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Basic electronics question - help me settle a bet

  1. Nov 21, 2011 #1

    Femme_physics

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    I was arguing with a friend (it's been a while since she studied electronics), that you can't this turn circuit:

    http://img689.imageshack.us/img689/5395/r123y.jpg [Broken]

    Into the following circuit by combining R2 and R3 as though they're parallel

    http://img510.imageshack.us/img510/3555/rthat.jpg [Broken]

    I explained to her that as long as there's voltage source on one of the legs in parallel you can't combined them (which I believe is the correct explanation). She think you can combine the resistors R2 and R2 in that case.

    So, can you help me settle this bet?

    Also, and this is just something that I've wondered about: are these two circuits basically the same calculations wise?

    I mean, can I combine the resistors in parallel despite the fact they're in different sides of the voltage source?

    http://img510.imageshack.us/img510/1245/resisresis.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 21, 2011 #2
    You are correct R2 and R3 in the first circuit are not in parallel.
    You are alos correct for the second circuit
     
  4. Nov 21, 2011 #3

    Femme_physics

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    Can you prove why? It appears my explanation of "because there is a voltage source there" is not satisfactory.
     
  5. Nov 21, 2011 #4
    The easiest way to think of parallel is that both ends of both devices have to be connected with no other electrical components between them (they have to have the same voltage across them).

    Since the "bottom" end of the R2 is connected to a voltage source and not to the "bottom" end of R3 they are not in parallel. Even if you replaced the voltage source with another resistor R2 and R3 would not be in parallel.


    There is no proof of this, the idea of series and parallel are concepts that make the math easier. Using the rule above for "parallel" you can look at a circuit and find parts that are easy to combine mathematically. But you could do away with the entire ideas of series and parallel and use something more rigorous like Kirchoff's circuit laws to solve a problem.


    For the second question, yes those two circuits are the same. It doesn't matter that one resistor is drawn on one side of the voltage source from the other. You can rearrange a circuit however you want graphically as long as you do not change any connections between parts.
     
  6. Nov 21, 2011 #5
    What you can do is convert this voltage source and that series resistor into current source and resistor in parallel.
    And then you are free to do whatever you want with your resistors, because they are free of any sources directly connected to them.

    visou.gif
     
    Last edited: Nov 21, 2011
  7. Nov 21, 2011 #6

    I like Serena

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    As Floid said, the ends of the resistors have to be connected, meaning they have to have the same voltage across them.

    The proof is that the total current Itot = V/R1 + V/R2.
    This means that the replacement resistor is:
    Rtot=V/Itot
    Rtot=V/(V/R1+V/R2)
    Rtot=1/(1/R1+1/R2)

    This does not work if there are extra components involved that make the voltage across the resistors unequal.
     
  8. Nov 21, 2011 #7

    dlgoff

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    If you use Norton's theorem to work a two loop problem, you can parallel R1 and R3 in order find an equivalent resistance r as shown here:

    nort6.gif

    For other methods to analyze two loop d.c. circuits, follow this link.

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dcex.html#c2"
     
    Last edited by a moderator: Apr 26, 2017
  9. Nov 22, 2011 #8

    Femme_physics

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    Thanks, that's a workable explanation IMO. And yes, Kirchhoff law does prove mathematically the fact that you can't combine them in parallel.

    Hmm, interesting, but from my thinking these are NOT equal, though, right? Or are they as well?

    http://img339.imageshack.us/img339/8729/noteqeqeq1.jpg [Broken]

    Thanks, but Norton theorem is beyond the scope of my studies as far as I know.


    Thanks as well Bass, ILS. :smile:
     
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  10. Nov 22, 2011 #9

    I like Serena

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  11. Nov 22, 2011 #10

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  12. Nov 22, 2011 #11

    I like Serena

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    Yup.

    The ends of the resistors are connected and they have the same voltage across them.
     
  13. Nov 22, 2011 #12
    I agree with ILS. The calculated current from the battery and through the resistors will be the same whichever circuit you use.
    The only (very subtle) difference is that the left hand circuit is 2 loops which produces 2 current loops. The current from the battery splits at the top and half goes to the right and half goes to the left. This means that those wires could be thinner than in the right hand circuit.
    This is the principle in a domestic 'ring main' circuit so that thinner wire can be used (cheaper)
     
  14. Nov 22, 2011 #13

    Femme_physics

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    And what about combining resistors out of order? Say, in this case, instead of combining R2 and R3 first, I combine R1 and R3 first. Is that ok?


    http://img268.imageshack.us/img268/5236/orderid.jpg [Broken]


    I can basically pick any combination I like, yes? In thios case, I can combine R1 and R2, and leave R3 uncombined as well: it still would be ok, yes?
     
    Last edited by a moderator: May 5, 2017
  15. Nov 22, 2011 #14

    Femme_physics

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    Excellent point!
     
  16. Nov 22, 2011 #15

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  17. Nov 22, 2011 #16

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  18. Nov 22, 2011 #17
    yes..... these are identical as far as doing calculations is concerned
     
  19. Nov 22, 2011 #18

    I like Serena

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    Is this still about settling the debt?
    What did you bet?
     
  20. Nov 22, 2011 #19

    Femme_physics

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    Thanks! :smile:

    Phew, and I'm out :wink: that was orgasmic.

    No, the last parts were purely for my own knowledge. The bet was strictly for the uppermost question at the beginning of this thread (whether you combine in parallel a resistor with a voltage source there).

    Believe it or not the bet says that if I win we do more physics together with my friend! :biggrin: (If I lost I'd have been forced to watch stupid TV shows! Thank goodness :wink: )

    Yea, we do silly bets. Should've asked for hours of back massage....since I knew I was right! But still....physics.....you know.... :wink:
     
  21. Nov 22, 2011 #20

    I like Serena

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    I'd lose on purpose for one or the other. ;)
     
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