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DC Circuit Analysis

Discussion in 'Electronic Basics' started by meyousikmann, Oct 1, 2006.

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  1. meyousikmann

    meyousikmann Guest

    I am currently studying Kirchoff's Current Law and Kirchoff's Voltage Law
    with respect to DC circuit analysis. I am an amateur just trying to learn
    the basics of electronics so go easy on me. According to the text that I
    have, if given this circuit
    http://i92.photobucket.com/albums/l14/meyousikmann/electronics.jpg I should
    be able to get the node voltages b and c using both methods KCL and KVL,
    which I can given the methods in the text. However, I was wondering if I
    could also solve this question using linear algebra methods.

    Going with KCL, I should be able to get two equations of the form:
    (...)Vb + (...)Vc = Kb (node b)
    (...)Vb + (...)Vc = Kc (node c)

    Going with KVL and using the definitions (loop a = V0,R1,R4) (loop b =
    R1,R2,R3) (loop c = R4,R2,R5), I should be able to get three equations of
    the form:
    (...)I2 + (...)I4 + (...)I5 = Ga (loop a)
    (...)I2 + (...)I4 + (...)I5 = Gb (loop b)
    (...)I2 + (...)I4 + (...)I5 = Gc (loop c)

    I am fine with solving the equations via linear algebra once I get them
    setup, but I am having a little difficulty getting the equations setup. I
    just need help setting up the equations as I don't want any answers with
    numbers. I just want to understand how to setup the equations. Anyone
    offer any assistance to an amateur electronics hobbiest?
     
  2. Alan B

    Alan B Guest

    I can understand that you are having problems setting up equations, because
    the "forms" that you give above don't make any sense to me. I don't know
    what you are implying by "(...)" and I don't know what kind of constants
    you are hinting at by "K" and "G."

    What also puzzles me is why you differentiate between Kirchoff's Laws and
    linear algebra, because the Laws are defined by linear algebra. Here's how
    I understand it:

    "loop a:" 0 = V0 + VR1 + VR2
    "loop b:" 0 = VR1 + VR2 + VR3
    "loop c:" 0 = VR2 + VR4 + VR5

    Using Kirchoff's Voltage Law, and using Superposition to open I0. So, how
    does that relate to the Ga Gb and Gc equations that you use above? I don't
    want to go through a lot of substitutions and solving without knowing what
    it is you are trying to accomplish. How do we get from here to where you
    want to be?
     
  3. Tom Biasi

    Tom Biasi Guest

    Another way to look at it is:

    In a closed loop the voltage drops must equal the voltage rises.
    In other words, the source voltage ( for that loop ) will equal all the
    drops ( I times R ). Draw the flow and label all directions and values.

    When you hit a node the current will split, label them. When you are all
    done some resistors will have more than one current flowing through them.
    Same direction, add. Opposite direction, Algebraic sum. (subtract). I times
    R is the voltage for that resistor. Set up a formula for a closed loop.

    Source = I1(R1)+I1(R2)+I1(R3)...etc.

    Sometimes the expression will be more complex such as Source = ( I1+I2-I3)
    (R1 )....

    All currents entering a node will leave the node.
    Sometimes you will have current leaving and can identify one entering and
    another one of unknown value entering.

    Tom
     
  4. Just to put the schematic into the thread:
    I've labeled the nodes. They are a, b, c, and 0 (gnd).

    The way I set this up is:

    Node a: Va=V0
    Node b: 0 = -G1*Va + (G1+G2+G4)*Vb - G2*Vc - G4*V0
    Node c: 0 = -G3*Va - G2*Vb + (G2+G3+G5)*Vc - G5*V0 - I0
    Node 0: V0=0

    Here, I've replaced R1..R5 with G1..G5 as the conductances.

    If you take a look at node b, for example, the way I look at this is
    that, (1) there is a current spilling into the node (I take spilling
    _in_ as a negative value) from Va through R1, so I write -Va/R1 or
    else -G1*Va; and, (2) there is a current spilling into the node from
    Vc via R2, so I write -Vc/R2 or -G2*Vc; and, (3) there is a current
    spilling into the node from V0 via R4, so I write -V0/R4 or -G4*V0. Of
    course, V0=0, so that term will drop out. For all this, there is also
    a current spilling out of the node which is Vb*(G1+G2+G4). This
    spilling in and spilling out must be equal but opposite in sign (or in
    other words, total to zero.) Makes sense this way?

    If not, you could set it up the traditional way, that there is a
    current through R1, which is (Vb-Va)/R1; a current through R2, which
    is (Vb-Vc)/R2; and a current through R4, which is (Vb-V0)/R4. The sum
    must be zero. This makes 0=Vb/R1-Va/R1+Vb/R2-Vc/R2+Vb/R4-V0/R4.
    Rearranging that also gives 0=Vb*(1/R1+1/R2+1/R4)-Va/R1-Vc/R2-V0/R4,
    which is exactly the same as the other way of looking at it that I
    just mentioned in the last paragraph.

    Either way.

    Anyway, as Va is assumed known, you only need solve for Vb and Vc and
    you have two equations with which to do that.

    Since V0=0, they are reduced slightly to:

    0 = -G1*Va + (G1+G2+G4)*Vb - G2 *Vc
    0 = -G3*Va - G2 *Vb + (G2+G3+G5)*Vc - I0

    A solution for Vc works out by hand to:

    Vc = (I0+Va*(G3+(G1*G2)/(G1+G2+G4)))/(G2+G3+G5-G2^2/(G1+G2+G4))

    I didn't want to mess around with more algebra for Vb, so using my
    calculator to solve instead, I got these:

    z = 1 / (G1*(G2+G3+G5)+G2*(G3+G4+G5)+G4*(G3+G5))
    Vb = z * (Va*G1*(G2+G3+G5)+G2*(Va*G3+I0))
    Vc = z * (G1*(Va*(G2+G3)+I0)+(G2+G4)*(Va*G3+I0))

    Set
     
  5. meyousikmann

    meyousikmann Guest

    Wow! I think I am actually starting to understand. Since I am just an
    amateur learning this stuff, I am still working through your explanation but
    it is starting to make sense. Thank you so much for putting this into
    English for me.
     
  6. meyousikmann

    meyousikmann Guest

    One question I do have is looking at node b you mention this:
    Since the diagram shows I4 pointing away from node b, wouldn't the current
    actually be spilling out of node b through R4?
     
  7. I never did put down anything called I4 nor did I write a direction.
    So I'm not really sure I understand your question. But it is very
    likely the case that the image I have in my head isn't in yours and
    you are asking about something in your head that isn't in mine. In
    other words, we are at cross-purposes.

    So let me try and explain my thinking better.

    Look at node b. There are only three paths arriving from three
    external voltages:
    I like to imagine the various voltages as ponds, so that Va, V0, Vc,
    and Vb are all ponds. At different water levels, though. So I then
    imagine that pond Va flows towards pond Vb via R1's conductance, that
    pond Vc flows towards pond Vb through R2, that pond V0 flows towards
    pond Vb through R4. I treat flows inward towards pond Vb (since we
    are only considering pond Vb at this point) as negative flows. Inward,
    is negative. As G1=1/R1, G2=1/R2, G3=1/R3; the sum of these is pond
    flows towards pond Vb is -G1*Va-G2*Vc-G4*V0. Meanwhile,
    simultaneously, there is a similar flow away from pond Vb towards the
    other three ponds. Outward is positive (though any convention you
    keep to is okay.) This is simply Vb*G1+Vb*G2+Vb*G4, or Vb*(G1+G2+G4).

    Whatever happens, pond Vb's height must be in equilibrium. If it has
    one particular height for all time, that is, it must have that same
    height all the time. So the sum of these flows in and out must be
    zero, else the node Vb's height (voltage) would be changing. And it's
    not, by assumption (DC question, yes?) So the sum is zero. Thus:
    -G1*Va-G2*Vc-G4*V0+Vb*(G1+G2+G4)=0

    Nowhere in there is I4 mentioned directly. It is inferred indirectly
    because G4 is mentioned twice and so if you know what to look for you
    will find that it is G4*(Vb-V0). But I don't need to mention it,
    explicitly.

    There is no single "pointing" direction in the mental model I just
    mentioned. Instead, you treat things as if they point both ways.
    First arriving from the outside, pointing inward. Second, leaving
    from the inside, pointing outward.

    I like not having to worry about keeping some assumed direction in
    mind.

    Jon
     
  8. redbelly

    redbelly Guest

    Something I have just realized, reading this thread, is how Kirchoff's
    laws automatically force the number of unknowns and equations to be
    equal, allowing a unique solution via linear algebra. Well, somehow I
    knew that had to be true, but this thread has made me think more about
    it.

    There is a current equation for each node, and a voltage equation for
    each loop, so the number of equations is the sum of the nodes and
    loops.

    Likewise for the number of unknowns (voltages and currents): each node
    has a voltage, each loop a current.

    Mark
     
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