How does Millman's theorem Works?

Discussion in 'Math' started by bbhkk, May 22, 2015.

  1. bbhkk

    Thread Starter New Member

    May 22, 2015
    Hi guys, I am new here.
    I don't quite understand the material about the Millman's theorem presented on the Online Textbook of All About Circuit.

    Here is an example given on the page:

    Up till here, my understanding of the theorem is:
    Potential Difference across all branches = Total Current into the branches / Equivalent Conductance of all branches together

    Whereas the
    Total Current into the branches = Sum of Current flowing in each branch
    Current flowing in a branch = EBi/Ri, i=1,2,3...

    However, in the next step, the emf of the battery is directly plugged into the EBi, which I don't understand why.

    Why the current of a branch can be calculated as if it is separated (see attachment: 001.png)?
    Did I missed something? Please help.

    Thank you.
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  2. MrAl

    Distinguished Member

    Jun 17, 2014

    The simplest way to look at this is probably using Thevenin and Norton theorems. That can put things in parallel right from the start. You can also view this through the concept of duality.

    For example, for B1 and R1, we can use the theorem:
    "A voltage source in series with a resistance is equal to a current source in parallel with a resistance"
    So for B1 the current source is:

    and that has to be put in parallel with a resistance equal to R1. Once we do that, the circuit is simplified. On the left side we end up with a current source I1 and a resistor equal to R1 in parallel with R2. Doing the same with B3 and R3, we'd get two current sources in parallel with three resistors, and the voltage at the top of the circuit would be:
    V=(I1+I3)*(parallel combination of all three resistors)
    and the parallel combination is:

    and so the voltage is:

    and again I1=B1/R1 and I3=B2/R2, so if we rewrite this all out fully we'd get the same equation you posted.

    The reason why all this works is because of linearity.
    Last edited: May 22, 2015
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
    The key to understanding the Ebook presentation of Millman is to note that further on in the presentation the Millman equations shows there is minus 20 volts across R1.

    That is the current is flowing the other way in this branch and subtracting from the total current.

    And well done for providing a proper link to the page in question.
  4. bbhkk

    Thread Starter New Member

    May 22, 2015
    Thanks a lot for the helps.
    The theorem gets straightforward to understand after converting things to parallel.
    However, I now hit another obstacle in arriving the same result without converting things to parallel in the first place.

    I was approaching the problem this way:

    First, based on superposition, I should be able to analyze the effect of individual voltage source one by one, then get the final result by summation.
    Therefore, per voltage source, I will have the following circuit.

    The parallel resisters can be replaced by a single equivalent resister.

    Then I redraw the circuit into a form of voltage divider.

    The potential difference across all branches in the original circuit should be equivalent to the summation of the potential difference across Reqi.
    Based on circuit 'C', the potential difference across Reqi is:

    Then by superposition, the potential difference across all branches (Vo) is:
    The problem is, I cannot reduce the above equation to the form as the Millman's Theorem.
    Did I have my equation wrong from the beginning or it is just a matter of algebra to reduce it?

    Please kindly give me some advice, thank you.
  5. MrAl

    Distinguished Member

    Jun 17, 2014

    Well doing it that way means looking at different forms of the same thing.

    For the original formula, we have:
    and so:
    V=(I1+I3)/(1/R1+1/R2+1/R3) [Equation 1]

    Now this can also be put into the form:
    V=(I1+I3)*Rp [Equation 2]

    where Rp is the parallel combination of all the resistors, and the parallel combo is:

    and if this is expanded it comes out to:

    and that is the equivalent parallel combination of three resistors R1, R2, and R3. So you see we have two forms of the same thing here.

    Now looking at Equation 2, if we multiply that times 1/Rp we should get I1+I3.
    So first we calculate the two required parallel resistances we need to do it your new way:

    and now we can calculate the contributions from each supply source:

    and so the total superposition sum voltage is:

    and this equals:

    and this factored a little we get:

    Now we can 'get rid of' the parallel resistance by multiplying this times the reciprocal of Rp, so we do:

    and using the expanded form of Rp we get:

    and if we expand this a little we get:

    and we recognize this as:

    and now to get the second form again we just have to multiply this by the other form of Rp which was:

    so we get again:

    which is again:

    and there we have the required form.

    This is just one way of doing it as there may be other ways. This way makes use of the two different forms of three resistors in parallel. If there were more resistors, like four, then we'd just use the alternate form for four resistors in parallel:

    and this is just the product of all N resistors divided by the sum of products of all resistors taken N-1 at a time, and of course this is equal to:
    Last edited: May 23, 2015
  6. bbhkk

    Thread Starter New Member

    May 22, 2015
    Hi, I am sorry for my lateness of response, and thank you for your detailed explanation.
    I followed your demo and successfully derived the theorem. It works great, thank you.

    However may I ask what does it mean by " 'get rid of' the parallel resistance" ?
    Algebraically, I understand the effect the multiplication of (1/Rp), but what is the idea behind doing so?

    Please advice. Thanks a lot!

  7. MrAl

    Distinguished Member

    Jun 17, 2014

    By that i just meant that when you have a current I1 times a resistance R that equals a voltage, because V=I1*R, and that resistance is in parallel to the current source, so to get rid of the parallel resistance means to remove any influence of that resistance, and that means we multiply by 1/R and that leaves us with a current:

    so we 'got rid of' the R on the right side which is part of how we change the entire form. Not the best way to describe this concept i guess.
  8. bbhkk

    Thread Starter New Member

    May 22, 2015
    Thanks, it's all clear now.