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
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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?
On Sun, 1 Oct 2006 12:40:36 -0500, in message , "meyousikmann" scribed:
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:
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?
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.
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.
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.
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.
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.
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