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Mesh Analysis KVl, need help

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data

    KVL.jpg

    2. Relevant equations


    3. The attempt at a solution

    I'm not getting the correct answers, this is what I tried

    Loop 1: 150 - 14*i1 - 6(i1+i2) = 0

    Loop 2: -24 + 3*i3 = 0
    Therefore i3 = 8A

    7A = i2 + i3
    Therefore i2 = 7-8 = -1A

    Plugging that in equation 1, I get i1 = 7.8A
     
  2. jcsd
  3. Feb 22, 2015 #2

    gneill

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    Staff: Mentor

    Your Loop 1 looks good. Can you explain your Loop 2 equation in more detail? What components are in the Loop 2 path, and are you summing potential drops or potential rises?

    The arrows on your figure hint that you are traversing the loop counterclockwise, and along with the assumed direction of I3 that you've indicated would make the change in potential across the 3 Ohm resistor positive, which your equation indicates, but it also shows the contribution from the 24 V supply to be negative. So, a contradiction there. There should be at least one more term in the equation, because you've just passed the 3 Ohm resistor and haven't made it back to the bottom of the 24 V supply yet...
     
  4. Feb 22, 2015 #3
    What is I2 in that diagram? Is it the current in the branch with the current source?
     
  5. Feb 22, 2015 #4
    I2 forms part of the 7A current. The 7A splits into I2 and I3, so 7A = I2 + I3
     
  6. Feb 22, 2015 #5
    I am summing potential drops across the resistors

    I thought the voltage would be negative on the 24V battery because I am moving opposite to the current flow. By one more term, do you mean I should include the 6 ohm resistor in loop 2 ?, or maybe their should be 3 loop equations and I am missing one ?
     
  7. Feb 22, 2015 #6

    gneill

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    Staff: Mentor

    But you haven't closed the loop. You have a term for the 24 V supply and another for the 3 Ohm resistor. How do you get back to the 24 V supply?
    Always count the potential change across a voltage supply in terms of your direction of travel around the loop, not the current direction. Current direction will influence the sign of the potential change across a resistor, but voltage supplies are fixed regardless of the current direction.

    You could include the 6 Ohm resistor in the path, or continue on around via the 14 Ohm resistor and 150 V supply. Either is fine, the important thing is you must close a loop for the sum to be zero.
     
  8. Feb 22, 2015 #7
    Ok, thanks. I included the 6 ohm resistor in the path and here are my equations now

    Loop 1, same as before
    150 - 14*i1 - 6(i1+i2) = 0

    Loop 2: 24 + 3*i3 - 6(i1+i2) = 0 (I subbed 7-i2 for i3)

    Solving these two equations I get i1 = 7.5 and i2 = 0

    Therefore the 150V source = i1 = 7,5A
    The 6 ohm resistor = i1+i2 = 7.5A
    The 24V source = i3 = 7-i2 = 7A

    Yay, got the correct answers, thanks for your help :)

    One more question, I sometimes get confused with choosing the current directions and this leads to the incorrect signs. I heard the current directions are arbitrary, is that the case ?
     
  9. Feb 22, 2015 #8

    gneill

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    Staff: Mentor

    Yes, current directions are arbitrary (except for current sources which are defined to have a particular direction). You must, however, be consistent in their use once you've chosen them.
     
  10. Feb 22, 2015 #9
    Cool, thanks
     
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