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Elementary Linear Circuits - Dependent Sources

  1. Aug 27, 2014 #1
    1. The problem statement, all variables and given/known data

    the problem is a circuit analysis. I can try to make the circuit but the image i created shows all the resistors and voltage sources.

    DE_1_10.png
    imgurl

    Find i,v,is,vs

    2. Relevant equations

    [itex]\sum[/itex]loop vi=0
    [itex]\sum[/itex]node ij=0

    3. The attempt at a solution

    I can not make a voltage loop as of yet. That would be my first attempt. Each node that I make an equation for results in more variables than what I have equations for. I am stuck on a very fundamental level here I think.
     
    Last edited: Aug 27, 2014
  2. jcsd
  3. Aug 27, 2014 #2

    NascentOxygen

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    Hi Devtycoon. Welcome to the Physics Forums.

    You don't have the right URL to display your pic. Let me fix that .....

    DE_1_10.png
     
  4. Aug 27, 2014 #3

    NascentOxygen

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    What are some of the equations you've been writing down?
     
  5. Aug 27, 2014 #4
    KCL for each node. I know there is a amperage going into/out of the 3 ohm resistor that I can not account for. I uploaded some work I have done with better images this time.
     
  6. Aug 27, 2014 #5
    I also can't find a good place to put my ground. Placing my ground in different places will change my equations.
     
  7. Aug 27, 2014 #6

    NascentOxygen

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    We don't need a whole panoply of images! One would be ample, with relevant notations added.

    First step: how do you decide the direction to mark the current in the 4 Ω resistor?
     
  8. Aug 27, 2014 #7
    My convention I want to stick to is assume that current flows from positive to negative terminal
     
    Last edited: Aug 27, 2014
  9. Aug 28, 2014 #8

    NascentOxygen

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    Well, you won't get far if you don't assume that! :smile:

    So, in what direction will the current be through the 4Ω resistor?
     
  10. Aug 28, 2014 #9

    NascentOxygen

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    You don't need a ground, but you can consider the lowest horizontal wire as ground if you wish.
     
  11. Aug 28, 2014 #10
    I will have the current going from the top of the 4Ω resistor (positive terminal) down through the bottom (negative terminal)
     
  12. Aug 28, 2014 #11
    So the positive terminal of the battery is connecting to the ground? Sorry, I am new to circuits.
     
  13. Aug 28, 2014 #12

    NascentOxygen

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    Sure! It doesn't care where you call ground.

    You don't need ground, so maybe leave it out if it's going to confuse you.

    Now, some of these equations .....?
     
  14. Aug 28, 2014 #13

    NascentOxygen

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    Correct. I asked because one of the pics you posted earlier seemed to show current going the other way.
     
  15. Aug 28, 2014 #14
    Ya I messed that up. I will begin at the node with 2i going in and i going out. It seems that I overlooked this situation. If 2 i is going in and i is going out then the 3 ohm resistor must have 1i going into that to have a KCL law satisfied.
     
  16. Aug 29, 2014 #15

    NascentOxygen

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    Right.
     
  17. Aug 29, 2014 #16

    NascentOxygen

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    Right.
     
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