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Kirchoff's Laws

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Relavent information is provided in the picture. Trying to solve the currents, I1, I2, I3.


    2. Relevant equations
    I(1)+I(2)+I(3)=0


    3. The attempt at a solution
    Too messy to even show.

    Basically i'm having a massive trouble understanding kirchoff's laws and this problem in particular. It's not that I don't understand the fundamentals of it, just how to approach and solve problems such as these.

    EDIT: Direct link for image: http://tinypic.com/r/1znwkzt/5

    http://tinypic.com/r/1znwkzt/5
     
    Last edited: Sep 6, 2013
  2. jcsd
  3. Sep 6, 2013 #2

    UltrafastPED

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    Now use KVL to generate an equation for the two labeled loops ... recall that the sum of the voltage drops around each loop is zero, and use Ohm's law to write the voltage drops for each resistor.
     
  4. Sep 6, 2013 #3
    So when I do loop 1 per say; I have to take into account the 15V battery as well and -/+ it into my equation as well?

    Cheers for the response too
     
  5. Sep 6, 2013 #4

    NascentOxygen

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    http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    Yes, every element in the loop; the battery is just another element.
     
    Last edited by a moderator: May 6, 2017
  6. Sep 6, 2013 #5
    Thanks will give it a try now, I can see myself ending up around a lot.. so many questions.. seems physics problems are infinite. ;)
     
  7. Sep 6, 2013 #6
    Feel so stupid that I can't even write the equations, either im really having a huge blank, or im really dumb. To be honest, need a step by step on how to write them even remotely properly.
     
  8. Sep 6, 2013 #7

    tiny-tim

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    Hi IronHide! Welcome to PF! :wink:

    Try KVL for loop 1.

    You must have some idea, so show us which bits you can do. :smile:
     
  9. Sep 6, 2013 #8
    I did, im just not confident.

    Obviously the +12V but then im confused as you go down the middle section, -15V, and the 50 ohm and 20 ohm resistors, use what current? And then I've got other stuff to worry about (Subject wise). Just the kind of the guy who will find out the method and see how stupid i've been and get all embarrassed cause it was so simple!
     
  10. Sep 6, 2013 #9

    tiny-tim

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    There's only three currents, I1 I2 and I3.

    Mark clearly on your diagram, with an arrow, which current is going through which resistor, and in which direction: that's the current you use for that resistor (and it will be + or - depending on whether the current is going the same way as that circular arrow).

    So what do you think the KVL contribution should be for the 50Ω and 20Ω ? :smile:
     
  11. Sep 6, 2013 #10

    CWatters

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    -15V is wrong.

    Consider.. when "you" went through the 12V battery you went from the -ve to the + terminal and gained +12V.

    When you go down through the 15V battery you are also going from the -ve to +ve terminal so it should be +15V.

    Then for the resistor use ohms law. V = IR or in this case I3*50. However if you look at the direction of I3 you will see that there must be a voltage drop as you go through the resistor so the voltage across the resistor will be -I3*50.
     
  12. Sep 6, 2013 #11
    So Loop One: 12+15-I3*50-I1*20?

    Well I get the right answer for that part of the question.

    Just like to say, a thanks to all who helped. I'm so grateful for the swift replies, and no one made me feel that dumb, and now I feel like how did I not realize this earlier. Thanks also for not just plainly writing the answer, and really making me work for it.

    Cheers to all,
     
  13. Sep 6, 2013 #12
    Also I have another question like this, so can I send my working out to you Tiny-Tim, and make sure you think its all okay?
     
  14. Sep 6, 2013 #13

    tiny-tim

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    Hi IronHide! :wink:
    No, private advice is against PF policy …

    just start as many new threads as you like! :smile:

    (a separate one for each question)
     
  15. Sep 6, 2013 #14

    tiny-tim

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    Looks good. :smile:

    Perhaps you'd better show us your loop 2 equation also, just to check. :wink:
     
  16. Sep 6, 2013 #15
    http://i40.tinypic.com/34ytcvq.jpg

    Heres the question, so basically you treat it as a normal question? not even a junction really, and just work through it. One of the questions is the current through the 5k ohm resistor, and another is the potential diff across the 8k ohm resistor. So just using V=IR, you can work it out? or just have to use the junction rules etc.?
     
  17. Sep 6, 2013 #16

    tiny-tim

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    You'll always need to use both KVL and KCL (for KVL, you'll need one less equatioin than there are loops, eg if there were 3 loops, you'd need 2 KVL equations).

    And yes, for the potential difference across the 8 kΩ resistor, you just use IR. :smile:
     
  18. Sep 6, 2013 #17

    UltrafastPED

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    Yes - you use Ohm's law to get voltage drops for the resistors in terms of the currents, or currents in terms of the known voltage drops. You use KCL and KVL to generate enough equations to solve for the unknowns.

    As your course progresses you will learn additional techniques which speed you through this process - including transformations between equivalent circuits. Most of the techniques are for linear systems.
     
  19. Sep 6, 2013 #18

    CWatters

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    There are different ways to solve circuits but if you have just been taught kirchoff's laws then that will probably be what they expect you to use. The general idea is to use KCL and KVL to write simultaneous equations which are then solved. If you have done it correctly you can solve for any unknown they ask for, eg the current through or voltage on any particular branch or component.

    In the two cases you have posted there are three loops you could apply KVL to and two nodes (with more then two branches) that you can apply KCL to. So 5 equations.

    In the latest example I would probably simplify it first by replacing resistors in series with one.
     
  20. Sep 6, 2013 #19
    Thankyou for all quick responses. I don't have answers, (not looking for them), but just wanting to check mine against them when I do it all. And I've been taught to use cramer's rule to solve, but that just seems like a massive waste of time sometimes.
     
  21. Sep 6, 2013 #20
    Loop 1 (4v one): 4-I1*6000-I3*8000=0

    Loop 2 (2v one): 2-I2*4000-I3*8000=0

    Look about right? Then just use cramers rule to solve?
     
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