1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kirchoff's Laws

  1. Sep 14, 2006 #1
    [​IMG]








    a) What is the current I1 which flows through R1?

    I1 =

    (b) What is the current I3 that flows through R3?









    3 Equations

    I1-I3 = 0

    I1R14 + I1R2 - E1 =0

    I3R2 + I3R3 +E2 = 0



    is that even right?
     
    Last edited: Sep 14, 2006
  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.
     
  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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Kirchoff's Laws
  1. Kirchoff's Laws (Replies: 25)

  2. Kirchoffs law (Replies: 1)

  3. Kirchoff's Law (Replies: 5)

  4. Kirchoff's Laws (Replies: 26)

  5. Kirchoffs Laws (Replies: 6)

Loading...