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Another circuit verification

  1. Dec 3, 2006 #1
    1. The problem statement, all variables and given/known data
    The problem is attached. The question asks: "find the current through the inductive element". I'm starting to wonder how many mistakes are in this text book. But it might be a mistake I made. Could someone check my Mesh Equations?


    2. Relevant equations



    3. The attempt at a solution

    SuperMesh 1 & 2: 10V<0 + 1kohm(I1 + 6mA <0) + 4kohm (I2) + j6kohm (I2) = 0

    SuperMesh Supplemental: I1 + I2 = 0.1 Vs

    Direct Source Supplemental: Vs = (I1 + 6mA <0) * 1kohm

    if anyone wants to take a stab at the final answer, I got 7.74mA <171.5, and the book got 1.378mA <-56.31. Way different than what I'm getting
     

    Attached Files:

  2. jcsd
  3. Dec 4, 2006 #2
    Your supermesh equation is wrongly written.

    Assuming that voltage gains are positive values, then in:
    10V<0 + 1kohm(I1 + 6mA <0) + 4kohm (I2) + j6kohm (I2) = 0
    the part "+ 4kohm (I2) + j6kohm (I2)" is incorrect as clearly there is a voltage drop (according to the direction of I2) across the 4k resistor and 6k inductor.

    The correct supermesh equation should be:
    10V<0 + 1kohm(I1 + 6mA <0) - 4kohm (I2) - j6kohm (I2) = 0
    which should give you the desired answer.
     
  4. Dec 4, 2006 #3
    So in a supermesh, you treat it as going in one direction the entire way around? I was assuming you treated it as each direction showed

    I was taught you add the voltage if there is a voltage drop (voltage drop in relation to the direction your reference mesh current), not subtract as you're showing. And if I'm wrong as you say, wouldn't the other part of the equation (involving the 1k resistor) be incorrect as well? I guess that would really depend on my first question as well
     
    Last edited: Dec 4, 2006
  5. Dec 4, 2006 #4
    Yes, because in mesh analysis you are actually applying the Kirchoff Voltage Law which equates all voltage drops (or equivalently voltage gains) around the loop to zero. And yes, you can only traverse the loop in either the clockwise or anticlockwise direction, not both. Supermesh is no different from mesh analysis except that you have now to consider more than one mesh current.
     
  6. Dec 10, 2006 #5
    thanks, I completely understand
     
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