Circuit with coupled coils.

Discussion in 'Homework Help' started by cdummie, Sep 26, 2015.

  1. cdummie

    Thread Starter Member

    Feb 6, 2015
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    Find the capacitance C1 such that module of complex current through the R is maximal. Known values are:

    R1=20Ω, R3=3Ω, L1=8mH, L2=4mH,C3=259uF, ω=10^3 rad/s k=sqrt(2)/2

    Circuit:

    kolo sa spregnutim kalemovima.jpg

    This is how i solved it:

    Since i have to find the current through the R, that means it's the same current as current through C1, so i could leave C1 and E alone, and find equivalent impedance for the rest of the circuit. If i choose the direction of currents such that both currents in branches with inductors are going 'in' the points since k is positive, so, basically i have a circuit with L1, L12 and R in series, and L2 + L12 in parallel with C3 and R3.

    Since i know k i can find L12 and its L12=4mH. S impedance Z1= jωL1 + jωL12 + R and Z3=R3 - j/ωC and Z2=jωL2 + jωL12 so Ze=Z1 + (Z2||Z3) .


    When i compute all of this i get Ze=(22.7 + j12.75)Ω.

    So now i have a circuit with E, C and Ze so current

    I1=E/(22.7 + j12.75 + 1/jωc)

    I1=E/(22.7 + j12.75 - j/ωc)

    I1=E/(22.7 + j(12.75 - 1/ωc))

    Current will be greatest if denominator is lowest possible, and when looking at C, it will be lowest if 1/ωC=12.75
    which means that C=78.4 uC.

    Is this valid approach, and is this correct?
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You can't ignore C1. Perform a full mesh analysis on the circuit and find the value of C1 that minimizes the impedance of the loop which includes C1 and R.
     
  3. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
    1

    I haven't ignored C1, i just found equivalent impedance for the rest of the circuit (E and C1 were not included) so i had a circuit looking like this:
    kolo sa Ze.jpg

    Then i know that I=E/mod(Ze - j/ωC1), and I will be bigger if mod(Ze - j/ωC1) is lower, i have done that in my first post.
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You can't leave C1 until the end. You'll notice that the circuit consists of two loops (meshs)--there is a larger one including C1; designate that loop number 1. Designate the smaller loop number 2. Designate the current in the larger loop as i1, and in the smaller loop as i2.

    Because of the coupling between the two inductors, some of the impedance in loop 2 is coupled into loop1, so you can't just calculate what you have called (Z2||Z3) and take that to be in series with L1, C1 and R.

    Do you know how to set up the mesh equations for a circuit? That's what you should do here.
     
    cdummie likes this.
  5. cdummie

    Thread Starter Member

    Feb 6, 2015
    104
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    I used Kirchhoff's laws and solved this, i got C1=50uC.
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I don't know what capacitance 50 uC is. The units of capacitance are farads, which are usually designated with the letter "F", not "C". Do you mean 50 uF?

    I don't get 50 uF for the result, so either you made a mistake, or I did. I can't help you find your mistake unless you post your work.
     
  7. cdummie

    Thread Starter Member

    Feb 6, 2015
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    Oh, i meant C1=50uF, my mistake, this is my work:
    I accidentally wrote C instead of F, but anyway, this is my work:

    Screenshot_1.jpg Screenshot_2.jpg
     
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    The branch current method is not the best method for this problem, so get rid of I1, I2 and I3. Relabel S1 as i1 (the current in the upper mesh), and relabel S2 as i2 (the current in the bottom mesh). It's conventional to show the mesh currents circulating in the same direction, all clockwise or all counter-clockwise.

    You only need two mesh equations to solve this problem.

    There are 4 inductances in this problem, L1, L2 (the two self inductances) and L12, L21 (the two mutual inductances). As it happens, L12=L21 so you only need to use one of them. I like to use the letter M to stand for the two mutual inductances.

    Your mesh equation for the upper loop is missing some terms. The voltage across L1 has 3 terms: jωL1*i1+jωM*i1+jϖM*i2. The term in red is due to the current i1 in L2.

    The voltage across L2 has 3 terms: jϖL2*i1+jϖL2*i2+jωM*i1. This term in red is due to the current i1 in L1. Because the voltage across each inductor also has a component due to the current i1 through the other inductor, the term jωM*i1 appears twice in the total equation for the upper mesh. There are other terms in that equation for R1 and C1, of course.

    Try writing just the two mesh equations.


    Your next problem is that you can't substitute numerical values in your equations before you solve them. You need to solve the two mesh equations as a pair of simultaneous equations. You will get complicated expressions for i1 and i2. Then you can do what is needed to maximize i1.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    I'm still wondering what "module of complex current" means?
     
  10. The Electrician

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    Oct 9, 2007
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  11. WBahn

    Moderator

    Mar 31, 2012
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    Ah, thanks. I was figuring he meant the magnitude. I had completely forgotten that "modulus" was another name for the norm or the magnitude of a complex number.
     
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