are both bisected by R5, with R1 and R3 both connected to battery + and R2 and R4 connected to battery -. Any extra resistance from the 2 short jumper wires is considered negligible and ignored.But perhaps I'm missing something? Care to expand on your reasoning that they're different?? I can see where if the schematic were re-drawn with horizontal/vertical lines it would be 2 square loops side by side. But the battery is still thrown in there almost as an afterthought sort of.

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 05:16:12 am
content: In Figure 7.1a it appears as if R1 is connected between points A and B, whereas in Fig. 7.1b R1 is connected between points A and C. Similarly R5 goes between B and D in the first figure, but it goes between C and D in the second figure. I agree the circuits have the same topology, but the locations of the numbered resistors are different (or maybe we could just say that the positions of nodes B and C are swapped). I guess that doesn't hurt as long as you ignore the letters.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 05:26:57 am
content: Quote from: IanB on March 07, 2013, 05:10:58 amUhh... I guess not? Like I said, our book does a horrible job of stating this stuff. Here, I'll re-write what the book says.. This is "Kirchoff's Laws" section in it's entirety, plus 1 quick example that doesn't show hardly any math on how to solve a simple one as I described.Kirchoff's LawsCharges moving through a circuit gain energy and lose energy in resistors. If we go completely around a circuit, thechangesin every energy per unit charges - that is, increases and decreases in voltage - sum to zero. This isKirchoff's loop law, and holds for any closed loop even if it's part of a more complex circuit:The sum of voltage changes around a closed loop is zero.The loop law is essentially a statement of energy conservation for circuits.In analyzing parallel resisotrs, we saw that the current flowing into point A in Fig 25.10 (just a battery with 2 parallel resistors) had to equal the current flowing out. That's because charge is conserved, and in a steady state charge can't be accumulating anywhere in a circuit. A junction of three or more wires is anode. If we count the currents flowing into a node as positive and those flowing out as negative, then we can stateKirchoff's node law: The sum of currents at any node is zeroI can see how that there's 2 unknown voltages at the two points C and D though... but not sure how I'm supposed to apply kirchoff's "loop law" of voltage then to find just those 2 points?

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 05:27:45 am
content: I think the main problem you have is you are forgetting that the resistors play a crucial part in how the circuit behaves. If you are not writing Ohm's law for each resistor in the equations you are trying to solve you will be running round in circles. You can't solve just currents or just voltages and not include the way the resistors relate the two.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 05:37:57 am
content: I think I'm getting even more lost here!I get that V=IR, I've used that equation for years. But neither the sole book example, nor the quick example in class made use of ohm's law until after we'd solved for the currents, then used the known current and resistance to calculate what the voltage across the resistor should be. And that was only done in my professor's quick example, the book doesn't even go that far with it, it simply solves for the currents.

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 05:39:10 am
content: Quote from: swooshcmk on March 07, 2013, 05:26:57 amSo, for instance let's look at Fig 7.1b.The current law at node C says:I1+ I5- I2= 0Now, for instance,I1= (VA- VC) / R1That allows you to eliminate I1. Proceed in the same way and you can eliminate all the other currents.Do you see where that takes you?

author: Harvs
author_rank: Super Contributor
timestamp: March 07, 2013, 05:41:07 am
content: The problem that I see in the way you are approaching this, is your method of attacking the maths.Looks like you've realised that I4=I5 and I1 = I2.Now with the equations you've got, you've got enough information to solve. You just need to refactor your equations to get a single equation that has only one variable.It's telling that you keep missing the =0 off your equations, that is you're not re-arranging your equations to try to solve for something.Pick one of the variables, say I1, and one of the equations. Then refactor and substitute the other equations until your left with I1 = a constant. Then same thing again but obviously it gets easier each time.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 05:52:06 am
content: Quote from: IanB on March 07, 2013, 05:39:10 amI think I see where that takes it, but then you would have to use the known voltage across the resistor to solve it. I believe the purpose of the lab is to start with nothing but known battery voltages and known resistor values and then calculate what the voltages and currents are supposed to be. The table of data is for us to later compare our calculated vs measured values. I'm not sure how I could find delta V for R1 to use that equation off hand.I agree that we've been somewhat shoddy at writing in the = 0 at the ends. This was towards the end of working at it for nearly 3 hours and we decided to write something down/take a picture and then go try and get some help. But even so, I guess we were missing how/where to factor out and substitute new values. We still don't to be honest.

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 05:54:17 am
content: Quote from: Harvs on March 07, 2013, 05:41:07 amYou can't solve a network using only currents with no other parameters included. The currents in the various branches of the circuit depend on the applied voltage(s) and the resistances of the various paths. Any solution to determine the currents must involve those other quantities.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 05:56:30 am
content: But isn't that why we substitute the current laws into the loop laws? Isn't that in effect expressing the voltages in terms of currents I guess?edit:Sorry, meant to say "isn't that why we substitute the current (node) formulas into the loop (voltage) formulas"...

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 06:03:07 am
content: A spot check of your measurements passes muster, which is good.For instance at node C:I1+ I5- I2= 03.1 + 1.7 - 4.8 = 0And around the loop ACDA:V1- V5- V3= 03.151 - 1.635 - 1.516 = 0Good measuring

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 06:04:42 am
content: Quote from: swooshcmk on March 07, 2013, 05:56:30 amNo, you choose either current balances around nodes, or voltage sums around loops, and then substitute Ohm's law into either.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 06:17:37 am
content: LOL Hey at least we got that part of it right!I didn't like how the lab's test lead probes made the readings fluctuate so much, so I brought in my own set of meter probes and alligator clips to improve our measurement qualityDo you have any known good online resources for this topic that I can go through? Perhaps one that shows an example problem step by step for something more similar to what these two problems are? I obviously am missing something and need to start at square 1 again apparently. I'm game for a thorough youtube video as well. We were under the impression that you started with the node and loop formulas and from there played around with it by substituting, multiplying/adding formulas until you managed to cancel it down to a single current variable and solved it. But we could never get it to work out that way. Perhaps the simple examples we were shown in class etc that worked on (1 ohm, 2 ohm and 4 ohm resistors) and this requires an extra level of understanding we just weren't shown?A huge thank you for taking your time with me. I'm still lost at it, but at least I know there's more to it than just playing around with the numbers until it worked!

author: Harvs
author_rank: Super Contributor
timestamp: March 07, 2013, 06:45:32 am
content: You just need to get your maths right. This is an essential skill to learn for future studies (they will already expect you to be able to do this.)Attached is my take at your equations to solve for I3 (equ 1, 2 & 3 are yours from the top of the board.) Hopefully this is enough to demonstrate the technique without giving away the whole result.

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 06:46:29 am
content: Sorry, I don't know of any resources I can point you to. But I can give you a few more hints about Figure 7.1b and see if anything starts to make sense.If you want to analyze a circuit you will need to have some known pieces of information that you begin with.In the case of this circuit, you might start by knowing the battery voltage and the resistance values. Then you can calculate the unknown voltages and currents in the circuit in terms of these known quantities.A good way to start is to assign zero volts to some point in the circuit and express the other voltages relative to these. We could say the battery negative VBis the zero reference for example. Then, since we know the battery voltage, we immediately know VA:VA- VB= VBATTThat leaves us with only two voltages we don't know: VCand VDPoints C and D are at nodes in the circuit where we can apply Kirchhoff's current law. Conveniently, this will provide two equations which can be solved for the two voltages. All that remains is to write out the two equations and solve them.For example, at node C we have this equation:I1+ I5- I2= 0A similar equation can be written at node D involving I3, I5and I4.Once you have these equations, Ohm's law will do the rest. The problem is solved.

author: Harvs
author_rank: Super Contributor
timestamp: March 07, 2013, 06:50:13 am
content: BTW just one more point. With equations like this, be very careful with rounding, you've got big numbers minus big numbers leaving very small numbers. Where possible I like to leave things as whole numbers right until then end, then I'll just put the whole lot into a modern scientific calc and avoid transcribed rounding errors.

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 06:52:01 am
content: Incidentally, all this work is being done to get a good understanding of the principles involved. You should definitely keep at it until you get what is going on. Later on some elegant and easier techniques will be presented for solving this kind of problem that will reduce all the complicated algebra. But Ohm's law will still be sitting there right in the middle of it all.

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 06:57:37 am
content: Quote from: Harvs on March 07, 2013, 06:45:32 amI'm not sure what you are trying to do there? That is to say, what are your givens and what are your unknowns you are trying to find?I can't make sense of what destination you are heading towards.

author: The Electrician
author_rank: Frequent Contributor
timestamp: March 07, 2013, 06:59:55 am
content: One big problem you've got going here is that you have too many unknowns. The currents I1 and I2 are the same. The currents I4 and I5 are the same. Don't do that!! Don't give different names to the same current! That's asking for trouble; you will end up with too many unknowns and not enough equations. For the direction you've shown I3 it is equal to I2-I1; you don't need I3 anyway. You only need two loop currents to solve this problem.The current in the left hand loop should be labeled I1, and the current in the right loop should be labeled I2.You have adopted the common convention of choosing your currents to flow in a clockwise direction around the loops.Now, the voltage across R1 is just I1*R1 volts, and the voltage across R2 is I1*R2 volts. If you're traversing the left loop and writing a KVL (kirchoff's voltage law) equation for that loop, the current in R3 will be I1-I2, so the voltage across R3 will be (I1-I2)*R3 volts; this changes polarity when you're traversing the right hand loop.Starting at the point labeled "B" at the middle bottom and proceeding clockwise around the loop, the voltage drops you encounter will be -1.5 volts at the battery, 981*I1 volts at R1, 6*I1 volts at R2, and 326*(I1-I2) volts at R3. According to KVL those voltages have to add up to zero.So, your equation for the left hand loop will be:-1.5 + 981*I1 + 6*I1 + 326*(I1-I2) = 0You can rearrange this equation to be:(981+6+326)*I1 - (326)*I2 = 1.5or:1313*I1 - 326*I2 = 1.5Now, see if you can come up with the equation for the right hand loop with current I2.I see that 4 more posts have been added while I was working on this.The two standard methods for solving networks like this are the nodal method and the loop (or mesh) method. IanB is describing the nodal method, while I'm giving you hints about the loop method. It's your choice.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 07:06:37 am
content: Interesting! I very much appreciate the help from both of you guys!Harvs - This is exactly the process we were going at. How did you make it look so easy?! lol I think the mistake we were making is we could get it down to 2 variables, but we weren't separating it to be a specific current = (formula). I guess we were more expecting to be able to cancel out the 2nd term. I see it really just entails another level of solving and substituting?IanB - The method he's after (as far as I understand from the examples I'd seen) is that you can then use your known value to plug into other formulas and keep knocking out the other currents. In class we would find all of the currents in this method, then apply ohms law to get the voltage across each resistor.I get ohm's law. I get the principles driving kirchoff's laws. I admit that math is not my strong point. I've always been more visual, and I've always had a harder time connecting the dots mathematically. The math is easier if I can "see it" in application, but i have to connect the dots first. The connections can be what takes me some time to make so I can "see" it. For example - I worked my butt off for a C in calculus 1. I've gotten straight A's in Chemistry, Physics 1, and so far in Physics 2.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 07:14:53 am
content: Perhaps it can be summed up somewhat as this:IanB is solving primarily using the node methodThe Electrician is describing the loop/voltage methodHarvs is combining the node and loop methods together to solve it by substitution and factoring.I guess the main thing is I didn't realize that the node and voltage methods were *different* ways to attack it. I was under the impression that the node and loop methods had to be combined and work together as Harvs has done, which is a somewhat more obscure method to attack it?

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 07:33:04 am
content: Quote from: swooshcmk on March 07, 2013, 07:06:37 amWell, yes, you can do that, but it helps if you can apply your visual insight to see over the foothills and get to the mountain.Let's try it (Fig. 7.1) the hard way. We'll use I1for the current through R1, V1for the voltage difference across R1, and IBATTfor the current through the battery.So we have:IBATT= I1+ I3I1+ I4= I5I3= I5+ I4I2+ I6= IBATTThat gives us 4 equations involving 6 variables, so there are two too many degrees of freedom to solve them. Two more equations are needed.We could try two voltage loop equations involving the battery:VBATT= V3+ V4VBATT= V1+ V2But oh dear, we have introduced four more variables (we know VBATTso that isn't included).All is not lost though, since we can write another four equations using Ohm's law:V1= I1R1V2= I2R2V3= I3R3V4= I4R4So we have now in total 10 (!) equations involving the following 10 variables: I1, I2, I3, I4, I5, IBATT, V1, V2, V3, V4We could solve these equations by a process of rearrangement, substitution and elimination.But doesn't that seem like a lot of hard work compared to the method I outlined above?I can't say what you are being expected to do, but I assume you are in college here, and you need to be able to see ways to cut problems down to size and solve them in the most efficient way. Writing 10 equations in 10 variables and then laboriously simplifying them down is not that way.

author: IanB
author_rank: Super Contributor
timestamp: March 07, 2013, 07:41:24 am
content: Quote from: swooshcmk on March 07, 2013, 07:14:53 amYes! This.Look at the problem and choose whichever way appears simpler.In our example there are four node voltages and six branch currents. Four is smaller than six, so go with nodes. By inspection, two node voltages are already known (the battery), so only two unknown voltages remain. Write two independent equations for these nodes (KCL) and you're done.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 07:53:28 am
content: A HUGE Thank You! guys!Yes, we were headed down the 10 variable route, and yes it was looking quite daunting to tackle! Realizing that that these laws can be used in combination with (and not just prior to) Ohm's Law is a huge step in understanding. Also, realizing you can approach it from the node method, the loop method, or the combined node/loop method helps as well.Time to get a fresh sheet of paper and give it another try!

author: The Electrician
author_rank: Frequent Contributor
timestamp: March 07, 2013, 08:56:56 am
content: I just realized thet IanB has been talking about a solution to the figure 7.1 circuit, whereas my advice was directed to the whiteboard circuit.The loop method can also be used with figure 7.1, of course.

author: swooshcmk
author_rank: Contributor
timestamp: March 07, 2013, 10:53:53 am
content: No problem, I caught that right away as you said "on the board". I've got the one we were trying to solve on the board all figured out now - it makes complete sense, the values came out to what I measured (or quite close as we didn't measure each individual battery, so we just used 1.5v for each), but the 7.1 circuit is still kicking my butt. There's got to be a better way to do this as IanB was referring to, in order to avoid all the substitution (it gets quite messy quite quick). But when I tried writing out all the I's as V/R form, I didn't really see anything to tip me off on a better method either.I see that I1 = (Va-Vc)/R1 and that I2 = (Vc-Vb)/R2, where Va = the battery voltage, and vb = 0V (batt -) and it looks like you can use those and the battery voltage to solve for Vc, except R5 throws a wrench in it all with it being I5 = (Vd-Vc)/R5. So then we're back to 2 unknowns...What am I missing? This thing is pure evil for being the first assigned kirchoff's diagram to solve. I need more bread crumbs to follow..

author: Harvs
author_rank: Super Contributor
timestamp: March 07, 2013, 11:01:43 am
content: I wasn't putting down anyone's methods by the way. Just the OP has been instructed on a methodology to use, so your best chance of passing the course is to learn the methods that have been taught. I just took the work they've already done, and finished a portion of it.I use to tutor this back in the day to first year EE students. This was taught when they first came in, their understanding of electronics was at the level that some couldn't tell which end of the battery to stick on their tongue (no offence to the OP BTW, I'm not referring to you). So, they'd be taught a methodical process (KVL and KCL circuit analysis), which they used to get a set of equations that were sufficient to fully describe the circuit. From there, they could apply high school algebra which is a pre-requisite for Oz EE courses. Then from there, they'd hit the labs and maybe start to understand what was going on.So at the stage of teaching this stuff, no intuitive understanding of what's going on was assumed at all.But what that meant also for a student, is by being able to correctly label nodes and currents, then build the correct equations, was > 50% of the marks.

author: The Electrician
author_rank: Frequent Contributor
timestamp: March 07, 2013, 12:05:21 pm
content: Ordinarily when you use the loop method, each one of your equations is based on KVL, and when you use the nodal method, each of your equations is based on KCL.When using the nodal method, you add up all the currents going into a node and set the sum to zero. If the sources are current sources, then it's easy to add their current to the other currents into a node because the current coming from a current source is known--it's just the value of the current source. But when you have a voltage source, you doin't know what the current coming from it is (until you solve the network), so it isn't always as straightforward to deal with a voltage source when using the nodal method (well, it really isn't too hard once you learn what to do).In your 7.1 problem, you only have two unknowns--the voltages at nodes D and C; you're letting node B be your reference, so you don't need to solve for the voltage there--it's zero by definition, and node A is the battery voltage.You need two equations to solve for two unknowns. Write a KCL equation for node C. The current in the three resistors is calculated as three fractions and their sum is set equal to zero.When you're working on the equation for a particular node, assume that the currents flowing away from that node are positive. This means that you assume the voltage at the node you're working is greater than all the other nodes, so the numerator of your fraction will be of the form (Vn-Vx), where Vn is the working node, and Vx is another node that a resistor is connected to. In other words, your working node comes first in the expression in parentheses.So, we have:(Vc-Va)/R1 + (Vc-Vb)/R2 + (Vc-Vd)/R5 = 0You should be able to get the equation for node D in similar fashion. But, wait, you'll say. We have 4 unknowns but only two equations! Not a problem. Note that Va and Vb are not unknown. Va is the battery voltage, and Vb is the reference voltage (zero, that is). Substitute those values into the two equations, and you're left with two equations in two unknowns which should be easy to solve. Of course, you also have to substitute the actual numerical values of the resistors into the equations.Hint: the second equation can start out with:(Vd-Va)/R3 +......

author: Harvs
author_rank: Super Contributor
timestamp: March 07, 2013, 02:01:59 pm
content: As suggested by The Electrician, you're probably better off looking for other equations.But, the equations that you did get do actually describe the circuit. Except for you got ADCB wrong. You've gone around the wrong legs in something like ACBD which clearly isn't what you want.If you've got everything, then you can just hit it with the brute force method of solving the equations. I know people are going to say it's the really hard way of doing it. But it's not really that hard once you're use to solving these types of problems, and it can be a lifesaver in exams when you've got something completely new that you don't know how to approach. Just go back to first principles, get enough equations to describe the system, then bash through them.Keep in mind that my working out is very long hand. With a standard graphics calculator these days and a competent operator it can just about do everything for you.Again, attached is a partial working out using your equations except for the one you got wrong.

author: swooshcmk
author_rank: Contributor
timestamp: March 08, 2013, 04:12:43 am
content: I just wanted to say a HUGE thank you to all who helped! It's making MUCH more sense than when I first posted. I should be pretty well set to go for the homework we will be getting assigned on it on likely Monday or Tuesday. Unfortunately our labs are currently just over a week ahead of our lectures, so we've been getting some work to solve in lab that we've not officially covered in the lectures, which can make things a bit rougher to solve. But you guys have definitely cleared up some misunderstandings I'd had, and have certainly helped me to be able to visualize the problems better! At least for now I can rest easy because I've got the lab assignment done now, and I've still got 16 hours before it's due.Plus I got news today that the problem with my oscilloscope was found (it broke 3-4 hours after I got it, used Tek 2465 from e-bay, horizontal display driver went out) and it should be repaired by tomorrow and getting calibrated. Overall a good day now!I owe you guys a round!

author: Online Simulations
author_rank: Newbie
timestamp: March 11, 2013, 12:21:09 pm
content: You don't need mathematician magic, once when you write your equation you can calculate them online in:http://www.cirvirlab.com/simulation/kirch_2_real.phpTell me if it was useful