AC Cicuit Problem

Discussion in 'Homework Help' started by BlackSuede, May 8, 2013.

  1. BlackSuede

    Thread Starter New Member

    May 8, 2013
    9
    0
    Hi guys,

    I have been having problems with the attached question for my first year electrical systems class. I have tried by mesh and nodal analysis (combining the Rx and Lx into one parallel load) and have had no luck in getting close to an answer. I am also trying to do a delta - star connection transformation on the top of the circuit but am having trouble with this as well. Any suggestions on how I can proceed to solve this?

    Cheers
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Normally one is expected to provide partial or attempted solutions when posting to the homework forum.

    If there were no current flowing in the load as suggested what would the voltage or potential be at the ZL node [w.r.t. ground]? Using this knowledge you might consider recasting the circuit with that condition in mind.
     
  3. BlackSuede

    Thread Starter New Member

    May 8, 2013
    9
    0
    I've attached work on the circuit taking into account the condition but I think I might be going about this wrong - both I can get two simultaneous equations but both would generate a solution containing the source voltage/impedance.
     
  4. screen1988

    Member

    Mar 7, 2013
    310
    3
    I have looked it and I think it is good. Now you only need to solve them.
     
  5. BlackSuede

    Thread Starter New Member

    May 8, 2013
    9
    0
    I've had a go at solving these but I just can't get a formula that isolates just the Zx in terms of just R, C and ω - my answer contains I1 and I2 for example on my mesh analysis solution of:

    Zx = Zc - I2/I1(R+3Zc)

    Is there any other suggested method which I could try and determine this?
     
  6. screen1988

    Member

    Mar 7, 2013
    310
    3
    Hi,
    I think from your last equation, you can express Zx in terms of I1, I2, Zc.
    Zx = -I2/I1 * Zc (1)
    In the second equation, you have:
    I2/I1 = Zc/(R + 2Zc) (2)
    From (1) and (2), you get Zx.

    I believe that there are many ways to do it but I am also new to electronics too.
    Let wait for some experts here.
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Under the heading "KVL around loop 1", you have an equation involving I3. You then collect terms and rewrite the equation, but I3 disappears.

    Just because you want I3 to be zero ultimately doesn't mean that you set I3 equal to zero in your equations.

    You must leave I3 in there (in all 3 equations) and solve for it. You will end up with an expression with R, C, Vs, Zx and so forth. Then once you have that expression, you set it equal to zero and then solve for Zx.

    Try doing this and show your work. Be forewarned; the expressions for I1, I2 and I3 will be cumbersome. Once you have the expression for I3, you need only set the numerator equal to zero to get your Zx.

    You will need to use 1/(j ω C) instead of Zc for the impedance of C in your final expression for I3.
     
    Last edited: May 8, 2013
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    As I suggested in my earlier post I believe one can look at the unique condition when the load current is deemed to be zero and modify the circuit accordingly. See the attachment.
     
  9. screen1988

    Member

    Mar 7, 2013
    310
    3
    Hi,
    I want to know if what I am doing is right or not.
    When there is no current through the load, I cancel the load branch(it is like the load is not connected to the circuit) and in this case the equivalent circuit seems different with the one you did.
     
  10. BlackSuede

    Thread Starter New Member

    May 8, 2013
    9
    0
    Am I still able to combine the Lx and Rx into a single impedance in the re-cast circuit and attempting to solve - although if I wanted the individual current through each component I would have to use current divider law?
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    One needs to be a little careful in this approach. Firstly this applies only in the situation where the load current is deemed to be zero. In other words recasting the circuit for all operating conditions is not a valid approach.
    How do we treat the load branch in this circircumstance? If the load current was very close to but not quite zero what would the load voltage be? Very close to zero as well. So at zero load current the load voltage must be zero relative to ground - or at ground potential. So we may treat the load as a short rather than an open circuit. Again, we can only do this when the load current is zero and the analysis with the circuit so rendered applies only at that condition.
     
    Last edited: May 9, 2013
    screen1988 likes this.
  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    You were doing fine with your original equations. If you simply leave I3 as a variable, you will have 3 equations in 3 unknowns. Just solve them and get a result for I3. Set that result equal to zero and you will end up with a simple equation in Zx to solve.

    Or, if you want to take t_n_k's suggestion, you can use the nodal method instead of the mesh method, and end with only two equations in two unknowns (Va and Vb). Then you can derive expressions for the current in R and in C, set them equal and solve for Zx.

    But, if I were you, I would forget about trying to use voltage divider formulas and so forth. Just go with a mesh or nodal analysis.

    Like I say, you were very close with your original equations.
     
    Last edited: May 9, 2013
    BlackSuede and screen1988 like this.
  13. screen1988

    Member

    Mar 7, 2013
    310
    3
    Ah, thanks!
    My mistake is that I have never thought about the voltage across it. Each time I see that I = 0, I usually consider it open.
     
  14. BlackSuede

    Thread Starter New Member

    May 8, 2013
    9
    0
    Thanks everyone, I have been able to get a solution now in terms of Zx with the real part representing the resistance and the imaginary the inductor of:
    Zx = - Zc^2 / (R+2Zc)
    Simplifying this down is horrible with all the converting between rectangular and polar form but it is still a solution as required in terms of just Lx, Rx and ω. When I do get a solution for this, I will post up - our lecturer has said it is a problem that can be solved in a couple of lines but I just can't see that...
     
  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    FYI, here's my solution of the original circuit using mesh analysis as you were doing to start with:

    [​IMG]

    Here's a solution of t_n_k's recast circuit using nodal analysis:

    [​IMG]
     
    screen1988 likes this.
  16. screen1988

    Member

    Mar 7, 2013
    310
    3
    Hi, The Electrician,
    I used the equations of BlackSuede by set I3=0 then solve them.
    I got the same result. Is this method really wrong?
     
  17. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    I really can't answer that question unless I see what you did. If you post your work I can answer.
     
  18. WBahn

    Moderator

    Mar 31, 2012
    17,718
    4,786
    This is correct, and it can be done in just a few lines. See the attached image.

    The key is to note that, when balanced as described, the node above the load acts as both an open and a short. It acts like a short in that the voltage has to be 0V in order for there to be no load current, but it also acts as an open because there is no load current, meaning that the current in the resistor HAS to loop back and go through the right hand capacitor.

    The source voltage and impedance is a not factor because, once balanced, it could be replaced with an ideal voltage source set to whatever the voltage on the left side of the left capacitor is and you couldn't tell the difference.

    With all this in mind, you have:

    (1) Ix = Ic + Ir

    But we also know, by inspection, that

    (2) Ix = VL/Zx

    (3) Ir = V1/R

    (4) Ic = (V1-VL)/Zc

    So this already gives us

    (5) VL/Zx = (V1-VL)/Zc + V1/R

    But we also know that

    (6) Ir = -VL/Zc = V1/R

    Thus

    (7) V1 = -VL(R/Zc)

    Substituting (7) into (5) gives us

    (8) VL/Zx = (-VL(R/Zc)-VL)/Zc + -VL(R/Zc)/R

    Note that the VL factors are common to all terms and hence go away

    (9) 1/Zx = - ((R/Zc)+1)/Zc - (1/Zc)

    Multiply both sides by -Zc

    (10) -Zc/Zx = (R/Zc) +1 + 1 = (R+2Zc)/Zc

    Flipping and taking the -Zc back to the other side

    (11) Zx = -Zc^2 / (R+2Zc)

    Which is what you got. Many of the steps above can be combined and done mentally pretty easily, but I tend to be a bit more explicit. Even so, when I did it I was still only six lines because, by hand, I can write fractions more clearly than I can do it inline.

    From here, you are only about three more lines away from having Rx and Lx in terms of just ω, C, and R (and Lx does not involve R).

    There is no need, at any point from start to finish, to do a single rectangular to polar conversion.
     
    screen1988 likes this.
  19. screen1988

    Member

    Mar 7, 2013
    310
    3
    I did it in post #6. I used the result of BlackSuede and solved it.
    Can you check it again?
     
  20. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Attached is my approach with the final few steps removed - in deference to the homework rules.

    It's a stretch of credibility to expect a student to solve this "in a few lines".

    To me few would be less than five. Subjective as that might be.
     
Loading...