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Discussion in 'Homework Help' started by moskan48, May 18, 2014.

  1. moskan48

    Thread Starter New Member

    Jan 25, 2013
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    Hello!!!!!
    Friends How To Solve This Circuit By Mesh Current Method:confused::confused:
     
  2. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    Label your meshes(indicate the same direction for all of them - ccw/cw), write the KVL equations for each loop, and post your work/attempt.
     
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  3. moskan48

    Thread Starter New Member

    Jan 25, 2013
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    note : the upper left is i1 and upper right is i2 and the lower mesh current is i3.
    1) 120 = 32i1 − 20i2 − 7i3
    2) −80 = −20i1 + 25i2 − 1i3
    3) What will be the third equation ??????????
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,392
    497
    That is not going to work.

    Assume clockwise mesh current.
    1) 5(i1)+20(i1-i2)+7(i1-i3)=120

    Now let see you do it.
     
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  5. moskan48

    Thread Starter New Member

    Jan 25, 2013
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    I take the current direction clockwise and i write the two equation.
    1) 5(i1)+20(i1-i2)+7(i1-i3)=120 is the same equation which i write but there is a problem in the third equation.
    What will Be the Third Equation???
     
  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,392
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    i3=-4 is it.
     
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  7. WBahn

    Moderator

    Mar 31, 2012
    17,748
    4,796
    What does it HAVE to be in order to satisfy the current flowing in the bottom branch?
     
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  8. moskan48

    Thread Starter New Member

    Jan 25, 2013
    12
    0
    Thank You Friends I solve It.
    The Equation Is:
    32i1 - 20i2 +0V = 92
    -20i1 + 25i2 + 0V = -84
    -7i1 - 2i2 + V = 32
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    While you basically got it, your work is hard to follow in part because your first two mesh equations have "0V" terms and the origin of these is not at all apparent. Then, your third mesh equation has a "V" term that is also out of the blue. Does this "V" have any relation to the two "V"s in the other equation? It is a variable? Is it supposed to mean "1V"? If it's a variable, what is it's definition? It the voltage form what to what? What is the polarity?

    This is the kind of stuff that gives graders fits -- and costs you points.

    You are close enough and have put in enough effort that I don't see any harm in showing how I would recommend working the problem.

    First, provide an annotated diagram showing the definitions of currents (both which current is which and what the direction of each current is) and other variables not in the original problem statement. This will not only help the grader (whom you really, really, want to make happy!), but it will help prevent you from making the kind of silly mistakes that we all are prone to make.

    [​IMG]

    Next, develop your mesh equations and present them in a form that makes it easy for someone familiar with mesh analysis to look at your diagram and at your equations and verify that they are, indeed, correct. Remember, everything after this point is math. ALL of the circuit analysis is contained in these equations. Most of the time, if you get these right you will get the lion's share of any partial credit that is available. So make it obvious that you got these right.

    <br />
\begin{array}{lcrcrcrcr}<br />
Mesh_1: \ & & (32\Omega) i_1  &-&  (20\Omega) i_2  &-&  (7\Omega) i_3 \; &=& \;120V \\<br />
Mesh_2: \ &-& (20\Omega) i_1  &+&  (25\Omega) i_2  &-&  (1\Omega) i_3 \; &=& \;-80V \\<br />
Mesh_3: \ &&&&&& i_3 \; &=& \; -4A<br />
\end{array}<br />

    Next, provide the solutions, showing work and intermediate results as appropriate.

    <br />
\begin{array}{lcrcrcrcrcrcr}<br />
Mesh_1: \ & & (32\Omega) i_1  &-&  (20\Omega) i_2  &=&  120V &+& (7\Omega)(-4A) \; &=& \;120V-28V &=& 92V\\<br />
Mesh_2: \ &-& (20\Omega) i_1  &+&  (25\Omega) i_2  &=&  -80V &+& (1\Omega)(-4A) \; &=& \;-80V-4V &=& -84V \\<br />
\end{array}<br />

    Solving, we have:

    <br />
i_1 \; = \; 1.55A<br />
i_2 \; = \; -2.12A<br />
i_3 \; = \; -4A<br />

    Then, perform one or more checks to validate your answers. You'll be surprised how often you find a mistake. In fact, I did when I first solved this problem and the only reason I caught it was because I did a check and the check didn't pan out.

    A fairly obvious check here is to compute the voltage across the current source two different ways and seeing if they agree. Note that the first annotated diagram I uploaded did not have Vcs shown on it. But since I used it for the check, I went back an updated the diagram.

    Going across the middle:
    <br />
Vcs \; = \; (7\Omega)(i_1-i_3) \; + \; (1\Omega)(i_2-i_3) \; = \; 40.73V<br />

    Going across the top:
    <br />
Vcs \; = \; 120V \; - \; (5\Omega)i_1 \; - \; (4\Omega)i_2 \; - \; 80V \; = \; 40.73V<br />

    Another check that is usually very good is to compute the power delivered and absorbed by each element and show that the circuit exhibits energy balance. In most engineering problems (even out in the real world), you can verify the validity of your answers from the answers themselves. Get in the practice of doing this as a matter of routine. Not only will it save you lots of points now, it may save you your job later and possibly catch a mistake that, left uncaught, could end up injuring or killing someone. As a practicing engineer, you have a moral, ethical, and legal responsibility to exercise due diligence in ensuring that your work is error-free.

    Finally, notice how I tracked my units throughout the work. I didn't just tack them on to the end results. Most mistakes you make (not all) will mess up the units. If you are truly tracking them, then you will catch those mistakes and be able to find and correct them very quickly before you have invested an hour or so working a problem that was guaranteed to yield a wrong result from the very beginning because of something you left out of your very first equation. This is also part of the due diligence I mentioned earlier.
     
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