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Passive sign convention.

  1. Sep 25, 2009 #1
    1. The problem statement, all variables and given/known data

    1) Q. Which of the elements follow the passive sign convention? (psc)

    [PLAIN]http://img401.imageshack.us/img401/3092/pic1r.gif [Broken]

    v1 = +5 (V)
    i1 = +2 (mA)

    v2 = -10 (V)
    i2 = +500 (µA)

    v3 = +1 (V)
    i3 = -3 (mA)


    2) Q. Which of the elements follow the passive sign convention?

    [PLAIN]http://img401.imageshack.us/img401/3092/pic1r.gif [Broken]

    v1 = -5 (V)
    i1 = +2 (mA)

    v2 = +10 (V)
    i2 = +500 (µA)

    v3 = +1 (V)
    i3 = +3 (mA)





    2. The attempt at a solution

    For number 1, I said elements 1 and 3 follow the psc, because the current is moving from the + end to the negative end.

    And number 3 is same as number 1, except the +/- signs change for the number 3. I feel like you have to do something like KCL or KVL to figure out the actual flow of those currents. After figuring out the actual flow of the current (the reference marks may be right or wrong), then you sort of determine which elements follow the psc.

    So I'm not sure if my initial stab at the problem is actually right( which I highly doubt ><). and if isn't right, then I don't know how else to determine the psc for these elements.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 25, 2009 #2

    Redbelly98

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    Something is wrong here. From KCL we should have i1+i2=i3, but that is not the case. From KVL we should have v2=v3, but that isn't the case either. It is impossible for this circuit to have these values; were they part of the problem statement or did you determine their values somehow?
     
    Last edited by a moderator: May 4, 2017
  4. Sep 25, 2009 #3
    "if the values of voltages + current are as follow, then which elements follow psc", So, I'm not sure if that word "IF" changes stuff. I was working out the KVL and KCL, and the equations weren't working out.
     
  5. Sep 26, 2009 #4

    Redbelly98

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    Okay, it looks like we are to ignore KVL and KCL ... for which I'm critical of your teacher or textbook or whoever came up with this problem. But that's not your fault.

    Let's look at one element, #3 in problem 1. In what direction (up or down) does the current flow? Hint: don't just go by the direction the arrow is pointing, consider the given value of the current.
     
  6. Sep 26, 2009 #5
    hm... for element three, using P=VI, you get -3, I'm guessing it goes down.
     
  7. Sep 26, 2009 #6

    Redbelly98

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    Can you show in detail how you got that result? What is the value of P for element three?
     
  8. Sep 26, 2009 #7
    For element 3, P=VI
    using that given values for element 3, [1(v)][-3(mA)] = -3(mW)
     
  9. Sep 26, 2009 #8

    Redbelly98

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    Your responses show an inconsistency, or something, so I would like your help in clearing things up before proceeding.

    Note: in the above post, I am asking about the current I.

    Note: in the above post you imply that, in order to calculate I, you already knew what V and P were, and could use V and P to figure out what that I=-3 mA.

    This is puzzling, because from your first post of the problem statement it looked like I=-3 mA was a given quantity. But if that were true, you wouldn't need to use P=VI to calculate I; you would just quote the given value.

    Note: here I am puzzled as to how you knew what P was before I. Perhaps I misunderstood your original Post #1, and those V and I values were actually the result of some calculation you had done, using information that you did not post.

    Ah, but here you imply that you did actually know what I was before you knew P, contradicting the earlier statement that P=VI was used to figure out what I was.

    -----

    At this point, perhaps we should just backup to where I asked the following question:

    Element #3, Problem 1: in what direction (up or down) does the current flow? And, how do you know?
     
  10. Sep 26, 2009 #9
    ahhh, I'm so confused..alright.
    The reference marks for i3 points down, so based on the reference mark, the current is flowing down.
     
  11. Sep 26, 2009 #10

    Redbelly98

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    The reference arrow defines the current flow when the current is positive. However, the current does not necessarily flow in the direction of the reference arrow.
     
  12. Sep 27, 2009 #11
    so in #1, element 1 follow the psc, since the current is positive, while element 2+3 don't follow the psc, since the current are not positives?
     
  13. Sep 27, 2009 #12

    Redbelly98

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    Not quite, you haven't accounted for whether the voltages are positive or negative. It looks like we should review what it means for a circuit element to follow psc.

    For an element that follows psc (like a resistor, for example), the current flows from a higher potential to a lower one within the device.

    So, if we have:
    +[→]-
    for some device, then we can think of 4 possibilities:

    1. V>0, I>0:
    The potential is higher on the LEFT (+ terminal), and current flows to the RIGHT.
    Since the current flow is toward a lower potential, the device follows psc.

    2. V<0, I>0:
    The potential is higher on the RIGHT. (Since V<0, the - terminal is actually at a higher potential.)
    Current flows to the RIGHT towards a higher potential, so the device DOES NOT follow psc.

    3. V>0, I<0
    Potential is higher on the LEFT (+ terminal).
    The current also flows to the LEFT (since I<0, the current is in the opposite direction as the arrow).
    Since the current flows towards a higher potential, it DOES NOT follow psc.

    4. V<0, I<0.
    This is left as an exercise for the reader.

    So the key is, for each element:
    1. Figure out which terminal is at a higher potential
    2. Figure out in which direction the current flows
    3. If current flows toward a lower potential within the device, it follows psc.
     
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