Mesh Current method help

Discussion in 'Homework Help' started by TripleDeuce, Oct 19, 2010.

  1. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    Determination of vTh = voc by mesh current method: Consider the
    circuit shown below. Our aim in this problem is to determine the open circuit voltage voc at the terminals A and B by utilizing the mesh current method.

    [​IMG]

    I drew the mesh currents in the appropriate windows but cannot get the right answer when I start writing my KVL's.

    From the diagram, I said 1.2Vx = i3-i2.

    I also wrote a super mesh equation that combined the windows of i2 and i3. Starting from the node directly under the current source

    1ohm(i2-i1) - (i2)0.5ohm - i3(0.625ohm) = 0

    working out the algebra yields

    0.5i2 - 0.625i3=i1

    Then I calculated the loop of i1 starting from ground

    20 -(0.5i2 - 0.825i3)1 - 1(i2-(0.5i2 - 0.625i3) = 0

    This gave i2 = 20A

    Looking at the diagram, you can see that Vx = i2*0.5ohms. I substituted this into the equation 1.2Vx = i3-i2.

    Rewriting gives 1.2(i2*0.5ohms)+i2 = i3 or 1.6i2 = i3

    so i3 = 1.6(20) = 32A

    I think I wrote the correct equations but the math just doesn't add up. Voc is 5V and in order for it to be 5V, i3 has to be 8A...
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    You have:

    "I also wrote a super mesh equation that combined the windows of i2 and i3. Starting from the node directly under the current source

    1ohm(i2-i1) - (i2)0.5ohm - i3(0.625ohm) = 0"

    This should be:

    1ohm(i1-i2) - (i2)0.5ohm - i3(0.625ohm) = 0

    How does the rest of the problem work out if you make this change?
     
  3. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    Can you explain why it is i1-i2 and not the other way around? The direction of i2 is up while i1 is down at the 1 ohm resistor
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    For consistency.

    You were following a clockwise path around the i2 and i3 meshes. When you pass a resistor, you can choose the voltage drop to be positive or negative for currents that are going in the same direction as your path around the loop, but you must do the same thing all the time.

    Look at the subexpression "- (i2)0.5ohm". The current i2 is flowing in the same direction as your direction of travel (clockwise). You used the convention that this is a negative voltage drop.

    But, look at the subexpression "1ohm(i2-i1)". You have made the voltage drop due to i2 positive even though i2 is flowing in the same direction as your direction of travel. That is inconsistent, and gives an error.
     
  5. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    0
    When I do it your way, I get i2 is -8.3A and i3 is 13.28A and i1 is -4.15A
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    This is incorrect, but since you haven't shown your work, I can't help you determine where you went wrong.

    If you want more help, show your work in detail.
     
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