Help Finding Dependent Current in a One Port

Discussion in 'Homework Help' started by Oschmid, Oct 26, 2014.

  1. Oschmid

    Thread Starter New Member

    Oct 26, 2014
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    0
    Hi, I really need help solving this problem. What I have done so far is label the node between the 2 and 8 ohm resistors v3 and then use nodal analysis as follows.

    v2 = 2*v1

    (v2)/4 + (v2-v3)/2 = 0

    (v3-v2)/2 + (v3-6)/8 + (v3)/2 = 0

    Which results in v2= 12/19 which is not the correct answer. Can someone please tell me what I am doing wrong and show me how to get the right answer? Thanks!

    [​IMG]
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,726
    4,788
    This appears to be a perfect example of what I am always preaching about -- namely the need to track units properly -- although it is hard to tell because all your equations are a mash.

    There is no point in labeling the node between the 2 Ω resistor and the 8 Ω resistor V3 (regardless of which 2 Ω resistor you happened to be talking about) because it is already labeled V1.

    Your first equation appears to be based on the dependent current source (I'm assuming that's where you got the 2*V1 from), but that is a CURRENT source and V2 is a voltage -- you can't just set them equal.

    The coefficient on the dependent voltage source MUST be such that when you multiply it by a voltage you get a current, meaning it must have units of amps/volt. Here we have to make assumptions, which is never a good thing, that it is 2 A/V (as apposed to 2 mA/V or 2 A/mV or something else entirely). If you had had the proper units, you would have ended up with an equation that stated:

    V2 = (2 A/V)*V1

    which, upon inspection, gives you volts on the left and amps on the right and you would have KNOWN it was wrong before you ever went any further. Even if you hadn't caught it then, your answers would have ended up with the wrong units and that would have told you that they were wrong.

    For your second equation, you only took into account the current going downward from node V2 through the 4 Ω resistor and the current going to the right from node V2 through the 2 Ω resistor. What about the current going downward from node V2 through the current source?

    So take a step back and try again. Also, try annotating your set-up equations so that it is easier for us (and the grader) to follow your work. For instance, your final equation should be written something like:

    Node V1: (V1-V2)/2Ω + (V1-6V)/8Ω + (V1)/2Ω = 0
     
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,387
    497
    Your first mistake was creating v3.

    Equation for v1
    (v1-v2)/2+(v1-6)/8+v1/2=0

    Equation for v2
    (v2-v1)/2+v2/4+2v1=0

    Two unknown (v1, v2), two simultaneous equations. That is all you need to solve the thing.


    I got:
    v1=6/17=18/51
    v2=-12/17=-36/51
     
  4. Oschmid

    Thread Starter New Member

    Oct 26, 2014
    7
    0
    Hi WBahn and shteii, thank you for your help! I just revisited this and solved it. I got confused with the labeling of V3 because I wasn't sure about where the voltage would go across the strip of wire between where I labeled V3 and V1. This was a handwritten problem from my professor and so I also thought that the V1 at the end of the one port and the V1 that is part of the current source were different because his hand writing changed. I will definitely start keeping the units on my work from now on because I can see how that would be a useful check.
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,726
    4,788
    It's key to realize that the voltage anywhere on a node is the same (as long as the size of the node is much, much smaller than the wavelength of the signal). The current in different parts of the node may not be, but the voltage is.

    Please tell me your instructor didn't write that... particularly writing the answers without any units.

    Tracking units is very probably THE most single most effective error detection tool available to an engineer and, as far as I'm concerned, not using it consistently and religiously amounts to gross negligence. Most mistakes we make -- and we all make them on a regular basis -- will screw up the units. The things that we do affect lives and while a doctor that messes up can kill someone, they are generally limited to killing one patient at a time; when an engineer makes a mistake, they can and do kill people in job lots. The other highly effect error detection tool is always asking if the answer makes sense and, as part of that, always estimating what the answer should be (whenever practical, which it often isn't) and comparing that to the answer that is arrived at.
     
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