Tricky circuit analysis

Discussion in 'Homework Help' started by JoyAm, Sep 20, 2015.

  1. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    Hello everyone, i had a really hard time with this one because i cant find the angles. I know the angle of the source voltage as zero and the angle of the empedance of the line to be 86.64. I can also say that the angle of the coil current is -90 of the angle of the coil voltage.But after that i am suck.The currents seem to form an orthogonal triangle but i dont know how to use that fact ( if its even helpful)
    cir.png
    thanks in advance
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Too many words.
    Too few digits.
     
  3. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    What do you mean ?
     
  4. shteii01

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    Feb 19, 2010
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    Source voltage?
    Impedance?
    Circuit layout?
     
  5. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    did you check the picture i have in thumbnail ? the only impedance i know is the one of the line
     
  6. shteii01

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    I don't see picture. I don't see thumbnail.
     
  7. JoyAm

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    Aug 21, 2014
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    Idk why i can see it myself try this link ( its a screenshot) http://prntscr.com/8iip3r
     
  8. shteii01

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    Are A1, A2, A3 current sources?
     
  9. JoyAm

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    Aug 21, 2014
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    I thought it is so because it has cos(Omega*t+0)
     
  10. shteii01

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    Feb 19, 2010
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    You are right, sorry, I was a bit dense.
    Should it not be 240*1.41? Since the magnitude has the square root of 2?

    I only have two questions.
    Are A1, A2, A3 independent current sources?
    Do you have value for L or XL (I assume it is inductor)?
     
  11. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    No they are amp meters or whatever they are called
     
  12. JoyAm

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    Aug 21, 2014
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    Yes it is an inductor but i do not have a value ( all i know is that this element is an inductor)
     
  13. shteii01

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    Ok. If A3 is 3 A, then Iload is 3 A.
    I know it sounds too simple. But... given the information you provided it is that simple.
     
  14. shteii01

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    What about square root of 2 in the amplitude of Vn? That would make it 339.41 volts @ 0°.
     
  15. shteii01

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    Overall, the solution looks like this:
    -You know A1.
    -Find the voltage drop across the impedance that is series with Vn (I will call it Z1).
    -ZL and Load are in parallel, that means that voltage across ZL and Load is the same. So take Vn substract voltage across Z1. That will give you voltage across ZL and Load since it is the same. This will give you values in complex form.
    -Now do Inverse Fourier Transform to get value in time domain.
     
  16. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    i divide it by sqrt(2) because i have the rms value on the phasor and yes i agree with what you say Iload will be 3 but what will be the angle ?(because if you dont take angles into consideration and do a kvl you get something like 4=5+3 )
     
  17. shteii01

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    Feb 19, 2010
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    Are you telling me that current meters in the circuit give you magnitudes only?

    If the load is purely resistive (a resistor), there will not be a phase angle to the current.
    In previous post I showed how to get Vload. You can take it and either add 90° or subtract 90° to get the phase angle of the Iload. To add or to subtract depends on the load. Is it resistive? Is it iductive? Is it capacitive?
     
  18. JoyAm

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    Aug 21, 2014
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    How can i find the voltage drop on z1 since i dont know the angle of the curent I1?
     
  19. JoyAm

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    Aug 21, 2014
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    Yes, magnitude only and i am not given information about the load so i guess it can be anything
     
  20. shteii01

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    Feb 19, 2010
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    Then you have too many unknowns. I don't think it can be solved.
     
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