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Determine currents Ia, Ib and Ic

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine currents Ia, Ib & Ic in the network?

    2. Relevant equations

    FIRST EMF

    RT = R1 + (R2 x R3) / (R2 + R3)

    RT = 47 + (j100 x ( -j75)) / (j100 + (-j75))

    RT = 47+ 300j (ohms)




    Ohms Law then states that:

    I1 = E1 / RT

    I1 = 12∠0º / 83-84.48j

    I1 = (996+1013.76j) / (14025.87)

    Using the Current Division Laws leaves:

    I2 = I1 x ( R3 ) / (R2 + R3)

    I2 = ( ((996+ 1013.76j) / 14025.87) x ( -j75 )) / (j100 + -j75)

    I2 = ( (2988-3041.28j) / 14025.87) ∠0º





    I3 = I1 x ( R2 ) / (R2 + R3)

    I3 = I1 x ( j100 ) / (j100 + -j75)

    I3 = ((3984+4055.04j)/ 14025) ∠0º



    SECOND EMF

    RT = R3 + (R1 x R2) / (R1 + R2)

    RT = - j75 + ( 47 x j100 ) / ( 47 + j100 )

    RT = ? Ω


    Ohms Law then states that:

    I4 = E2 / RT

    I4 = 10∠0º / ?

    I4 = ?

    Using the Current Division Laws leaves:

    I5 = I4 x ( R2 ) / (R1 + R2)

    I5 = ? x ( j100 ) / ( 47 + j100 )

    I5 = ? x ?

    I5 = ? Ω



    I6 = I4 - I5

    I6 = ? _ ?

    I6 = ? Ω


    So:

    IA = I1 - I5

    IA = ? - ?

    IA = ?∠ ?º Ω



    IB = I3 - I4

    IB = ? - ?

    IB = ?∠ ?º Ω



    IC = I2 + I6

    IC = ? + ?

    IC = ?∠ ?º Ω

    Can someone please let me know if the information provided is answering the question correctly? Thanks!

    In the diagram R3 is j75. However, it is suppose to read -j75.
     

    Attached Files:

    Last edited: Oct 16, 2013
  2. jcsd
  3. Oct 16, 2013 #2

    gneill

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    Staff: Mentor

    Wow, looks like a lot of work taking the superposition approach. Why not write a pair of mesh equations? Two of the currents will pop out as the mesh currents, and the third is just the sum of the other two.
     
  4. Oct 16, 2013 #3
    I spent all day calculating all theses, and everytime i have a different figure and they dont look good. I have a simillar example but with much easier numbers ( no j).
    A pair of mesh equations?
     
  5. Oct 16, 2013 #4
    Ok i will have a go!


    -12 +100j (I1 + I2) + 47I1= 0 and -100j (I1 + I2) +10- (-75j)I2 = 0

    Am i right?
     
  6. Oct 16, 2013 #5
    The answer for I2 = 0.4- 4jI1 and for I1= 147jI1+400I1=12-40j.

    Which looks rubish and very not correct!
    I need to finish this asap my time is running out!
     
  7. Oct 16, 2013 #6

    gneill

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    Staff: Mentor

    I'd choose my mesh currents to correspond to the currents Ia and Ib . Then "walk" both loops in the directions of the mesh currents and sum the potential changes:

    Loop1:
    +12 - Ia47 - (Ia + Ib)j100 = 0

    Loop2:
    +10 - Ib(-j75) - (Ia + Ib)j100 = 0

    arriving at what appears to be the same equations that you found. When I solve for Ia and Ib I get reasonable values.

    If you think carrying through all the complex manipulations while you solve the equations might be tripping you up, try leaving the component values as symbols and solve the equations symbolically. Plug in complex values in the final expressions for Ia and Ib and then reduce.
     
  8. Oct 16, 2013 #7
    12- Ia47 - (Iaj100+Ibj100) =0
    12- Ia47- Iaj100- Ibj100=0
    Ia47 - Iaj100= Ibj100-12
    Ia(47-j100)=Ibj100-12

    Ia= (Ibj100-12) / (47-j100)

    Is it right so far?
     
  9. Oct 16, 2013 #8

    gneill

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    Staff: Mentor

    Watch your algebra!
     
  10. Oct 16, 2013 #9
    And for loop 2:

    10-Ib (-75j)-(Ia+Ib)j100=0
    10+Ibj75-Ia100j-Ibj100=0
    Ibj75-Ibj100-Ia100j=-10
    Ibj(-25)- Ia100j=-10
    Ibj(-25)-100j((Ibj100-12)/ (-47-j100)) =-10

    Ibj(-25) -((10000Ibj2 - 1200j)/ (-47-j100))= -10

    Ibj(-25) - ( 10000(-1) Ib -1200j ) / (-47-j100) =-10

    Ibj(-25) +(( 10000 Ib + 1200j) / (-47-j100)) = -10

    And what next?
     
  11. Oct 16, 2013 #10
    ooops, thank you!
     
  12. Oct 16, 2013 #11
    235bj +1500b= (-11280j + 24000-2 ) / (47-100j)

    b= ( (-11280j + 24000-2 ) / (47-100j) ) / (235j +1500).

    Its after midnight will jump to bed and do some checking on my algebra.
    thank you
     
  13. Oct 16, 2013 #12

    gneill

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    Staff: Mentor

    Continue banging away until you've solved for Ib. Then use that Ib to go back to a previous expression for Ia and solve for Ia.

    It's a lot of work carrying out all these complex operations for every manipulation of the equations. I would suggest solving symbolically first, then plug in the values for the impedances in the final expressions. Then it's just two expressions to evaluate with complex numbers, rather than a whole series of complex expansions and shuffles over and over.

    I'll think about whether there's another approach that would be less time consuming.
     
  14. Oct 16, 2013 #13

    gneill

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    Staff: Mentor

    Okay, I'm thinking that a nodal analysis approach might be a bit more streamlined. If you take the top middle node and call its potential Vx, and if you leave the complex impedances as R2 and R3, then the node equation becomes:

    ##\frac{Vx - 12}{47} + \frac{Vx}{R2} + \frac{Vx - 10}{R3} = 0~~~~~## whence:

    ##Vx = \frac{R2(12 R3 + 470)}{R2 R3 + 47 (R2 + R3)}##

    Plug in values for R2 and R3 and solve for Vx (the only really messy expression to deal with), then:

    ##Ia = \frac{12 - Vx}{47}##

    ##Ib = \frac{10 - Vx}{R3}##

    ##Ic = Ia + Ib##
     
  15. Oct 16, 2013 #14
    You are torturing yourself!

    Once you have learned how to do calculations with complex arithmetic, and have practiced for a while, it's time to move on to modern methods using a computer or calculator. As you can see, it's easy to make mistakes when doing such an involved series of complex calculations by hand.

    Your school should have Matlab or Mathcad or something similar on the school network, which can do these sorts of calculations.

    Also, you can find applications for free on the web that can do complex arithmetic. You can even find complex matrix solvers.

    If you plan to continue your studies in a field that requires these sorts of calculations, you might consider buying a handheld calculator that can do complex arithmetic.

    Hewlett-Packard has recently introduced a new calculator, the HP Prime, and this introduction has led to a discounting in price for the older calculators. The older HP50G can do complex arithmetic, including complex matrix calculations, and can be had for under $100. This is a very powerful calculator and you would be able to use it for your entire academic career.


    Edit: Here's an online calculator for complex numbers using the RPN format of HP calculators:

    http://www.alpertron.com.ar/CALC.HTM
     
    Last edited: Oct 16, 2013
  16. Oct 17, 2013 #15
    Ok,

    Vx= 12.67 + 4.28j
     
    Last edited: Oct 17, 2013
  17. Oct 17, 2013 #16
    Thanks, i checked it out, looks ok but quite complicated for me. but for sure i will try to purchase software to help me with calculation asap!
     
  18. Oct 17, 2013 #17
    Ia = -0.014-0.09j
    Ib=( -2.67-4.28j) / -75j

    I didnt calculate Ic yet cause a amd be doesnt look very good!
    Can you check it out for me?
     
    Last edited: Oct 17, 2013
  19. Oct 17, 2013 #18

    gneill

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    Staff: Mentor

    Doesn't look right. Calculate the numerator and denominator of the expression separately and show your values. Maybe we can pin down where the slip occurred.
     
  20. Oct 17, 2013 #19
    Vx= 12.67 + 4.28j
    Ia = -0.014-0.09j
    Ib=( -2.67-4.28j) / -75j

    I found a mistake, i changed it. Is it better now? if not i will post my calculations :-(
     
  21. Oct 17, 2013 #20

    gneill

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    Staff: Mentor

    The numbers look good. Might as well express Ib in decimal form too.
     
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