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Homework Help: Kirchoff's Laws, nodes and current

  1. Sep 14, 2006 #1
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Sep 14, 2006 #2
    Not quite.
    Label the node in between R1, R3, and R2. What can you say about this node in terms of current?

    From kirchoff's law you know that
    [tex] \sum i_{in} = \sum i_{out} [/tex]

    The trouble you are having is that there is another current. Can you see why?

    Looking at that node we talked about.

    [tex] I_1 = I_2 + I_3 [/tex]
    where [itex] I_2 [/itex] is defined as leaving the node.
     
    Last edited by a moderator: May 2, 2017
  4. Sep 15, 2006 #3
    so is this how?

    I1 = I2 + I3
    I1R14 + I2R2 - E1 =0
    I2R2 + I3R3 +E2 = 0
     
    Last edited: Sep 15, 2006
  5. Sep 15, 2006 #4
    For the last equation, you want to make sure you keep the signs right. You can indicate a drop of voltage as a result of resistors as NEGATIVE, while an increase in PD as a result of a cell as POSITIVE or vice versa, just keep it CONSISTENT
     
  6. Sep 15, 2006 #5
    So

    I1 = I2 + I3
    I1R14 + I2R2 - E1 =0
    -I2R2 - I3R3 +E2 = 0


    i tried that and it is not the right answer
     
  7. Sep 15, 2006 #6
    Ok so im guessin it is something like that
    I1 = I2 + I3
    -E1 + I1R1 + I2R2 + E2 + I1R4 = 0

    wat bout the 3rd equation? Thanks
     
  8. Sep 15, 2006 #7
    try writing KVL for the other loop
     
  9. Sep 15, 2006 #8
    Actually
    I1 = I2 + I3
    -E1 + I1R1 + I2R2 + E2 + I1R4 = 0
    E2+R2I2+R3I3
     
  10. Sep 15, 2006 #9

    HW is due in 2 hrs and i tried everything any help is appreciated :redface:
     
  11. Sep 15, 2006 #10
    The two loop equations are:

    [tex] -\xi_1 +I_1(R_1) +I_3(R_2) +\xi_2 +I_1(R_1) = 0 [/tex]
    [tex] -\xi_2 -I_3(R_2) +I_2(R_3) = 0 [/tex]

    Note: That is two equations with THREE unknowns. So you need to come up with another equation to solve.
     
  12. Sep 15, 2006 #11
    Thanks Alot
     
  13. Sep 15, 2006 #12
    Try to do this when you are coming up with the equations.

    1) First draw the currents (you can choose whatever direction you want).
    - Remember that current does not change when passing elements in a series. - It only changes when it reaches elements in parallel.

    2) Traverse a branch (ie make a loop).
    - When you get to an element (such as a resistor) if the current direction is pointing into the element then write that voltage term (ie V=IR, where I = current going into the element, and R = the resistance of the element) as POSITIVE.
    - If you are going around the loop and you reach an element but the current is pointing in the opposite direction of the element, then write that term as NEGATIVE.
    - Set all of these terms equal to 0. Take note that the voltage supplied (such as from a voltage source) is equal to the voltage drops (such as a resistor takes away a certain amount of voltage).

    You really need to do a few examples for all of this to sink in.
     
  14. Sep 16, 2006 #13
    ok sorry but this equation for some reason did not work, i know one equation should be


    I1 = I2 + I3
    -E1 + I1R14 + I2R2 + E2 = 0



    I just still cant get the third equation for the left loop

    i was thinking it was

    -E2 + R2I2 + R2I3 + R3I3

    any help, thanks
     
  15. Sep 16, 2006 #14
    Woops sorry buddy. I interchanged I2 and I3.

    The correct equations for this network are:

    -E1 +I1R1 +I2R2 +E2 +I1R4 = 0
    -E2 -I2R2 +I3R3 = 0
    I1 = I2 +I3
     
  16. Sep 16, 2006 #15
    lol :( nope not it i have the answer, it is the 2nd equation that is giving me trouble to get. i need to understand this for coming exam. I asked my TA and he said it is not a straight forward thing
     
    Last edited: Sep 16, 2006
  17. Sep 16, 2006 #16
    Are you saying that what I posed (the last one) is not the right answer?

    If that's the case, let me know what the right one is.


    It is VERY straightforward... I have no idea why your TA is saying that. You basically have to remember two things.

    1) the voltage around a loop is equal to zero.
    2) current in equals current out.
     
  18. Sep 16, 2006 #17
    Never mind, for some reason it wouldnt accept my answer but i kept clickin ok and all of a sudden it took it. I had the equation from the start way before i did this thread. It wouldnt just accept my answer. Damn computers. Thanks for ur help
     
  19. Sep 16, 2006 #18
    I'm almost positive the last answer I put down is correct... if I'm wrong, I'm either having a huge space out (I did already with the switching of I2 and I3) or I have no idea what I'm doing... (which I hope is not the case ;)

    oh...
    so what is the answer?
     
    Last edited: Sep 16, 2006
  20. Sep 16, 2006 #19
    Answer was .031 but i had .03064435 even though it says on top to keep all sig figs and to aviod rounding
     
  21. Sep 16, 2006 #20
    ur equations were right
     
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