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Superposition for dependent sources and Pspice

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem is uploaded it is problem number 2


    2. Relevant equations



    3. The attempt at a solution
    I really just want confirmation that my work is ok, and the command for a dependent current source in PSpice. The things that I found that are dependent, are very confusing looking things instead of diamonds. I have tried E and F but I don't know how to use them?
     

    Attached Files:

  2. jcsd
  3. Mar 5, 2013 #2

    gneill

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    Staff: Mentor

    Check your work for problems 2,3, and 4.

    I'm not sure about the syntax or format of the dependent current sources for PSpice; I make use of LTSpice where there are several versions of dependent sources.
     
  4. Mar 5, 2013 #3
    ok is there something wrong with them I will upload my work for 3 and 4
     
  5. Mar 5, 2013 #4
    Ok my work for 3 and 4 are uploaded
     

    Attached Files:

  6. Mar 5, 2013 #5
    3 is on the left and 4 is on the right
     
  7. Mar 5, 2013 #6
    2 was already uploaded is there something I specifically did wrong?
     
  8. Mar 5, 2013 #7

    gneill

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    For #2 I just note that at least one of the current values you've found doesn't agree with values I find. I don't recall having seen your detailed work on this problem.

    For #3 it looks like you have a different answer this time, although I'm not sure if I'm interpreting your formatting correctly. What, as a number, do you get for the maximum current for Ix?

    For #4, it looks like you're attempting a Δ-Y transformation (or two?). If so, the numbers don't look right to me. Can you describe the steps you're making?
     
  9. Mar 5, 2013 #8
    for number 2 that is one of ther first I uploaded. For number 3 I find for Ix' using the max power and resistance of each resistor that gives me an Ix' for each resistor. Then I find a Ix'' using voltage and total resistance. Which I then put into an Ix'+Ix''=Ix equation for both the 100 ohm and 25 ohm resistor. When I have done this I get 200mA for the 100 ohm and 100mA to 25 ohm. I then set up the inequalities to solve for the maximum current that should flow before ohmic heating destroys the limiting resistor. The values on the other side of the inequality are found from current dividers formula I25=Ix(Rseries/R100) and I100=Ix(Rseries/R25) the Ix given for I100 the current for the 100 ohm resistor and the 25 ohm resistor only using the Ix for 25 ohm resistor. So when I multiply .2(125/25) I get 1 A or 100mA so Ix<100mA and for the other I get 125 mA the lower number is the limiting so I choose it. For 4 R1=RaRb/(Ra+Rb+Rc), R2=(RbRc)/(Ra+Rb+Rc), and R3=RaRc/(Ra+Rb+Rc). I have marked what I am using for Ra Rb and Rc. I then converted the bottom two resistors 2 ohm and 3 ohm to 32/16 and 48/16 respectivley so that it would make since when I add the R1 and 2 ohm resistor together and the R2 and 3 ohm resistor together. When I do this I get 72/16 for the 2 ohm and R1. 63/16 for the 3 ohm and R2 resistors. The R3 goes on top of the 72/16 and 63/16 which are in parallel. I then put the R3 in series with the 72/16 and 63/16 resistances which gives me a total of 2.1A.
     
  10. Mar 5, 2013 #9
    For number for I am doing a delta-y transformation
     
  11. Mar 5, 2013 #10

    gneill

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    Wow, a single massive paragraph to wade through...

    If so I must have missed it. Sorry. Still, the results that you posted on the "answer sheet" don't match what I'm seeing when I solve the problem.
    I don't follow what your Ix'' represents. When you apply an Ix to the circuit, both currents (through the resistors) change. This is because the node voltage where the resistors meet changes. Instead of looking at the maximum currents for each resistor, consider the maximum allowed potential drops across them. That will fix the maximum allowed potential at that node, from which you can then calculate the current Ix that it corresponds to.
    I think you mean 2.1Ω. But that is not correct. Your method looks fine, but something went wrong in your reductions, perhaps in the determination of the parallel resistance.
     
  12. Mar 5, 2013 #11
    ok so for number 3 I use the .1A from before and mutiply that by 100 ohm to get 10 volts. and .2A also from before and muiply that by 25 ohm to get 5 volts. so the total potential at the node would be 5 volts because 10-5=5 volts. Am I good so far?
     
  13. Mar 5, 2013 #12

    gneill

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    Not quite, but close. The maximum allowable drop across the 100Ω resistor is 10V, and across the 25Ω resistor is 5V. If the drop across the 100Ω resistor is 10V, that would mean the node potential would have to be 2.5V (since there's a 12.5V potential at the other end of the resistor, and 12.5V - 2.5V = 10V across the resistor).

    So you have a choice of either 2.5V or 5V for the node voltage. The question then is, which one is produced by the maximum Ix. You should be able to work out Ix for both cases and choose the maximum. Consider applying nodal analysis to determine Ix, where in this case you're starting out knowing the node potential and Ix is the variable.
     
  14. Mar 5, 2013 #13
    For number 4 it should be 3.6 ohm correct?
     
  15. Mar 5, 2013 #14

    gneill

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    Yes, that looks better :smile:
     
  16. Mar 5, 2013 #15
    Ok so doing the nodal analysis for 25 ohm Ix-(5/25)=0 so Ix=.2A. for the other 100 ohm resistor Ix-(10/100)=0 so Ix=.1A. I fear I have done the nodal analysis wrong! Should it be Ix-(2.5/100)=0 so Ix=.025A for the 100 ohm resistor?
     
  17. Mar 5, 2013 #16

    gneill

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    You have to include the effect of both sources in your nodal analysis. Suppose Vx is the node voltage (which you'll set to either 2.5V or 5V later), then just write the node equation.
     
  18. Mar 5, 2013 #17
    Also I can re upload my work for problem 2 so I can get that fixed up as well?
     
  19. Mar 5, 2013 #18

    gneill

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    Sure.
     
  20. Mar 5, 2013 #19
    Ok so I hope I am not irittating you with my ignorance but here goes my second attempt 2.5+100Ix+25Ix-12.5=0 so Ix=.08A for the 100 ohm. For the 25 ohm resistor 5+125Ix-12.5=0 so Ix=.06A. Once again I fear it is incorrect
     
  21. Mar 5, 2013 #20

    gneill

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    Just write the node equation with variable Vx as the node voltage. Solve for Ix. Then you can plug in 2.5V or 5V for Vx and compare the values of Ix.
     
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