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Circuit Analysis - Resistors and a battery

  1. May 15, 2014 #1
    Circuit Analysis -- Resistors and a battery

    1. The problem statement, all variables and given/known data

    Hello, if you see the attached I am trying to find the current I4.My plan was to find Voltage across R4 and use Ohm's law.



    2. Relevant equations

    V=IR




    3. The attempt at a solution

    I have simplified the circuit down(attached pic), I thought I could use the voltage divider equation to find voltage across R4||R6 which would be equal to the voltage across R4 however I'm getting a different answer

    I got 50*(3.197/(3.197+4.7)) =20V making I4= 2mA

    http://i.imgur.com/vjVbcEi.png
    http://i.imgur.com/6FwW7QG.png
     
    Last edited by a moderator: May 15, 2014
  2. jcsd
  3. May 15, 2014 #2

    phinds

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    So you believe that absolutely no current is flowing through R3? How do you come to that conclusion?
     
  4. May 15, 2014 #3
    Hmm that makes sense, I was thinking that going by Kirchoff's voltage law that the sum of the loop on the right would=0 in which case I could use the potential divider equation.
     
  5. May 15, 2014 #4
    Though it makes sense as Voltage across R3 would be the same as R2 yet if you did the voltage divider equation then you would get a different solution.

    So the only way would be to put R2||R3 then do the voltage divider equation? Or is there another way?

    Thanks
     
  6. May 15, 2014 #5

    phinds

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    This whole "voltage divider" thing is, to me, a brain-dead way of going about it. Don't get me wrong, I know people use it but I just find it a distraction that leads to the kind of error that you got. Just do a full loop analysis. It will take a lot more steps than what you were trying to do but that's because what you were trying to do was wrong.
     
  7. May 15, 2014 #6
    It was the way the question was worded, it guided you towards doing it that way.Personally I prefer loop analysis although my university will only let us use nodal, it gets marked incorrect if we do a loop analysis.
     
  8. May 15, 2014 #7

    phinds

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    Gads, I HATE that kind of crap. I had a prof once who counted my solution to a calculus problem as wrong because I used a simple method out of Thomas and he wanted us to use a more complicated method out of the book he made us use.

    Well, node analysis in this case is about the same as loop analysis, so do that :smile:
     
  9. May 15, 2014 #8
    Thanks, I think you're spot on, I've done it the way he wanted but I'll double check it with loop analysis to see which way is quicker, I think everybody has "their" way of doingthings and that should be acceptable.
     
  10. May 15, 2014 #9

    NascentOxygen

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    Staff: Mentor

    You can use the potential divider approach, basing it on resistor R3 and all the other resistors combined into a single resistance.

    What answer do you get?
     
  11. May 16, 2014 #10

    I did it using the divider approach by combining R2*R3/(R2+R3) This gives 1.45kΩ

    Then 50*(1.45/(1.45+4.7)

    =11.79V

    Then 11.79/10kΩ

    =1.18mA


    Thanks for the help and suggestions
     
  12. May 16, 2014 #11

    NascentOxygen

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    Staff: Mentor

    Looks right.
     
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