kirchhoff's Question

Discussion in 'Homework Help' started by full, Sep 3, 2014.

  1. full

    Thread Starter Member

    May 3, 2014
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    Hi
    kirchhoff's law
    [​IMG]
    I can do first loop in circuit but in second loop I not understand how I can do loop!
    can you help me and explain how I can do second loop .

    this is not homework,

    thanks
     
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  2. full

    Thread Starter Member

    May 3, 2014
    225
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    loop1:
    KVL
    10V - 2I1 - 6I3 - 4I1 =0

    loop2:

    -15V + 8I2 + 6I3 =0

    why source is negative in loop2?
    and resistors are positive in loop2?
    thanks,
     
    Last edited: Sep 3, 2014
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    I3 = I1 + I2

    10 = 2I1 + 6I3 +4I1 = 2I1 + 6I1 + 6I2 + 4I1 = 12I1 + 6I2

    15 = 6I3 + 8I2 = 6I1 + 6I2 + 8I2 = 6I1 + 14I2

    Can you solve it from there?
     
  4. full

    Thread Starter Member

    May 3, 2014
    225
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    can you explain how you are do loop 2 (kirchhoff's rules) ?
    15 = 6I3 + 8I2
    thanks
     
  5. full

    Thread Starter Member

    May 3, 2014
    225
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    why this circuit is different for first circuit
    [​IMG]
    in first circuit the loop 2 is {-15V + 8I2 + 6I3 =0}
    but in second circuit the loop2 is { -2 + 8 +3Ix =0}

    please can you explain what differences ?!

    thanks ,
     
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  6. full

    Thread Starter Member

    May 3, 2014
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  7. Jony130

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    Feb 17, 2009
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    There are many methods to write KVL.
    For example in first circuit for the voltage sources if you go from "plus to minus" you add " -" and if you go from "minus to plus" you have "+".
    And for the resistors. If the resistor current is in the same direction as loop. We give "-", but if current flow in opposite direction the the loop, we give "+".

    First circuit loop 1
    We start at the bottom of a 10V voltage source (point E). And we go from minus to plus so we have +10V. Next we go from point A to B. I1 is in the same direction as our loop so we give "-". +10V - I1*2. Next we go from B to D. And again I3 is in the same direction as loop, so again "-".
    +10V - I1*2 - I3*6. And we have the same situation for 4 ohm resistor.
    And this is why we end up wit this
    10V - I1*2 - I3*6 - I1*4 = 0
    kir.png
    Now the second loop.
    We start at point B and we go from B to C (from "plus to minus" ), so we add -15V.
    Next from C to D. Notice that now I2 and I3 currents opposite our loop "current", so must give "+". So finally we end with this. -15V + I2*8 + I3*6 = 0

    For the second circuit the book use different method (convention).
    They first label the voltage drops across resistors.
    Loop 1

    ZZ.png

    Next they start at point A and go to point B. At point A we firs we see a "minus", so we write
    -12V. As we go from B to C , at point B we see "plus", so -12V +v;. And form C to A we also see "plus".
    And this is why we have
    -12V + v + 2V = 0

    For the second loop we again start at A and we go from point A to point C. And first what we see At point A is a "minus" sing -2V.
    At point C we see "plus" -2v +8V and at point D we also see "plus".
    So we end up with -2V + 8V + 3Ix = 0

    We can use this method to analysis the first circuit.
    Current flow from "+" to "-" so we label the voltage drops based on the direction of currents. In resistors current must enter the positive voltage polarity node.

    kir.png

    Loop 1 - start at point A

    I1*2 + I3*6 + I1*4 - 10 =0

    Loop 2 - start at D

    -I3*6 + 15 - I2*8 =0

    And sorry for my poor English
     
    Last edited: Sep 3, 2014
  8. full

    Thread Starter Member

    May 3, 2014
    225
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    thank you very much sir, this is very kind of you ...:):)
     
  9. full

    Thread Starter Member

    May 3, 2014
    225
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    I use 2 methods in this circuit but the result not same
    I think the first method is true and second method is wrong
    please sir can you explain how I knew currents direction in any circuits?

    werf.png
    method 1:
    -15 +1I +2Vx +5I + 2I=0


    Capture.PNG
    method 2:
    15 - 1I -2Vx +5I + 2I=0

    sorry for my poor English

    thank you,
     
    Last edited: Sep 4, 2014
  10. MikeML

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    Generally, it doesn't matter. Assume a direction. Since voltage drops each carry a +- sign, if you guess wrong, the algebra will take care of it.

    btw- I think this problem is a trick question. I compute the current to be zero.
     
  11. Jony130

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    Feb 17, 2009
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    No, in this circuit you have only one loop.
    And to avoid confusion try to stick to one "sign convention"

    http://hades.mech.northwestern.edu/index.php/Kirchhoff's_Current_and_Voltage_Laws#Kirchhoff.27s_Voltage_Law_and_Loop_Analysis
    http://web.engr.oregonstate.edu/~traylor/ece112/lectures/kvl.pdf
    http://www.niu.edu/iteams/documents/ueet602/KVL and KCL.pdf
    Very strange because I get I = 5/6A = 0.833A
     
    Last edited: Sep 4, 2014
  12. MikeML

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    Jony,

    Do you agree that if I = 5/6, then V(x)-V(y) = I/2 = (5/6)/2 = 0.416667V?
     
  13. Jony130

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    Isn't 2Vx is a voltage controlled voltage source?
     
  14. MikeML

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    I took it as being a voltage-controlled current-source with a transconductance of 2. That is why I got I = 0A.

    If I take it as a voltage-controlled voltage-source, then I get 0.83333A flowing from X to Y.
     
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