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Homework Help: Impedance Matrix of a Potential Divider.

  1. Apr 20, 2013 #1
    1. The problem statement, all variables and given/known data
    This question should be quite easy, I'm really struggling though.

    http://img32.imageshack.us/img32/5241/potdiv.png [Broken]

    1) Write down two simultaneous equations for this circuit that describe the voltage as a function of current

    2)Write down the impedance matrix that relates voltage to current

    3)Determine the matrix equation that relates current to voltage

    2. Relevant equations


    3. The attempt at a solution

    Vi = I1Ra + (I1-I2)Rb
    Vo = (I1-I2)Rb


    [itex] \left[ \begin{array}{c} Vi \\Vo \end{array} \right] = \left[ \begin{array}{cc} (Ra+Rb) & -Rb \\ Rb & -Rb \end{array} \right]\left[ \begin{array}{c} I1\\I2\end{array} \right][/itex]


    I don't understand what is being asked here, and it's worth a lot of marks. But so is the last section, so it can't just be the same thing again. What do you think it is asking?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 20, 2013 #2
    An impedence matrix seems like a very sophisticated way of talking about a potential divider...

    Put "simply" the voltage experienced at the load is:





  4. Apr 20, 2013 #3
    Another good point to make is that unless the terminals on the right are connected no current I2 will flow.. as there is no path to ground. And if they are shorted then the load is in parallel with no load... and so all of the current will flow through RA only. And if the load on the right is non-zero -> then the question is rather different isn't it. Its rather late here and it's been a long day so I may have fundamentally misunderstood the question or misapplied circuit analysis. Will have another look tomorrow.
  5. Apr 20, 2013 #4
    Its okay I've got it now. I misread the question, part 3 is "relates current to voltage" and part 2 is "relates voltage to current".

    So for part 3 I had to invert the impedance matrix to get the admittance matrix A . And then

    I= A * V

    A matrix does seem like an overly complicated way to find the currents in a potential divider, I guess it could be used to create simulations or something though. I think this was just an arbitrary example to test our matrix knowledge.

    And yes I've also been thinking that no current will flow through I2 if the terminals are not connected. This confused me a bit but I think my answer is still correct.

    I think that Vo will be connected by a load, e.g a 10 ohm resistor, but my matrix equation will still be true for any load as the question seems to imply that you already know all the voltages and currents and you just want to work out the resistances.

    Like I said though, It's just an arbitrary matrix question.
    Last edited: Apr 20, 2013
  6. Apr 20, 2013 #5
    Im glad that you've solved it. This is a strangely mathematical way of talking about circuit analysis. It is very probably true and yet slightly weird at the same time!
  7. Apr 20, 2013 #6

    The Electrician

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    Gold Member

    In your schematic you have the I2 current going the wrong way according to the convention usually used. See this page:


    Notice how the currents are all going into the network (two-port), not out. The way you have done it the 2,2 element of your matrix is negative. That element should be the resistance looking into the network (from the output) with the input open circuited, and it shouldn't be a negative resistance.

    Did you draw that schematic yourself, or did you paste from the actual homework assignment? I would check with the instructor about this.
  8. Apr 23, 2013 #7
    Thanks for the link, its good stuff :)

    The diagram was taken straight from a past exam paper.

    I think however the impedance matrix is still correct, or at least it is correct in the eyes of my tutor. Like I mentioned earlier this was from my maths exam (as part of an electronic engineering degree), so the question is just an arbitrary matrix question.

    I think that arrow showing [itex]V_{0}[/itex] is the wrong way around, as I think that [itex]V_{0}[/itex] is not a voltage source, but a voltage drop across a load between the two nodes of this port. In this case I2 is going in the correct direction, and I will just end up with a negative value for [itex]V_{0}[/itex] if I am to calculate it.
  9. Apr 23, 2013 #8

    The Electrician

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    Gold Member

    Your tutor makes the rules for homework problems, so you need to do what gets you the good grade.

    But, once you're out of that class, be aware that the elements on the main diagonal of a Z matrix should all be (by convention) positive when the elements of the circuit are just plain old resistors.
  10. Apr 23, 2013 #9
    Thank you I will keep that in mind.
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