Mesh Currents and Node Voltages

Discussion in 'Homework Help' started by missxlian, Jul 10, 2012.

  1. missxlian

    Thread Starter New Member

    Jul 10, 2012

    Could someone guide me with some steps to solve this problem? I'm having trouble doing this with both the node voltage and mesh equations

    For node voltages I got the equations:
    vb - vc = vs
    va( 1/96 + 1/32) - vb( 1/96 + 1/120) - vc(1/32) + vo = 0
    is - (va - vc)/32 - (va-vb)/96 = 0

    I don't know where to go from there..
  2. panic mode

    Senior Member

    Oct 10, 2011
    how did you get second equation? if the numbers are resistances, this s not dimensionally correct either (Vo is voltage, all other terms are current).

    i'd say
    Vc=Vo; Vb=Vc+Vs;
    Is=Vb/120 + Vc/30
    Is=(Va-Vb)96 + (Va-Vc)/32
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    What may not be immediately obvious is that the 96Ω & 32Ω resistors play no role whatsoever in the final outcome.

    I suggest the simplest method of solving this problem would be through the use of superposition.
  4. WBahn


    Mar 31, 2012
    You can't go anywhere from there because, without even looking at the problem, I can tell you that the second equation is wrong (I am assuming that the '96' and the '32' are resistance values).

    va( 1/96Ω + 1/32Ω) - vb( 1/96Ω + 1/120Ω) - vc(1/32Ω) + vo = 0

    The first three terms are currents (voltage/resistance). The last term is just voltage. You can add them. But, because you don't track your units, you missed this and would have gone right on plugging numbers into equations and come up with an answer and, perhaps, then tacked some units onto the final answer.

    Units tracking is perhaps the single most effective way to catch errors as they are made. Most errors (not all) will impact the units, but you can only catch them if the units are there to be impacted and if you are actually paying attention to them.

    If you start tracking your units consistently and aggressively, you will probably see not only a marked improvement in your grades, but will be far less likely to kill a bunch of people once you are doing this in the real world.

    I'll examine your problem in a subsequent post (probably later today).
  5. WBahn


    Mar 31, 2012
    I would tend to agree that it is not immediately obvious. I haven't worked the problem yet, but I will be interested to determine if this is true (though I imagine you are correct) and, more important, why it is true.

    I would tend to agree. Anytime I see both voltage and current sources in the same problem, I keep superposition close at hand. But the OP has no say in the matter -- they are specifically told to solve it twice, once using mesh and once using nodal analysis. I fully recommend doing superposition as a check and to get a feel for when each of the three has the advantage.
  6. WBahn


    Mar 31, 2012
    Point #1: Don't make people guess what you mean by va, vb, and vc. Mark the nodes on the diagram with the labels you will use in your work. See the attached figure.

    You node voltage equations should end up in the form:

    ( )va + ( )vb + ( )vc = ( )

    Where each '( )' has some constant inside.

    Remember that Nodal Analysis is, primarily, just a very formalized application of KCL to a circuit. Having said that, the use of KCL in Nodal Analysis is a means to an end. The goal is to develop a set of N simultaneous equations that relate N node voltages to each other.

    You've recognized this (correctly) in your first equation by noting that

    Vb - Vc = 0

    In your third equation, you state:

    is - (va - vc)/32 - (va-vb)/96 = 0

    The first term is the the current flowing into the node from the current source, the second term is the current flowing out of the node from through the 32Ω resistor (and since you are subtracting this term, you do want the current flowing out of the node) and the third term is the current flowing out of the node through the 96Ω resistor. So this is fine.

    Now, what about that second equation? You state:

    va( 1/96 + 1/32) - vb( 1/96 + 1/120) - vc(1/32) + vo = 0

    Rearranging this (and adding the units) you would have:

    (va - vb)/96Ω + (va - vc)/32Ω + (0V - vb)/120Ω + vo = 0

    So the first term is the current flowing out of node 'a' through the 96Ω resistor and the second term is the current flowing out of node 'a' through the 32Ω resistor. But the third term is the current clowing into node 'b' through the 120Ω resistor. Why is that there? And the forth term is just a voltage. Why is it there?

    Are you trying to group nodes 'b' and 'c' into a supernode? I think that is the case and, if so, you came very close. In fact, the only mistake is precisely the one that the units error was screaming about. Each of the first three terms represent currents into the supernode through the 96Ω, 32Ω, and 120Ω resistors respectively. Now you just need the current into the supernode through the 30Ω resistor.

    Note that you have 'vo' and you have 'vc'. Thus you have four voltages and three equations. Your forth needs to relate vo to the node voltages.
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    I sometimes find re-casting the topology view is helpful...
    mlog likes this.
  8. WBahn


    Mar 31, 2012
    Yes, those are quite helpful recastings.

    Since I haven't used it for well over a decade, I had actually completely forgetten about the trick of duplicating supplies and then splitting the circuit accordingly. Thanks for reminding me -- I'm sure there have been times when that trick would have come in handy and hopefully I won't forget it again before the next time comes around.

    After solving the circuit, it became apparent how to 'talk away' the influence of the top two resistors based on viewing the problem from the perspective of Superposition.

    Now, here is an interesting extra credit question:

    Since the values of the 96Ω and 32Ω do not impact the value of vo, does this mean that we make them arbitrarily large (removing them in the limit) or that we can make them arbitrarily small (shorting them in the limit). If not, why not, since they exactly cancel out from the equations when solving for vo?
  9. mlog


    Feb 11, 2012
    In regards to the R1 = 32 Ω and R2 = 96 Ω resistors, it looks to me like you may have the following combinations (all of the units are Ω).

    R1 = 0
    R2 = 0
    R1 = ∞
    R2 = ∞
    R1 = 0, R2 = ∞
    R1 = ∞, R2 = 0

    You may NOT have the following combinations (all of the units are Ω).

    R1 = 0, R2 = 0
    R1 = ∞, R2 = ∞

    The problem with the 1st prohibited combination is that when both resistors are shorted, it nullifies Vs.

    The problem with the 2nd prohibited combination is that when borth resistors are opened, it nullifies Is.

    Since Vs and Is are essential to the transfer function, neither Vs nor Is may be nullified.
  10. WBahn


    Mar 31, 2012
    Yep. So why didn't this show up in the math? There should have been a step at which, in order to have the equations be valid, we should have had to place those conditions on it.

    The answer, of course, is that the math did tell us these things, we just tend to ignore them and move on with churning the algebra and only bother to go back and look when and if we get bit.