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Using Kirchhoff's Laws to Find Currents, Source Current and Power Dissipated

  1. Nov 2, 2012 #1
    1. The problem statement, all variables and given/known data

    In the circuit shown in Figure Q8 below, if Vs = 10V, use Kirchoff’s Laws to determine the currents i1, i2, i3 and the source current is. Calculate the power dissipated by the resistors in this circuit. Confirm that the power dissipated by the resistors is the same as the power supplied by the power sources.


    2. Relevant equations

    kirchhoff's current law
    kirchhoff's voltage law

    i1=i2+i3

    3. The attempt at a solution

    I've genuinely tried to get the answers to this but I just don't understand it. I want to understand the answer and how the answer was reached. I have another diagram which I want to do on my own from the help given here.
     

    Attached Files:

  2. jcsd
  3. Nov 2, 2012 #2

    tiny-tim

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    welcome to pf!

    hi cschear87! welcome to pf! :wink:

    you can get the ratios of i1 i2 and i3 by doing KVL round each of the two loops at the right (separately)

    and you can get the ratio of i5 and i1 by doing KVL round the loop at the left

    (and of course KCL gives you i5 = i1 + i2 + i3)

    show us what you get :smile:
     
  4. Nov 2, 2012 #3

    gneill

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    Can you show one of your attempts? How have you tried to apply Kirchoff's laws?
     
  5. Nov 2, 2012 #4
    Re: welcome to pf!

    Any chance you could phrase that as though I was say a 5 year old. lol. I'm just starting this class and feel a bit at the deep end.
    Before this post I tried
    1/i1 + 1/i2 + 1/i3 - 1/20 + 1/40 + 1/80 = 7/80 = 11Ω
    Total R. = 140Ω
    V/R = 11/140 = .08 (A?) -- .08 x 11 = .86 -- .86 x 140
    =.006

    I'm really really bad at math so it was my ABSOLUTE best guess and the book from college was written by the university and thus not great. I got through my binary questions and I'm stuck on algebra *cries*
     
  6. Nov 2, 2012 #5

    tiny-tim

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    hi cschear87! :smile:
    no, those aren't using Kirchoff’s Laws

    Kirchhoff’s Laws are the circuit (or loop) law (KVL), and the node (or junction) law (KCL)

    look them up in wikipedia or the PF Library :smile:

    (and then start by applying KVL to the right-hand loop, the one with only i2 and i3)
     
  7. Nov 2, 2012 #6

    So does i1 + i2 + i3 = 10v?
    I think I understand that voltage in = voltage out. So should the currents add up to 10v?
     
  8. Nov 2, 2012 #7
    I = current

    and the voltage stays the same in a parallel circuit
     
  9. Nov 2, 2012 #8

    tiny-tim

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    sorry, cschear87, but this is nonsense :redface:

    how can currents equal a voltage? :confused:

    and it isn't "voltage in = voltage out"

    it's "current in = current out"​

    cschear87, you clearly have no grasp of how Kirchhoff's laws work

    you need to read your book (or lecture notes) again, carefully, until you understand them

    (or wikipedia, or other websites)
     
  10. Nov 2, 2012 #9
    That's what I was saying. I don't understand it. And I've read the notes dozens of times. All two pages in the book I'm given about it. I've just started this course and I'm trying to understand but have limited literature for this subject (there isn't a text book for this class just notes provided by university). This isn't an excuse, I just don't understand this. This 10Q assignment I've gotten through the first 7 parts without help. I just don't understand based on the material provided. That's why I came here. I thought someone could help. It doesn't matter how many times I read that "the total current flowing into any junction equals the total current flowing out of it" I don't understand. It also doesn't matter how many times I read that "in any closed loop in a network, the algebraic sum of the voltage drops (PDs) around the loop is equal to the emf acting in that loop"... makes no sense. Looks like greek to me. Nevermind
     
  11. Nov 2, 2012 #10

    tiny-tim

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    ok, i believe you …

    but that doesn't help me explain it to you …

    if you don't follow it from the books, how am i going to make any difference? :confused:

    all i can suggest is that you think of electric current as being like the current (of water) in a river

    when the river divides into three channel (as in the diagram), the total water flowing into the juction has to equal the total water flowing out of the same junction​

    ... then re-read your book on the junction law (node law, KCL), and see if it makes any more sense
    sorry, but i have no suggestion for this, other than to point out that voltage is just another name for energy-per-charge, and KVL is basically an energy equation :redface:
     
  12. Nov 2, 2012 #11
    You will get it, maybe see a few different explanations. Tinytim's is above, here is another.

    KCL:

    Keep in mind what current is -- it is moving charges, inside a wire it is moving electrons. At a point where wires connect, each wire is carrying some moving electrons. If the number of electrons coming into that point did not equal the number of electrons leaving, you would have electrons accumulating there. This is not allowed**. So the number of electrons coming in must equal the number of electrons leaving. If you divide that by time you can speak of the rate at which electrons arrive must equal the rate that electrons leave. This becomes KCL: the sum of incoming currents = the sum of outgoing currents.

    *** If electrons were allowed to accumulate at a point, charge would build up and generate an electric field that would oppose and eventually stop the currents from flowing. We call those devices capacitors so when we want to model that we would put a capacitor in the circuit. A simple wire cannot perform this function. A capacitor is actually a broken wire with large plates connected at either end. Electrons do not flow across the broken wire but charges accumulate on the plates.


    KVL:

    As mentioned, this is an energy equation. But maybe you can understand it intuitively. If I take an charged particle at some point in the wire, it will have a certain amount of energy. If I then move this charged particle along the wire, its energy will change because it experiences a force (the electric field) that is summarized in a quantity known as a voltage. But if I keep moving this particle around the wire and come back to the point I started at, it should have the same energy as when I started.

    KVL means the change in energy as I travel around a closed loop is zero. The energy changes are measured as voltage drops across each circuit element (the wire itself is assumed ideal with no resistance and therefore no energy drop; if it were not ideal it would be modelled as a resistor on an ideal wire).
     
    Last edited: Nov 2, 2012
  13. Nov 3, 2012 #12
    OK, I think I understand a bit better now. CURRENTS in = CURRENTS out and voltage drops when contact is made with each circuit element. That seems to make sense. How do I put that information into an equation based on my diagram?
     
  14. Nov 3, 2012 #13

    tiny-tim

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    start by applying KVL to the right-hand loop, the one with only i2 and i3

    what do you get? :smile:
     
  15. Nov 3, 2012 #14
    For KVL, the sum of the voltages = 0. Right?
    For KCL current in = current out. So depending on the arrows in the diagram, and how they connect to the nodes, that would, I guess, dictate the equation. If generally i1 = i2 + i3, then iS = i1 + i2 + i3?
     
  16. Nov 3, 2012 #15

    tiny-tim

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    (let's leave KCL for the moment, it's a bit more complicated than it looks)

    Yes, the sum of the voltage drops has to be zero, because the emf (in that loop) is zero.

    ie, the RHS of the KVL equation is zero.

    So, in terms of I and R, what is the LHS? :smile:
     
  17. Nov 3, 2012 #16
    So is the current Voltage (I)/Resistance (R)?
    Voltage is 10V and would resistance be 20, 40 and 80?
    (LHS? Not far enough to know abbreviations yet :( )
     
  18. Nov 3, 2012 #17

    tiny-tim

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    LHS is left-hand-side, and RHS is right-hand-side (of an equation)! :biggrin:

    ok, i'll do this one for you …

    KVL for the right hand loop:

    the RHS is 0
    (because there's no emf in the loop)​

    the LHS is 80i3 - 40i2
    (because the voltage drop across each resistor is IR, and because if we go round clockwise then i3 is positive and i2 is negative)​

    so the whole KVL equation is 80i3 - 40i2 = 0,

    ie i2 = 2i3 :smile:

    (btw, if we'd gone anti-clockwise round the loop, it'd have been -80i3 + 40i2 = 0, which still gives i2 = 2i3, doesn't it?

    so it doesn't matter which way we go round ! :wink:)


    any questions about all that?

    if not, now you try the middle loop (the one with i1 and i2) :smile:
     
  19. Nov 3, 2012 #18
    2i1 - i2?
    = 2(20) - 40 = 0
     
  20. Nov 3, 2012 #19

    tiny-tim

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    that doesn't make sense :redface:

    write it in the standard form, i1R1 + (or -) i2R2 = 0
     
  21. Nov 3, 2012 #20
    I think where I'm confused is what exactly is R. There's no R in the diagram which is partially why I'm confused I think. And I thought that I had to use the current law to find the current for the first part of the question. I'm confusing myself :blushing: And you're very good to try and help. This course is all online with my university so I have very little interaction with other students and professors unfortunately which is why I haven't gotten an answer back about this from them. Online tutorial next week though. :smile:
     
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