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Need help to find I in a circuit (mesh analysis?)

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data
    http://tinypic.com/r/3478nle/6
    need to find the current I

    2. Relevant equations
    was thinking about using mesh current analysis, but our teacher hasn't gone over it yet.

    3. The attempt at a solution
    no clue, totally stumped on what to do. was thinking mesh analysis and i attempted that, but i had no idea what i was doing after a while.
     
  2. jcsd
  3. Feb 8, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Mesh analysis looks like a fine idea. Why don't you show how you got started so we can see how to help you when you get stuck?
     
  4. Feb 8, 2013 #3
    well i assigned currents I1, I2, and I3 to the right, left, and top loops respectively. moving in clockwise direction this what i got:

    right loop:
    3V+21(I1)-6(I2)-6(13)=0

    left loop:
    3V-6(I1)+21(I2)-6(I3)=0

    top loop:
    -6(I1)-6(I2)+21(I3)=0

    now the trouble i'm having is the part where i'm supposed to solve all 3 equations simultaneously. my method is getting I1=I2=I3 in the bottom two equations and then plugging them into eq#1 so i can solve for I1 (only current i need to find).

    cancelling out I3:
    3.5(3V-6(I1)+21(I2)-6(I3)=0)
    -6(I1)-6(I2)+21(I3)=0
    --
    10.5V-21(I1)+73.5(I2)-21(I3)=0
    -6(I1)-6(I2)+21(13)=0
    --
    -27(I1)+67.5(I2)=-10.5
    67.5(I2)=27(I1)-10.5
    I2=0.4(I1)-0.16

    cancelling out 12:
    3V-6(I1)+21(I2)-6(I3)=0
    3.5(-6(I1)-6(I2)+21(I3)=0)
    --
    3V-6(I1)+21(I2)-6(I3)=0
    -21(I1)-21(12)+73.5(I3)=0
    --
    -27(I1)+67.5(I3)=-3V
    67.5(I3)=27(I1)-3
    I3=0.4(I1)-0.044

    substituting into eq#1:
    21(I1)-6(I2)-6(13)=-3
    21(I1)-6(0.4(I1)-0.16)-6(0.4(I1)-0.044)=-3
    21(I1)-2.4(I1)+0.96-2.4(I1)+0.26=-3
    16.2(I1)=-4.22
    I1=-0.26A

    but i know i did this wrong...cause i got a negative current.
     
  5. Feb 8, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    Nothing wrong with a negative current, it just means that your assumed current direction for the loop is the opposite of the actual current direction. Your assumed direction was a clockwise mesh current, right? So the actual current is counterclockwise.

    Note the direction that was assigned to I in the diagram; It's the opposite of your assumed mesh current. So I = -I1 ...
     
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