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Circuit Analysis using Phasors

  1. Mar 19, 2015 #1
    signalsProb.jpg
    Do I have the right idea? I'm not getting what is in the solutions manual but it's been wrong on past chapter and I'm beyond frustration.
     
  2. jcsd
  3. Mar 19, 2015 #2

    SteamKing

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    Is C = 130 μF or 130 mF?

    This is why there is a PF rule against posting handwritten problem statements.
     
  4. Mar 19, 2015 #3
    It's milli
     
  5. Mar 20, 2015 #4

    donpacino

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    why do you have Vc = -Ic * Zc?
     
  6. Mar 20, 2015 #5
    Because the polarity of the capacitor in reference to the current flow. If the current flow Ic was in the direction towards the negative terminal of the capacitor then I wouldn't have added the minus sign to the current. Is this a wrong assumption on my part?
     
  7. Mar 21, 2015 #6

    gneill

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    You have two components with a common current that has a given defined direction. The expected potential drop is in the direction of the current flow, and that is what is indicated by the "+ -" on the capacitor. So in this case the labeling which defines how to interpret the potential across the capacitor is in accord with the assumed current direction.

    The bottom line is, the minus sign is not correct.
     
  8. Mar 21, 2015 #7
    Ok I see that now. I think I was confusing ideas from transients. My main problem is even when I didnt have the minus sign in there, I wasn't getting the book solution so I don't know if my process is flawed or not. The book solution is Ic=88.7sin(20t-27.5°)mA and Vc=2.31sin(20t+62.5°)Volts with the angles in degrees.
     
  9. Mar 21, 2015 #8

    gneill

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    Well then, the book solution is not correct for the given circuit. The total impedance is Z = 5 - 0.385j Ohms, so its angle is -4.4°. No way can you get -27.5° from there.
     
  10. Mar 21, 2015 #9
    Ok, I have been able to get that. Then I find the current in the loop using ohm's law I = V/R which gives me 0.997sin(20t+4.4°) Amps. Then I use the current through the capacitor which I just found and the Z of the capacitor which is 0.385∠-90° to find the voltage across the capacitor, using ohms law V=I*R which gives me 0.384sin(20t-85.6°) Volts.
     
  11. Mar 21, 2015 #10

    gneill

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    Yup. Looks good to me!
     
  12. Mar 21, 2015 #11
    Really?! Oh thank you thank you! I got that the first time I worked this out and couldn't figure out where I was going wrong! I hope these book publishers can get their act together before the 9TH EDITION geeez well thank you sir!
     
  13. Mar 21, 2015 #12

    donpacino

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    just something to think about josh. there are many different types of capacitors. the one used in your homework problem is an electrolytic capacitor. They are polarized and must be biased correctly (one side having higher voltage than the other. The is the reason for the plus and the minus. that being said when doing homework problems such as these, unless stated otherwise, assume all capacitors are ideal.

    http://en.wikipedia.org/wiki/Types_of_capacitor
     
  14. Mar 21, 2015 #13

    gneill

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    Don, the problem statement does not mention anything about the physical nature of the capacitor or its construction, only that it has a certain value of capacitance in millifarads. The + and - indicated on the diagram and labeled Vc defines how to interpret the capacitor voltage polarity (i.e., it defines Vc).

    I agree that there are practical considerations for real-life capacitors, particularly large-valued ones that take advantage of thin films in electrolytic capacitors. But there's no sign of needing any of these complications in this purely theoretical exercise.
     
  15. Mar 21, 2015 #14
    I appreciate the real world view Don, and the theoretical exercise as well, gneill. Maybe you two can clear up something that's slightly confusing for me. When I took circuit analysis 1 we would go around and label the polarity on components starting with the source. The source in this circuit has positive on the top side and we are taught that current flow is from positive to negative so I would label the left side of the resistor the opposite sign which in this case would be negative. I would then label the right side of the resistor as positive. Now coming to the capacitor, it has positive on the top side which wouldn't make sense to me unless we are at a state where the capacitor is fully charged and the source just turned off. The way the book labeled the capacitor makes it seem like current should flow from the capacitor in the upwards direction. Can anyone clear up this for me? Sometimes the book labels a voltage like Vc in the opposite polarity just to make us pay attention. That's why when I didn't get the book solution, I thought maybe I needed to assume that.
     
  16. Mar 21, 2015 #15

    gneill

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    Potential drops occur in the direction of the assumed current flow. The resulting change in potential is + to -, higher to lower.

    Fig1.gif

    One can choose to define a potential measurement in a fashion that is opposite to this, but it won't change the actual potential across the component, merely how one interprets it.
     
  17. Mar 21, 2015 #16
    but if the left side of v1 is the same polarity as the top of vs, then why would current flow through the wire at the top of the source and into v1 on the left side? starting at the source you have - then + then + then - then + then -. You have 2 +'s together....shouldn't it always go from + to - or - to +?
     
  18. Mar 21, 2015 #17

    gneill

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    The labels on the components designate the change in potential that you would measure if you were to place a voltmeter across them. Do a "KVL walk" from the bottom of the source, through the components, to the top of the source. How do those potential changes add up?
     
  19. Mar 21, 2015 #18
    ok I think that cleared up my confusion. The oscillation of polarities only occurs between components and not between components and sources.
     
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