Power/Current Problems

Discussion in 'Homework Help' started by freemindbmx, Mar 16, 2014.

  1. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    Power Picture-

    I need to find the power generated,the 18ohm resistor on top is highlighted and labeled R1.Forgot to edit it in sorry.. I get stumped by the 18,6,6 ohm resistors on top. Ive tried delta-y transformation but get stumped, then I tried a KVL but cant find a good spot to place a KCL. What would you recommend

    Current Picture-

    And a way to find the Current I in the circuit, I just need a way to start,I get stumped when trying to use a KCL
     
  2. NguyenDon

    New Member

    Feb 15, 2014
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    Have you learned the node-voltage or mesh-current method?
     
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    You are not making sense.
     
  4. WBahn

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    Mar 31, 2012
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    Your second diagram does not appear to represent the same circuit as the first. It appears you tried to do two source transformations and combine it into one but messed them up.

    Or are these two separate problems entirely? If so, pick one and let's focus on that and then we can turn our attention to the other.

    Take things one step at a time and show your work. We can then see where you are going right and where you go off the rails and help you get back on track.
     
  5. freemindbmx

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    Mar 5, 2014
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    Sorry for the inconvenience,ill make sure to post everything. But these are two completely separate problems, and ive only learned about mesh analysis. Once again I apologize :( But lets start with the first if you don't mind. The power problem, I know power is represented by P=IV and since its being supplied it'll be negative. When I try a mesh analysis I cant decide were to put what where. I can tell the first current say I1,and it goes through the 20v and the 1 ohm,but when I label the others I get confused. I like to label current going in as positive and current going out negative.But im guessing since the book has the top 18ohm resistor labeled in blue that this is the current we have to use to multiply V*I=P
     
  6. shteii01

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    And the circuit for this problem is where?
     
  7. freemindbmx

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    Mar 5, 2014
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    should be the second picture labeled power,I reedited my last post filling in some extra thoughts
     
  8. shteii01

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    [​IMG]




    This?
     
  9. WBahn

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    Mar 31, 2012
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    One of the key advantages to mesh and nodal analysis is that they seldom involve guessing. They are systematic ways of applying KVL and KCL, respectively, to a circuit.

    Look at your circuit and do you see how it resembles three "window panes"? Each pane is a mesh and the mesh current is a fictitious current running through the components in that pane's frame. So you have three such currents and, by convention, use assume that they are all either going clockwise or counterclockwise. Any frame element that is shared by two meshes and thus has two mesh currents flowing in it. The actual current in those components is the difference between the two mesh currents. For outside edges that only bound one pane the mesh current IS the actual current in those components.

    Performing a mesh analysis only involves going around each mesh and summing up the voltage drops (or gains, your pick) and setting them equal to zero (which is the case since it is a closed circuit). You just write the voltage drops/gains in terms of the mesh currents.

    So with this in mind, put together a diagram in which the mesh currents are clearly labeled and then write out the KVL equations for each mesh.
     
  10. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    I learned mesh just as you said, I label a independent loop and then label the current through it. such as the loop that involes the 20v,1ohm,6ohm,6ohm,I also like to label the current opposing it such as the 6ohm resistor on top(I3) as I calculate Loop1 (I1). Loop with I1 is:I remade the circuit picture, moving clockwise when finding voltages

    -20+1I1+6I1-6I3+6I1-6I2=0
    simplfy: 13I1-6I2-6I3=20

    Right
     
  11. shteii01

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    Now do the other two loops. Then use substitution to solve for whichever mesh current you actually need.
     
  12. freemindbmx

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    Mar 5, 2014
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    so for Loop two I get:

    6I2-6I1+6I2-6I3+18I2=0

    simplified: -6I1+30I2-6I3=0

    Loop 3:

    6I3-6I1+18I3+6I3-6I2=0

    simplified: -6I1-6I2+30I3=0

    solving gives I1= 2 ,I2=1/2, I3=1/2

    I just choose the current I3? and 20*(1/2)=10 but the back of the book gives 40w
     
  13. shteii01

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    Feb 19, 2010
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    Are you nuts?

    The current that leaves voltage source is I1, not I3. So power generated by the voltage source is voltage provided by the source times current though the source, which is I1. Why the Ef are you even looking at I3?
     
  14. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    im an idiot lol,I wasn't thinking just now,im watching the walking dead....my bad lol im facepalming pretty hard right now, but moving onto the next problem.
     
  15. freemindbmx

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    Mar 5, 2014
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    using a KVL; in clockwise direction

    Loop1(I1):
    21I1-6I2-6I3=3
    Loop2(I2):
    -6I1+21I2-6I3=3
    Loop3(I3):
    -6I1-6I2+21I3=0

    right
     
  16. shteii01

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    Feb 19, 2010
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    You missed minus signs.
     
  17. WBahn

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    Mar 31, 2012
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    You need to get in the habit of tracking your units throughout your work. Yes, most teachers and textbook authors are very sloppy about this -- but that's because few of them have ever watched someone die because they couldn't be bothered to track their units.

    You also need to indicated the direction for each of your mesh currents on your diagram -- don't make the people looking at your work guess.

    So your mesh equations should have been written as:

    I1: I1(9Ω+6Ω+6Ω) - I2(6Ω) - I3(6Ω) = -3V
    I2: -I1(6Ω) + I2(6Ω+6Ω+9Ω) - I3(6Ω) = -3V
    I3: -I1(6Ω) - I2(6Ω) + I3(6Ω+9Ω+6Ω) = 0V

    Which reduce to:

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

    Notice the symmetry of the equations. The currents on the diagonal are added while all of the off diagonal currents are subtracted. Furthermore, the off-diagonal coefficients are symmetric about the diagonal. Now, in this particular problem there is additional apparent symmetry that is purely a result of the resistor value choices in this problem.

    Continue tracking the units through the work -- don't just tack on units that you expect the answer to have at the end. Most mistakes you make will screw up the units and if you are tracking them you will catch the error almost immediately and know exactly where to look to fix it.
     
  18. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    I see what your saying WBahn,but I posted in my last couple posts that I was moving in the clockwise direction,although ill show it in the pictures from now on. But I will label my units(Ω and V,im guessing this is what you mean) from now on,just looks nicer and neat. But im alittle confused on whether I got the right KVL equaitons, Wbahn's equations seem like I got them right while Shteii01 says I missed some signs.

    Loop1(I1):
    (21Ω)I1-(6Ω)I2-(6Ω)I3=-3V
    Loop2(I2):
    -(6Ω)I1+(21Ω)I2-(6Ω)I3=-3V
    Loop3(I3):
    -(6Ω)I1-(6Ω)I2+(21Ω)I3=0V
     
    Last edited: Mar 17, 2014
  19. WBahn

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    Mar 31, 2012
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    As much as possible, you want your diagrams to be self-contained.

    Notice that my equations match Shteii01's while yours had the wrong side on the right hand side of the I1 and I2 equations.
     
  20. freemindbmx

    Thread Starter Member

    Mar 5, 2014
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    ahhh I see now, but after changing I just solve for all three unknowns and use the current that is (I2) as (I).Correct?
     
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