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Homework Help: Impedance and phase angle

  1. May 2, 2014 #1
    1. The problem statement, all variables and given/known data


    Given the circuit above.

    a) Find I1, I2 and I.

    b) Find the power used in branch 1 and branch 2. (please correct my "translation" if it's wrong)

    c) Find the resulting impedance and phase angle. (Is resulting impedance a term?)

    d) Find the voltage between node A and node B.

    e) How can we get the phase angle to be 0 in the circuit?

    2. Relevant equations

    3. The attempt at a solution

    The question I need help with, is c). I've done both a) and b).

    My impedance, ZT, which I used in order to find the currents, is 28.3Ω ∠-20.5°

    What's the difference between impedance and "result" impedance? My professor said that the magnitude of the impedance will still be the same (28.3Ω), but the phase angle will be different.

    The way he explained it, I'm thinking that something is affecting the impedance, with a positive phase angle. I'm thinking the answer should be around 10 degrees.

    Does anyone have any input on this problem? Please let me know if you want me to post my calculations for the first two problems. I have everything written down on paper so it won't take too long.

    Appreciate any feedback.

    Attached Files:

  2. jcsd
  3. May 2, 2014 #2


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    Staff: Mentor

    Recheck that phase angle. It appears to be about double what I'm seeing.

    My interpretation is that it implies the net impedance as seen by the voltage source.
  4. May 2, 2014 #3
    I actually like this. Below will be how I calculated it.

    Z1 = 30Ω + j[itex]\frac{15}{2}[/itex]∏Ω
    = 38.15Ω ∠38.15°

    Z2 = 20Ω - j31.83Ω
    = 37.59Ω ∠-57.86°

    ZT = [itex]\frac{Z1 * Z2}{Z1 + Z2}[/itex] = [itex]\frac{(38.15Ω ∠38.15°)(37.59Ω ∠-57.86°)}{50Ω - j 8.27Ω}[/itex] = [itex]\frac{(38.15Ω ∠38.15°)(37.59Ω ∠-57.86°)}{50.68Ω∠-9.39°}[/itex]

    ZT = 28.3Ω∠-10.32°

    I made a mistake when converting my Z1 + Z2 into polar form. Also, from my original attempt, I had an error with a - sign. This might still be wrong, though.

    How does this differ from the actual ZT? The voltage source is at 0°.
  5. May 2, 2014 #4


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    Staff: Mentor

    Yup. Looks much better now!

    It doesn't. Your ZT is the net impedance of the circuit which the voltage source "sees". Presumably the voltage source is providing the reference angle to which all other voltages and currents are to be compared. The angles associated with impedances are a separate matter: they are an intrinsic property of the components depending only on frequency of operation.
  6. May 2, 2014 #5
    Ok, so my ZT is my final answer. I must've misunderstood what my professor tired to explain.

    Perhaps it's an easier way, or at least a way, to figure out the currents in exercise a) - without using the impedance? If there is, then the question makes sense at least.

    Well, I'll be moving onto the next question. You might hear from me :surprised
  7. May 3, 2014 #6
    On my e) question: "
    e) How can we get the phase angle to be 0 in the circuit?" I've done this:

    I would like to add a capacitor in parallel with the rest of the circuit, since it will add a positive phase angle to my circuit, and my total Z is 28.3 at a angle of - 10.32 degrees. My thinking behind this is that the new total Z will be 28.3 + x magnitude from the capacitor, and -10.32 + y degrees = 0 degrees. Am I onto something here?

    Any input on this would be greatly appreciated. Thanks.
  8. May 3, 2014 #7


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    Staff: Mentor

    The question is pretty vague actually. It doesn't specify what the limits are, if any, to the changes you are allowed to make to the circuit. Adding a power factor correction component in the form of a parallel capacitor or inductor is a common approach though.

    Note that your starting impedance already has a negative phase angle. Does that make the load "look" capacitive or inductive to begin with? (what's the impedance angle of a capacitor? An inductor?).

    You might consider looking at the load in terms of admittance rather than impedance (Y = 1/Z), and do so in rectangular form rather than polar form. Admittances in parallel add in straightforward arithmetic fashion, so choosing an admittance to cancel the imaginary term can be done by inspection...
  9. May 3, 2014 #8
    That's very true, to be honest. However, I feel that, with what we've learned in out lectures in mind, that what they want here is me adding a reactance. I assumed, wrongly, that it should be a capacitor.

    As mentioned, I was mistaken when I assumed I had to add a capacitor.

    To answer your question, I think it makes the load look capacitive, thus I must add a inductance in parallel to the circuit to make the load "neutral".

    We haven't had any lectures about admittance, although we've talked about it. I will look into this, though, and see if I understand it.

    But the way I see it, I can add a reactance, which can be either a capacitor or inductor. Since the load is capacitive, I need to add an inductor. Is this a possible solution to the problem?
  10. May 3, 2014 #9


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    Staff: Mentor

    Yup. An alternative is a series reactance. Or, perhaps more impractically, altering the frequency of the source...

    After you do the parallel reactance calculation try the series version for extra credit :wink:
  11. May 3, 2014 #10

    I tried drawing the new series and setting up the expression.

    ZT = [itex]\frac{Z2*Z3}{Z2+Z3}[/itex] + [itex]\frac{Z23*Z1}{Z23+Z1}[/itex]

    where Z3 is the unknown inductor.

    What I get then is x ∠0° = [itex]\frac{(37.59Ω∠-57.86°)(xΩ∠90°)}{20Ω-j31.83+xΩ+j0Ω}[/itex] + [itex]\frac{(37.59*xΩ∠-57.86+x°)(38.15Ω∠38.15°)}{20+xΩ-j31.83+30Ω+j15/2 ∏Ω}[/itex]

    I'm not able to find x from this expression, and I'm afraid I've taken one step back heh.
  12. May 3, 2014 #11


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    Staff: Mentor

    No need to bring the individual impedances of the branches back into the problem. You can work with the net impedance of the load.

    To find the compensation component for a series connection use impedance. To find the compensation component for a parallel connection use admittance. Use rectangular form for the complex values.

    Call the existing load impedance Z and its equivalent admittance Y = 1/Z. Call the impedance of your compensating component z, and its admittance y. All you need is for z or y to cancel the imaginary component of Z or Y.

    Impedances in series add: Znew = Z + z

    Admittances in parallel add: Ynew = Y + y
  13. May 3, 2014 #12
    I'm still left with a similar expression as above. I'd rather not write down exactly what I've done but here's my thinking:

    I've written down my ZT as Z. I've converted my Z into Y. I've set up the expression Ynew = Y + y. A small example of that would be Ynew = [itex]\frac{1}{Z}[/itex] + [itex]\frac{1}{x}[/itex]

    Then I'm lost again... What am I missing here?
  14. May 3, 2014 #13


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    Staff: Mentor

    Just use Ynew = Y + y. Otherwise, why introduce admittance if you're just going to discard it and go back to impedance?

    What is the value of Y (numerical)?
  15. May 3, 2014 #14
    The magnitude is [itex]\frac{1}{28.3}[/itex] = 0.0353 S (with an angle of positive 10.32)
  16. May 3, 2014 #15


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    Staff: Mentor

    Rectangular form please.
  17. May 3, 2014 #16
    Of course. 10.32 S + j 89.8 S. How should I use this to proceed? :confused:
  18. May 3, 2014 #17


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    Staff: Mentor

    Hmm. Those components look too big to yield a magnitude of 1/28.3 . Check your calculation.

    To proceed (once you have the correct rectangular form for Y), you want to find a value for y that will cancel the imaginary component of Y when you add them (Y + y).
  19. May 3, 2014 #18
    I've been mixing polar and rectangular form. Hopefully I'm done with that now :redface:

    So the rectangular form: 0.0347 + j0.00632.

    So what I want is for y to be something -j0.00632. That something can be 0?

    y = 0 - j0.00632

    Ynew = 0.0347 + j0.00632 + (0 - j0.00632)
    = 0.0347 + j 0

    Now I'm interested i y, which is z = [itex]\frac{1}{y}[/itex] = [itex]\frac{1}{0.00632}[/itex] = 158.23 Ω

    I'm not sure if what I've written here makes any sense, but at least I have an answer which does make sense. If I add an inductor with 158.23Ω∠90° impedance the phase angle will be 0.
  20. May 3, 2014 #19


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    Staff: Mentor

    Yes, that's the right idea. What value of inductor will give a reactance of 158 Ohms?
  21. May 3, 2014 #20
    Didn't notice this, sorry.

    2∏fL = Ω -> [itex]\frac{2∏f}{Ω}[/itex] = [itex]\frac{1}{L}[/itex]

    [itex]\frac{2∏*50hZ}{158Ω}[/itex] = [itex]\frac{1}{0.5H}[/itex]

    [itex]\frac{1}{0.5H}[/itex] = 2H.

    I haven't been very accurate with decimals here.
  22. May 3, 2014 #21


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    Staff: Mentor

    You do strange things with your algebra :smile:

    If ##2 \pi f L = \Omega## then just divide both sides by ##2 \pi f## to yield ##L = \frac{\Omega}{2 \pi f}##.
  23. May 3, 2014 #22
    Sigh. In my defense, my method works! I just got confused by my own stupidity, and switched the Henry's.

    But, I'd rather not talk about it. :tongue:
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