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Solving for current in a circuit

  1. Sep 12, 2013 #1
    I have no idea what I a doing. I dont know how to deal with the two symbols (the one that looks like a diamond and a circle with an arrow).

    If those were replaced with a battery I would be able to solve it, but I can't solve it because I dont know which direction the current is flowing in each branch of the circuit.

    Anyways I did KVL for each loop and can someone tell me if I am correct?

    Loop 1:
    I0(6kΩ) - Vx + i1(8Ω) = 0
    Loop 2:
    Vvoltage over element of 5A - i1(8Ω) + Vx = 0
    Loop3: -3Vx - Vx = 0

    Out most loop: i0 - 3Vx = 0

    Can someone let me know if that is correct?

    Then what should I do next to solve for i0?
     

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  3. Sep 12, 2013 #2
    I just need to know which direction the current is going through each branch. does any one know?
     
  4. Sep 12, 2013 #3

    gneill

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    Staff: Mentor

    The circle with the arrow inside is an ideal current source. It will always inject exactly 5 A no matter what. The diamond with the arrow inside is called a controlled current source. It will produce 3Vx amps. That is, by some means not shown it "measures" the potential Vx across the 4 Ω resistor and then produces an amperage of 3 times the magnitude of that potential difference. If you wish, you can think of the coefficient "3" as having units of Ohms so then it becomes 3 Ω * Vx, which by Ohm's law results in Amperes.

    attachment.php?attachmentid=61739&stc=1&d=1379015000.gif

    Regarding your solution attempt, is there some particular reason you chose to use KVL loop analysis rather than nodal analysis? The reason I ask is that your circuit contains only current sources and one independent node, which makes it very amenable to nodal analysis.

    If you really, really want to use loop analysis, since the two current sources are in parallel I'd suggest combining them into a single controlled current source: I = 5 - 3Vx amps directed upwards. That will eliminate the third loop entirely.

    When you first write your loop equations, don't use Vx as a potential drop. Write it as -I1*4Ω. Use that for Vx everywhere; that will tie in the current I1 to the controlled source's current in your equations without needing a separate equation for Vx.
     

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  5. Sep 12, 2013 #4

    gneill

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    Staff: Mentor

    You won't know for sure until you've solved the circuit. If you guess wrong for a current at the outset, the value you find will be negative. No big deal, that just means it's really flowing in the opposite direction to your guess.
     
  6. Oct 8, 2013 #5
    edit..
     
    Last edited: Oct 8, 2013
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