Finding Current with Kirchoffs Laws

Discussion in 'Homework Help' started by hogesjzz30, Sep 3, 2013.

  1. hogesjzz30

    Thread Starter New Member

    Sep 1, 2013
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    I am really stuck on a question for my first year engineering physics course involving a circuit with 2 loops and 3 voltage sources. I have tried 6 times and gotten 6 wrong answers! Can anybody offer me any advice for where I am going wrong?

    Here is the question, my latest attempt, and the circuit in PHeT simulator

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  2. adam555

    Active Member

    Aug 17, 2013
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    In your first equation of loop 1 E2 should be added to E1, not subtracted.
     
  3. hogesjzz30

    Thread Starter New Member

    Sep 1, 2013
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    When I do that I end up with I2 = 10.36A....?
     
  4. adam555

    Active Member

    Aug 17, 2013
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    You got most signs in the first equation wrong (I guess in all equations).... you need to imagine a direction of the current (either clockwise or counterclockwise) on each loop and then go through each element looking on whether the sign of the voltage is in the direction of the current, or against it. If the the element has a negative to positive in the direction of the current, then it's a positive element on the equation; if not, then it's a negative; like this...

    E2 + E1 + (I1*R4) - (I2*R2) + (I1*R1) = 0
     
    Last edited: Sep 3, 2013
  5. hogesjzz30

    Thread Starter New Member

    Sep 1, 2013
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    So from the assumed direction of the currents given the current would flow anticlockwise in both loops. So that gives E1 + E2 in Loop 1 as you say, and which I have changed it to, and E2 - E3 in the second loop, which is what I have. Since voltage always drops over a resistor shouldn't all of the values be (-Ix)Rx..?
     
  6. adam555

    Active Member

    Aug 17, 2013
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    Sorry, like this....

    E2 + E1 + (I1*R4) - (I2*R2) + (I1*R1) = 0
     
    Last edited: Sep 3, 2013
  7. adam555

    Active Member

    Aug 17, 2013
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    You go element by element in a direction and check if the current goes in one or another direction; you had the E1 and E2 right, it's I2*R2 you got wrong.
     
  8. Raymond kunda

    New Member

    Aug 26, 2013
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    have got the correct answer now?am trying to analyze the question
    hey man if the the arrows are showing the assumed direction of current...............
    taking it like that is not right because current cant flow from negative to positive
     
    Last edited: Sep 3, 2013
  9. hogesjzz30

    Thread Starter New Member

    Sep 1, 2013
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    wait. so now I'm adding E1 and E2 again? I think I am ready to throw out this whole question, This is the 3rd piece of advice I've had and every time I keep getting told different things, and none of it matches up with my lecture notes or textbook >_<
     
  10. hogesjzz30

    Thread Starter New Member

    Sep 1, 2013
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    So then for loop 2 I should have E2 + E3 + (I3*R3) + (I2*R2) = 0 ?
     
  11. adam555

    Active Member

    Aug 17, 2013
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    E3 goes in the opposite direction, so it should be subtracted.

    I need some sleep now... just check the signs according to the directions you were given the currents.

    I'll be back later.
     
  12. hogesjzz30

    Thread Starter New Member

    Sep 1, 2013
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    No, still getting the wrong answer, and I can't figure out why
     
  13. adam555

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    Aug 17, 2013
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  14. Raymond kunda

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    Aug 26, 2013
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    its the thing of coming up with good equations and solve them simultaneously
     
  15. adam555

    Active Member

    Aug 17, 2013
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    This is the part you got wrong....

    Yes, voltage drops on all resistors; but this means that on each loop:
    E1 + E2 + ..... n = (I1*R1) + (I1*R2).... n
    or what is the same
    Et - (It*Rt) = 0
    But what we need to find out is the direction of the current to know whether or not we write a positive or a negative sign. For example: if the arrow of Ix goes in the same direction, then you write a positive sign, and if it goes in the opposite direction, then a negative sign. But you do it that way in the first equation I wrote; then change the signs when you pass them from side to side (the second equation up here).
     
    Last edited: Sep 3, 2013
  16. hogesjzz30

    Thread Starter New Member

    Sep 1, 2013
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    SO my equations should be:
    For Loop 1: E1 + (I1*R4) - E2 - (I2*R2) + (I1*R4) = 0
    For Loop 2: E2 - E3 + (I3*R3) + (I2*R2) = 0
     
  17. hogesjzz30

    Thread Starter New Member

    Sep 1, 2013
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    Yeah I know that's how you do it, I just have absolutely no idea how to get the right equations! I have been at this one stupid problem for 3 days now and I am starting to lose it!
     
  18. adam555

    Active Member

    Aug 17, 2013
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    Last post.... (you only got 2 wrong)

    E1 + (I1*R4) + E2 - (I2*R2) + (I1*R4) = 0
    E3 - E2 + (I2*R2) + (I3*R3) = 0

    I forgot to tell you something above: the arrows are for the I signs on each resistor, for the batteries it depends on the direction of the loop. So, if 2 batteries are on the same direction (the direction of the loop) they are added, and if they are in opposite directions they are subtracted; that's common sense.
     
    Last edited: Sep 3, 2013
  19. adam555

    Active Member

    Aug 17, 2013
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    I clarify... not for the resistor sings, for the current signs on each resistor.
     
  20. adam555

    Active Member

    Aug 17, 2013
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    And don't forget the internal resistors too; which I omitted in the examples.
     
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