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Direction/sign of current

  1. Sep 8, 2016 #1
    1. I know this is really simple but I'm missing some little piece of the puzzle

    No problem getting to vab=
    • +1.5V for a)
    • +1.5V for b)i.
    • -2.5V vor b)ii.
    • +0.5V for b)iii.
    upload_2016-9-9_4-8-23.png

    But what about current I?

    2. I=V/R

    3. so for IR2
    • b)i. I have 1.5V/1kΩ=1.5mA (sure I agreee with that)
    • b)ii. I have -2.5V/1kΩ=-2.5mA (hmmm)
    • b)iii. I have 0.5V/1kΩ=0.5mA
    does b)ii. make sense? If the voltage is reversed to -5V does the current also reverse? What if I just chose to draw the arrow for current the other way round? How do I account for that when analysing the circuit?

    Really simple I know, but it's doing my head in!
     

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  2. jcsd
  3. Sep 8, 2016 #2

    Bystander

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    What do you know about reverse current through diodes?
     
  4. Sep 8, 2016 #3
    Only that as far as this introductory course is concerned, it doesn't exist!

    I'm just trying to understand how to determine if the current through the resistor (#2) has a positive or negative value. Does it always have to be positive when entering the positive reference of the resistor?
     
  5. Sep 8, 2016 #4

    CWatters

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    So if no current is flowing through the reverse biased diodes would anything change if you removed them?
     
  6. Sep 8, 2016 #5
    No not in the case of V(in)=-5V. They're effectively an open circuit. What I'm trying to understand is if the voltage is negative so therefore V(R2) is negative (using the polarity references I have drawn in) does it then follow that I (being V/R) =-2.5V/1k-ohm in the direction given, or does some other convention apply?

    Or If I flipped the 'I' arrow around, how would that be reflected in the relevant formulas?

    Or to put it another way, can a reference direction of current be arbitrarily assigned or is it determined by other variables?
     
  7. Sep 9, 2016 #6

    NascentOxygen

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    On paper you can assign or assume any direction for the current for your calculations. But if your calculations give a negative current, then when relating that to the hardware you would say current is in the direction opposite to your assumed direction.

    In easy circuits where it is obvious which direction the current goes, it makes sense to assume that obvious direction right from the beginning (unless you just wanted to prove to yourself that it will all work out right in the end even when you assume the opposite direction to what you know is going to be right). :smile:
     
  8. Sep 9, 2016 #7

    CWatters

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    +1

    Note that the +/- marks next to R2 are probably intended to define what is meant by a positive voltage across R2. This is consistent with the direction defined for positive current I.

    Normally you define these things at the outset and don't change them just because Vin become negative.
     
  9. Sep 9, 2016 #8
    OK thanks for all the replies, that's exactly what I thought. Actually I think I've got it now, but here's my reasoning (prepared earlier) and my solution at the end.

    n.b, the references drawn in are mine, because we have been admonished enough times to give a magnitiude, unit and reference for every value :)

    Let's simplify the circuit to a 5V battery and a 2000Ω resistor, since the diodes are not relevant here. The current enters our positive reference of our resistor R

    VR is +5V and I is 2.5mA - simple.

    upload_2016-9-9_17-59-1.png

    Let's now say the battery voltage is -5V which can be expressed as

    upload_2016-9-9_17-59-19.png

    or more intuitively:

    upload_2016-9-9_17-59-39.png

    so now by KVL, +5V+VR=0, so VR is -5V. If I still enters the positive reference of R, do we still simply use I=V/R and get -2.5mA?

    My check here is to look at the power relationships. The sign convention I prefer, because it seems simple and universal, is to
    • Take the values given for V and I, be they negative or positive.
    • Multiply them to give a result (negative or positive),
    • which indicates energy 'travelling' in the direction indicated for conventional current.
    • A negative result simply means energy travelling in the opposite direction.
    • Then, a positive value of energy entering the positive reference of a circuit element is energy consumed by that element, and:
    • energy entering the negative potential reference is energy supplied by that element.
    • A negative value of energy entering the positive potential reference is negative energy consumed, which is interchangeable with energy supplied.
    • Likewise with 'negative energy' into a negative reference, we have negative energy supplied or positive energy consumed.
    So going back to the example, if VR is -5V and I=-2.5mA, P=VI=+12.5mW. So far so good, because while voltage and current are both -ve, our conventional current reference (as shown by 'I') enters the positive reference of R so energy is consumed as expected.

    But what if I got bored of the clockwise current reference and just switched it? It's arbitrary after all:

    upload_2016-9-9_18-0-5.png

    If I still =V/R we get -2.5mA into the negative reference of R so P=-2.5mA*-5V=+12.5mW supplied by the resistor. Clearly nonsense, so where have I gone wrong?

    Obviously in this case I could 'just' reverse the sign, because it is a trivial example I already know the answer to. But in a more general sense, in cases where the answer is less obvoius, can I still operate with my convention as described above and add the caveat that if the conventional current reference enters the negative reference then we use V=-IR and P=-VI?

    BRAINWAVE: can I use, instead of V as given, Vd; 'voltage drop in the reference direction of conventional current'? That would mean that in the last case, since I is anticlockwise, Vd would be -(-5V)=5V, which would satisfy my sign convention. Please tell me that's right, I like my convention above, it's simple like me :)
     

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    Last edited: Sep 9, 2016
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