The question asks to find "v", the voltage in the 4ohm resistor. could you take a look at the circuit found here http://i966.photobucket.com/albums/ae143/df4sale/mesh/mesh_zpsd5f5dffe.jpg have i marked it up correctly? and am I right in saying the following; mesh 1 = (5Ω+2Ω) Imesh1 - (2Ω) Imesh2 mesh 2 = (2Ω+1Ω) Imesh2 - (2Ω) Imesh1 mesh 3 = (4Ω) Imesh3 If this is correct what is the next step in solving the question? thanks in advance!
It's been a while since doing a mesh analysis, but my impression is that you are going to need at least one more mesh equation to solve the problem. You have 4 unknowns in the circuit which will require 4 equations to solve. Try to think of one more current mesh that you can write and you can then solve for the unkonwn currents.
What is "mesh 1", the quantity on the left hand side of your first equation. Given the contents of the right hand side, whatever it is has to have units of volts, but I have no idea what it is refering to. Remember, mesh equations are nothing more than somewhat formalized ways of writing down Kirchhoff's Voltage Law around a loop. Now, your figure shows three mesh currents but you have four meshes! So define Imesh4 to the the clockwise current in that fourth mesh. Having done that, what is v in terms of the resistances and mesh currents? Having done that, three of the four mesh currents you can solve for by inspection. Which three?
Does the question require you to use mesh analysis? Nodal analysis would only require 3 equations, and would give the voltage across the 4 ohm resistor directly.
True, but would you rather solve three equations in three unknowns in which all three are truly unknowns or would you rather solve four equations in four unknowns when three of the equations are trivial? In both cases, you end up having to solve one equation to get the answer, so it is somewhat six for one and half a dozen for the other.
Speaking for myself, the solution of a set of linear equations is never the stumbling factor because I do these problems on a calculator that has built-in matrix algebra capability. The time-consuming part of the problem (compared to the time required for solving) is the setting up of the equations. I don't know for how long the instructors require students to perform the solution by hand, and when they allow the student to use the many resources for solving linear equations. When I have taught individuals how to solve networks, I feel that the conceptual understanding was of paramount importance, and after a few times solving the equations by hand, I want the student to use a solver. This way, I can assign harder problems, and the algebra is left to a calculator or computer. You said in another thread to Jony130 "Remember, nodal analysis is just KCL, so in general you want to avoid nodal analysis when there are voltage sources for which you will have to get currents for. Problems like this are much better dealt with using mesh current analysis. " In a similar vein, mesh analysis is to be avoided when there are current sources involved. I've noticed that beginners who post these sorts of problems seem to have conceptual problems when doing nodal analysis with voltage sources, and mesh analysis with current sources. So, if there is no requirement to use a particular method, I suggest to them to avoid those things. I'm not sure what you mean by this. If a nodal analysis is done for this problem, the voltage across the 4 ohm resistor will be one of the solution elements; no additional equation need be solved. But, after a mesh solution, the current through the 4 ohm resistor results, and one more (trivial) equation must be solved, namely V(4) = I3 amps*4 ohms. Is this what you meant by saying "...you end up having to solve one equation to get the answer..."?
In general I agree. In this particular case, I don't see the set-up using nodal being easier than the setup using mesh, though. You only have to set up one equation and it is an extremely simple one. The other three (the three that are marked in the original diagram) truly are by inspection. Assuming that Jony130 has already worked it out, the first three are I1 = 4A I2 = 15A I3 = 5A and the fourth is simply: I4*(10Ω) = I1*(5Ω) + I2*(1Ω) + I3*(4Ω) I don't need a solver or even a calculator to come up with an answer for I4 of 5.5A and hence a current (top to bottom) in the 4Ω resistor of 0.5A and a voltage across it of 2V. I think setting up the three nodal equations is considerably more involved (not saying it's hard) and more likely to let errors slip through. But I also am more than willing to admit that this problem is somewhat of an extreme example. Additionally, I tend to look for things like this because I have almost never used equation solvers even in "real work", but I worked in a rather lunatic fringe environment in which each engineer had to wear many, many hats. As a result, the amount of time that would elapse between needing to do a particular task was frequently long enough that you had to come back up to speed on the tools needed to do the job. For stuff like solving equations, it was almost always quicker to solve the equations by hand than to find a calculator or software tool that could do it and then refresh your memory on the steps involved. To be sure, anything that I was doing on a sufficiently regular basis I would stay sufficiently up to speed on the best way to do it (given the tools at hand). The only problem I have with that is that you have to be careful to avoid having them start letting the tool do their thinking for them. There is a strong tendency to accept whatever the tool spits out as the answer and to stop asking if the answer makes sense. When doing things by hand, most people learn by experience to make that an ongoing part of the process. Not to say that it is inevitable that people forget how to estimate answers and do sanity checks if they use solvers and such, just that you need to be sure to take steps to prevent it otherwise it will tend to naturally move in that direction. Yes, and I almost made that exact point at that time and decided not to. In both cases, of course, there are exceptions. When I have current sources in exterior branches (or branches that I can make exterior) I want to use mesh. Similarly, if I can choose a reference node that gives me voltage sources tied between the reference and other nodes I want to use nodal. The classic "never say never" situation.... In this case, you've got three current sources, but they are all in exterior branches and therefore all three are trivial solutions to one of the mesh currents. Same here, except that in the lower level courses I think it is important to give them plenty of experience with these types of problems with the requirement that they use the techniques that typically evoke conceptual issues specifcially to help get them over those problems. I want them to avoid using certain techniques in certain situation because those techniques lead down more difficult paths to the solution, not because they don't understand or know how to walk that path. If we are looking for the voltage across the 4Ω resistor (which I think is the case, here) then yes, you have one non-trivial equation to solve in both cases and, using mesh, you actually have two pretty trivial equations to do after that. If the question were looking for the current in that resistor, then you have one non-trivial and one pretty trvial equation in either method.
thanks for the quick replies guys. In reply to "The Electrician" the question asks to use Mesh, Nodal and Thevenin's theorem to calculate the valve of V in the 4Ω resistor. Having looked at my obvious mistake a new "Mesh" diagram has been produced with 4 meshes to represent the for unknowns. http://i966.photobucket.com/albums/ae143/df4sale/mesh/mesh2_zps0f5bb9d3.jpg Therefore, Mesh 1 (5Ω+2 Ω)I1 (2Ω)I2 (5Ω)I4 Mesh 2 (2Ω+1Ω)I2 (2Ω)I1 (1Ω)I4 Mesh 3 (4 Ω)I3 (4Ω)I4 Mesh 4 -(4Ω)I3 (1Ω)I2 (5Ω)I1 + (5Ω+4Ω+1Ω)I4 = -5I1 I2 4I3 +10I4 Substituting in the known valves gives; 0 = (-5*4)-(15)-(4*5)+10I4 0=-20-15-20+10I4 10I4=55 I4=55/10 = 5.5A Therefore, V=4(I4-I3) V=4(5.5-5) V=4(0.5) V=2V Having worked this out and seeing WBahn's reply I'm going to take an educated guess that this is now correct?
Carrying on from this question to calculate the voltage V using nodal analysis. I am not getting the value of 2V which I should be. Please can someone point out where I am going wrong? thanks Circuit diagram: http://i966.photobucket.com/albums/ae143/df4sale/mesh/nodal_zpsa7cfdd9a.jpg At each node; Node a 0 = (Va-Vb)/2 + (Va/2) + (Va-Vc)/5 Node B -15 + 4 + (Vb-Va)/2) = 0 Node C 5 – 4 + (Vc-Va)/5 + (Vc/4) = 0 Are these correct? If so I must be calculating the next part wrong.
These are NOT equations. They are each, at best, one side of an equation. An equation has to set two things equal to each other (hence the name), hence each of the above expressions must be set equal to something. In the above cases, that something is NOT zero! Remember that mesh equations are nothing but KVL and so you have to take into account ALL the voltage changes around the entire mesh. You have completely ignore the voltage across the current sources in each mesh. Again, not an equation, but just one side of one. And track your units. Don't drop them and then tack something back on at the end as an afterthought. Now a 0 pops up that wasn't there before, finally making it an equation. You luck out because, for this mesh, their are no sources to ignore and hence the other side of the equation is, by coincidence, equal to zero. And since you are plugging in the "known values" you are bypassing your prior, incorrect, mesh "equations" and so the fact that they are wrong goes uncaught. This last part is correct (or would be, if you had tracked your units properly).
Thanks, must have been a typo. So next I have tried to solve the three equations for the value of Vc, which will represent the negative value of "v" in the 4ohm resistor. Node A 0 = (Va-Vb)/2 + (Va/1) + (Va-Vc)/5 Multiplying by 10 gives: 0=5Va-5Vb+10Va+2Va-2Vc (Call this equation 1) Node B -15 + 4 + (Vb-Va)/2) = 0 Simplifying gives: -9+(Vb-Va)/2 = 0 which goes to: Vb=18+Va (Call this equation 2) Node C 5 4 + (Vc-Va)/5 + (Vc/4) = 0 Simplifying gives: 1 + (Vc-Va)/5 + (Vc/4) = 0 (Call this equation 3) Sub equation 2 into equation 1 Va=(Vc+45)/6 (Call this equation 4) Sub equation 4 into 3 gives an answer of Vc=-55 ?! This may seem trivial but I've been trying to solve this for hours now
Why would Vc be the negative of the voltage across the 4Ω resistor? How can we tell how or why you got a particular wrong answer when you don't show the work that led to that answer? Consider the following, starting with your work and then putting in the units and cleaning up a bracing typo: ------------ Node A ------------ 0 = (Va-Vb)/2Ω + (Va/1Ω) + (Va-Vc)/5Ω Multiplying by 10Ω gives: 0 = 5Va - 5Vb + 10Va + 2Va - 2Vc (Call this equation 1) ------------ Node B ------------ -15A + 4A + (Vb-Va)/2Ω = 0 Simplifying gives: -11A + (Vb-Va)/2Ω = 0 which goes to: Vb = 22V + Va (Call this equation 2) ------------ Node C ------------ 5A – 4A + (Vc-Va)/5Ω + (Vc/4Ω) = 0 Simplifying gives: 1A + (Vc-Va)/5Ω + (Vc/4Ω) = 0 (Call this equation 3) ------------ Sub equation 2 into equation 1 ------------ 0 = 5Va - 5Vb + 10Va + 2Va - 2Vc (EQN 1) 0 = 17Va - 5Vb - 2Vc Vb = 22V + Va (EQN 2) 0 = 17Va - 5(22V + Va) - 2Vc 0 = 17Va - 110V - 5Va - 2Vc 0 = 12Va - 110V - 2Vc 0 = 6Va - 55V - Vc Va = (Vc + 55V)/6 (EQN 4) ------------ Sub equation 4 into 3 ------------ 1A + (Vc-Va)/5Ω + (Vc/4Ω) = 0 (EQN3) Multiply by 20Ω 20V + 4Vc - 4Va + 5Vc = 0 9Vc = 4Va - 20V 9Vc = 4[(Vc + 55V)/6] - 20V 54Vc = 4Vc + 220V - 120V 50Vc = 100V Vc = 2V Now, consider that this is NOT my homework, yet I found the time and the energy to track my units and to show my work. If I can do it, why can't you when it IS your homework? If you show your work, then when you discover that an error exists, it makes it much simpler to track down where the error occurred and to walk the changes forward. Perhaps that would have saved you some of those hours, meaning that perhaps the additional minutes it would have taken to show the steps would have been time well spent. Just suggesting the possibility. It also helps the grader discover where you made your mistake and whether it was a simply math blunder (which it was in this case) or something conceptual (which you also had in this case) and assign partial credit accordingly.
From starring at the problem for too long without a break and so resulted in making stupid mistakes... *sigh* found the mistake, thank you for being pedantic WBahn. thanks the electrician also! Before I try the next thevenin method Im going to have a lie down in a dark room....
I was actually just pondering the first question and, unfortunately, the answer is not an affirmative to the second. No, I have plenty of other things I should be doing, one of which is actually rather pressing (as in, I have to have it done in the next two hours). But tracking units and showing the steps is simply how I work, so if I hadn't typed it into the post, I would still have had to write it down on a piece of paper. Since the answer to this problem is already known, I saw it as an opportunity to lead by example -- though I would have organized my work a bit differently if I hadn't been deliberately trying to mirror the progression used by the OP.
I'm not sure how this remark was intended, since "pedantic" has several meanings at least one of which ("meticulous") is of a neutral tone, though almost all others, such as "overly concerned with formal rules and trivial points of learning" are derogatory. If you mean the former, then I happily plead guilty. If you mean the latter, then about all I can do is point out that people can and do die, sometimes in job lots, when engineers can't be bothered with the trivial points of their chosen profession.