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Circuit with two voltage source, need help

  1. Sep 10, 2008 #1
    In the picture u can see there are two voltage sources, Im confused one how to apply KVL to this. the book has this problem worked out and they get -12+4i+2v-4+6i=0 but i dont get this , i get this -12+4i+2v-4v-6i=0 and solutions isnt the same, what is my problem? For some reason there getting +6i and im getting -6i. WHy does this happen. Can someone please explain clearly, how to solve something like this, and for future circuits ?
     

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  3. Sep 10, 2008 #2

    tiny-tim

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    Hi th3plan! :smile:

    I can't see the picture yet, but I assume that you've drawn one arrow the wrong way round.

    In any circuit problem with more than one loop, you must draw arrows on each straight section (not just one for each loop), and mark each arrow with the (unknown) current, i1, i2 etc.

    Then look at every junction on the diagram, and check that KVL I is satisified.​

    If you do this for your problem, I'm sure you'll find there's a junction where it isn't. :smile:
     
  4. Sep 10, 2008 #3
    lets say its going clockwise the current -12+4i+2v-4v-6i=0 , i get this but i dont understand why its wrong the second battery its throwign me off
     
  5. Sep 10, 2008 #4
    Its a 1 loop tim
     
  6. Sep 10, 2008 #5

    tiny-tim

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    hmm … I suppose I'm just going to have to wait for the diagram … :redface:

    I wonder what the 12 is going to be? :smile:
     
  7. Sep 10, 2008 #6
    I uploaded the circuit again, please explain me why its +6i not -6i, and how do two batteries act on a circuit like this
     

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  8. Sep 10, 2008 #7

    NoTime

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    I can't see your second image yet, so assuming it's the same.
    I suggest you actually draw current arrows around the loop.
    Yes, even though it is a single loop.

    Without the rest of the circuit.
    If you have two batteries +(12)-+(4)-, how do they act?
     
  9. Sep 10, 2008 #8

    tiny-tim

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    Hi th3plan! :smile:

    ok, I can see the first picture now.

    It shows three voltages sources (two "facing" one way, and one the other way), and two resistors, all in series.

    (btw, why couldn't you have said that? :wink:)

    The book multiplies the current, i, by the two resistances, 4 and 6, and adds them: 4i + 6i.

    You want to subtract them: 4i - 6i. Why??

    The resistances themselves are neutral, and the current, i, is flowing the same way, since, as you pointed out earlier, there's only one loop.

    (that's why i assumed there were two loops …*it's the only way to get both i and -i)
    You just treat them like any other potential difference, in any loop in KVL 2 … if the potential differences are "facing" opposite ways, then of course you subtract them. :smile:
     
  10. Sep 10, 2008 #9
    im dumb, i figured it out heh
     
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