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Superposition Circuits problem, need guidance and help

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data
    http://i615.photobucket.com/albums/tt237/chaz0426/Superpositionproblem.jpg


    2. Relevant equations



    3. The attempt at a solution

    I know you have to analyze the circuit by turning off each independent current/voltage source and I've done several practice problems in my book but I've never run into this paticular set up. I spent all night trying to figure this one out. Any help or direction appreciated.

    This is where I got: shutting off the current source I got I = 0.116 Amps but I'm not sure where that current applys or if it splits up since the voltage source is in parallel. Is it right next to the voltage source or on the branch with all the resistors? Do I apply current division? I'm confused...

    Then shutting off the voltage source I used current division but I'm confused at this point: are the 15 ohm and 5 ohm resistors in series and then in parallel with the 10 ohm? I'm confused.
     
  2. jcsd
  3. Apr 1, 2009 #2
    Ok the book says the answer is 0.660Amps but I'm confused as to how.

    canceling the current source I set up the equation as -3.5 + 10i + 15i = 0 and got i = 0.114A around that one loop and ignored the 5ohm resistor.

    canceling the voltage source and using current division rules I got the current going through the 10 ohm resistor as being 1.2A and the current going through the 15 ohm resistor being 0.8A. That makes sense since 0.8+1.2 = 2A on the left side using KCL.

    0.8A - 0.114A = 0.660 A which is the book answer but I'm confused because when I go back to the entire circuit and plug in all the values, it seems as if the 0.8A current and the 0.114A current are going in the same direction which would be 0.8A + 0.114A = 0.914 which doesn't seem to be right. I guess it seems as if it would only work if I switched the 0.114A current the other way. This is confusing.

    Any insight. This problem shouldn't have taken the hours it did.
     
  4. Apr 2, 2009 #3
    Here's a tip for you how to determine the direction of the current. Voltage sources produce current going from their + -terminal to - -terminal and current sources to the direction in which their arrow point. I'm also curious, how you got .114A as the other current. .8A-.114A=.686A Maybe a typo?
     
  5. Apr 2, 2009 #4
    I believe I meant 0.140 sorry about that. Yeah I recently discovered about the +'s and -'s thing too my book barely mentions that but I guess it's assumed. Anyway I've given up on that one and just assumed my answer is correct unless somone tells me why its not. I have no idea how to get anything else.
     
  6. Apr 2, 2009 #5
    I'm getting a bit confused. If you know how to determine the directions of the currents, are you still saying the current is .8A+.14A=.94A?
     
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