circuit query - disregard

Discussion in 'Homework Help' started by chrischivers1, Apr 10, 2015.

  1. chrischivers1

    Thread Starter New Member

    Mar 23, 2015
    Last edited: Apr 16, 2015
  2. MrAl

    Distinguished Member

    Jun 17, 2014

    Can you post the question as a picture file like gif instead of pdf?
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Clearly the reactive elements in primary and secondary cancel at the operating frequency.
    The primary input resistance thus comprises 15 ohm plus the effective equivalent resistance of the 8 ohm transformed from the secondary.
    So your answer of 15 ohm would be incorrect.
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Are you familiar with circuit analysis of mutually coupled impedances?
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Then you will need to do some reading on the topic. You won't be able to proceed without that skill. Any good undergraduate text on linear circuit analysis should include a section on the topic of mutually coupled circuits. If you can obtain a copy, I would recommend "Linear Circuits" by Ronald E. Scott. Scott was a Professor of Electrical Engineering at Northeastern University and was a highly regarded teacher.
    Last edited: Apr 10, 2015
  6. MrAl

    Distinguished Member

    Jun 17, 2014
    Hello there,

    Try using the "T" equivalent for the transformer. You should be able to get the information for that on the web. That makes the analysis much more straightforward. Once you transform to the T equivalent, you'll get a circuit that is easier to analyze.
    If you still need more help i can get back here later with more info if no one else adds more help. Basically i would look this up myself too as i dont use this every day either anymore. It's not that hard to do though.

    As tnk says, it looks like the input and output are at a resonant frequency which looks very close to the drive frequency, so at the frequency:
    the two inductors react with the two respective capacitors to create a zero impedance, which then leaves just the two resistances and the K factor to consider. If that's really the case then this will be even more simple.

    Try it yourself and see what you can come up with first, i think you will find this interesting.
  7. MrAl

    Distinguished Member

    Jun 17, 2014
    Hello again,

    Since this problem involves mutual inductance maybe we better use the 'full' method to get the answer, then if you wish you can look for shortcuts later.

    First we calculate the mutual inductance:

    Next we can form two mesh equations, with currents flowing into the transformer from input and output, and V1 the source voltage:

    Solve this set of simultaneous equations for I1 (and I2) and then we have the input impedance:

    Solve that for real and imaginary parts, and if we assume that 159kHz is really 159155Hz then the imaginary part will be very close to zero. If we dont assume that then there will be a tiny imaginary part but still much smaller than the real part.
    So for this problem we have almost a pure input resistance Rin, but depending on how we view that 159kHz we may have a small imaginary part too. My guess is they want you to consider that 159kHz as being the exact resonant frequency of the two inductors and respective caps so that the imaginary part comes out to exactly zero.

    For the sake of brevity i skipped some details, so if this isnt clear just say so and we can go over it.
    Last edited: Apr 13, 2015
  8. MrAl

    Distinguished Member

    Jun 17, 2014

    Yes 's' is sometimes called the complex frequency or the Laplace variable, etc.
    To solve for sinusoidal problems we can then set s=j*w and find real and imaginary parts.
  9. MrAl

    Distinguished Member

    Jun 17, 2014
    Hi again,

    M looks correct, very good :)

    s gets replaced with j*w, or i*w, where j or i is the complex operator, which means s is a complex number.
    But yes w=2*pi*f. So you replace s with either j*w and do some more calc's, or you replace it with s=j*2*pi*f and do the calcs.
    I like to wait until the end to replace w with 2*pi*f, but that's up to you.

    s is a complex variable such as:
    3+5*j or 3+5*i

    where 'j' is the complex operator sometimes written as 'i' instead. So to calculate the impedance you have to use complex math. In the above, '3' is the real part and '5' is the imaginary part, and together they form a single complex number.
    If you are unsure how to go about doing this, we can go over that next. The rules for complex numbers are not that hard to learn even if you have never done this before.
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    If this is an assignment you are presenting for credit towards a formal course score, one would hope you are intending to acknowledge the substantial contribution of @MrAl to the finished product.
  11. MrAl

    Distinguished Member

    Jun 17, 2014
    Hi again,

    Ok sure, but give me a few minutes to get this together. I'll post right here again. I wanted to at least acknowledge your reply as it took me a while to get back here today.
  12. MrAl

    Distinguished Member

    Jun 17, 2014
    Hello again,

    First, thanks to tnk for that kind suggestion :)

    Now we'll run though the calculations, although you dont have to do it this way if you dont want to, you can start with the values for the components but that leads to a lot of messy numerical calculations. Some of these equations will seem a little long, but dont let that stop you as you can use an online tool such as at Wolfram.

    You have done some of this too now but i repeat here for quick reference and clarity:

    First we calculate the mutual inductance:

    we could have left that for later too though.

    Next we can form two mesh equations, with currents flowing into the transformer from input and output, and V1 the source voltage:

    and we will replace the impedances with zL and zC as needed to keep some of the results more clear:

    It may be simplest to solve the second equation for I2 and then insert into the first, then we can solve for I1.
    So starting with:

    and solving that for I2 we get:

    Substituting that into the first equation we get:

    Solving this for I1 we get:

    and notice this is in the form I1=V1*K, so to compute the impedance we find V1/I1 which is:

    where K is:

    we need the inverse of that (Zin=1/K) so we get:

    Now that we have a representation of Zin, we can replace all the z impedances with their actual impedances:

    and we get:

    and after expanding that we get (this is all one equation):

    Next we replace s with j*w and get:

    and after condensing all the powers of j and expanding we end up with:

    and now the imaginary part is any part that is multiplied by j, and the real part is any part that is not multiplied by j, so we can break this into real and imaginary parts:



    and from here it is just a matter of replacing every component with their value, and also replacing M with its value, then calculating the result for both parts. the imaginary part should be much less significant than the real part because of the match of caps and inductrances.

    As i said before though, you can start with the actual values if you like including that for w and see if it is easier for you. Many of these terms will reduce then.
    There are also matrix methods but i'd have to look that up again myself too.
  13. MrAl

    Distinguished Member

    Jun 17, 2014
    Hello again,

    Well that was for the input resistance, but you can see we also solved for I2 and I1 so that helps answer the other questions too once you apply those solutions.

    I realize i gave probably the least straightforward way of doing this, so maybe we can look at the numerical approach too. I like working symbolically whenever the solution isnt too overly complex (this isnt that complicated even though ti may look long) so that i can come up with a formula to solve for any such situation. With this kind of solution we can use it with any problem like this by just plugging in the numbers now.

    The numerical solution would run the same way, except before we start we would plug in all the values for the caps and inudctors and M and then start calculating, reducing as we go. We could even plug in 'w' as soon as possible to get a more reduced result right off, as long as it seems comfortable to do it that way.

    For example, we get I1 as:

    or in alternate form: