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Superposition, circuit analysis.

  1. Jan 12, 2014 #1
    Hi, could check if I have the right answer please. The question (Q 4b) is in the attached image below.

    My answer is:


    First for analysis the DC voltage source.

    i2 = 0.82A, i1 = 1.97A

    therefore V across 10Ω resistor = 11.5V.

    For the analysis of the current source I got a bit confused, however, I think I might have got it right. I shorted the voltage source and found the current across the 50Ω resistor by doing:

    Rt= (10//2) + 12 = (41/3)

    i50Ω = (Rt*is)/(Rt+50Ω) = 0.86A

    Then to find the current across the 10Ω resistor:

    Rt = 2Ω hence i10Ω = (2*0.86)/(2+10) = 0.14A.

    Thus V10Ω = 1.4V.

    Then adding the two separate source responses across the 10Ω resistor the total voltage V10Ω= 1.4 + 11.5 = 12.9V.

    Is this correct? :confused:
     

    Attached Files:

    Last edited: Jan 12, 2014
  2. jcsd
  3. Jan 12, 2014 #2

    rude man

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    Homework Helper
    Gold Member

    There is no 50 ohm resistor.

    First thing is to label the resistors and sources. Put in numbers only at the very end.
     
  4. Jan 12, 2014 #3
    Woops lol.I'll try again.
     
  5. Jan 12, 2014 #4
    First if we remove current source we have a simply voltage divider:
    Vx1 = 110V * (10||14)/(5 + 10||14)
    Next we connect a current source but we short a 110 voltage source we have a very simple current divider.
    http://en.wikipedia.org/wiki/Current_divider
     
  6. Jan 12, 2014 #5
    The circuit divider doesn't seem that simple to me. ?????????
     
    Last edited: Jan 12, 2014
  7. Jan 12, 2014 #6

    Attached Files:

  8. Jan 12, 2014 #7
    Thanks that's excellent, but how do you know to make the equivalent circuit like that? I was trying to draw it with the 4Amp going into the circuit splitting between the 5ohm resistor and the other branch.
     
  9. Jan 12, 2014 #8
    Hey what can I say, I simply look at the circuit and I see equivalent circuit after some thought.
    So all you need is some experience, that's all.
     
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