Help, finding current and voltage using Kirchhoff's Law

Discussion in 'Homework Help' started by Luan Cristian Thums, Sep 22, 2014.

  1. Luan Cristian Thums

    Thread Starter New Member

    Aug 29, 2014
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    What I tried, so far, is using both KLs, ended up with a bunch of equations and found I1 = -0.0367A, I2 = -2.94A and I3 = -3.19A. What I'm supposed to find are the currents I1, I2, I3 and the voltages V1 and V2.

    Can someone point me in the right direction here? What is the best approach to solve this problem?

    I'm attaching a image of the circuit.

    Thank you.
     
  2. wmodavis

    Well-Known Member

    Oct 23, 2010
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    Write three loop equations.
    Use standard mathematical techniques to solve for the unknowns I1, I2 & I3.
    With those you can use Ohms law to determine the voltage drops V1 & V2 across the two resistors.
     
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  3. Jony130

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    Feb 17, 2009
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    Notice that V1 = 20V + 7*I3 , and I3 = 5*V1
    So form there we can find V1 and I3
    V1 = 20V + 7*5*V1 ---->V1 = -20/34 = -0.588235V
    And I3 = 5*V1 = 2.9411A

    And V2 = V1 + 2*I3
     
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  4. shteii01

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    Feb 19, 2010
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    Your pic is 3 megabytes... I aint downloading that.
     
  5. Luan Cristian Thums

    Thread Starter New Member

    Aug 29, 2014
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    Thank you very much for your prompt answers, I really don't have time to focus on that now, but I'll redo the problems tomorrow and reply to this thread with more information.

    Sorry about that, it most certainly could have been resized.
     
  6. JoeJester

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    Apr 26, 2005
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    here's a more reasonable size ....
     
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  7. eduardo schmoller

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    Sep 23, 2014
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    I found: I1 = -0.04 A, I2 = -2.94 A, I3 = 0.22 A, |V1| = 0.64 V, |V2| = 6,6 V
     
  8. Jony130

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    Feb 17, 2009
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    Are you sure that your answer is correct ?
    From KVL we have V1 = 20V + 7I3 and additional I3 = 5V1 so if we solve this for V1 we get this result
    http://www.wolframalpha.com/input/?i=V_1 = 20 + 7*5*V_1
    And from KVL V2 = Va - 2I3 = 90/17 = 5.294V
     
    Last edited: Sep 23, 2014
  9. MikeML

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    Oct 2, 2009
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    207.jpg


    Here is what LTSpice says:
     
    Last edited: Sep 23, 2014
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  10. The Electrician

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    Oct 9, 2007
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  11. The Electrician

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    Here's what I get:
    Probx.png
     
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  12. Luan Cristian Thums

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    Aug 29, 2014
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    Thank you sir, I got the same results!

    Noticing that I3 = 5V1 and knowing that V1 = 20 + 7I3 I found I3 and V1, to later derive V2 from V2 = V1 - 2I3 (second loop).
    Having the voltage drop on both resistors and knowing their resistance I used Ohm's law to get both still unkowned currents.

    I believe that V2 = V1 - 2I3
     
  13. MikeML

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    Oct 2, 2009
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    Fixed.
     
  14. Luan Cristian Thums

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    Aug 29, 2014
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    There is 2 basics answers on this thread (I believe?!) one coming from V2 = V1+2I3 and the other from V2 = V1-2I3.
    Which one is correct?
     
  15. Luan Cristian Thums

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    Aug 29, 2014
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    Thank you very much for your help, specially for introducing me to this software. But R on the second resistor is 30 ohms.
     
  16. Jony130

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    Well because 2I3 is connect backwards the correct one is V2 = V1-2I3
     
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  17. MikeML

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    Also fixed:
     
  18. The Electrician

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    In your simulation, I(B3) has a different polarity than I(V3); this will give different results for those places that have a defining equation involving I3 as defined in the original circuit. It looks to me like the current direction of B3 is the one that agrees with I3.

    For example, your defining equation for the voltage source B1 is 7*I(V3), but I(V3) is going in the opposite direction compared to I3.

    What do you get if you change the definitions of B1 and B2 so they use I(B3) rather than I(V3)?
     
  19. MikeML

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    The current in B3 and V3 is going in the same direction. LTSpice's convention (conventional current, not electron current) is that the current that flows out of the + terminal of a voltage source carries a negative sign. See the attached circuit:

    207a.jpg

    Since V3 is inverted, then -(I(V3)) is positive.

    A radically different (but correct) solution. This is because of the way that LTSpice handles Behavioral Voltage and Current sources.

    207b.jpg
     
  20. The Electrician

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    In post #9, you've shown an LTSpice simulation, but you didn't say whether you believe it's correct. Do you?

    Looking at the original circuit, the left hand mesh shows that V1 = 20 + 7*I3. The right hand mesh shows that I3 = 5*V1. Making the obvious substitution, we get V1 = 20 + 5*7*V1; from this we get V1 = -20/34=-.588235

    This is so simple that surely it is not mistaken, but this isn't what LTSpice gets in post #9. Is there something I'm not understanding?
     
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