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Circuit Analysis

  1. Mar 15, 2006 #1
    Hi, I'm wondering if I made a mistake in getting the Kirchhoff equations to find all the currents.

    Here's what I got:

    Loop 1:
    [tex]10.0I_1 + 20.0I_2 + 30.0 I_4 = 15.0 V[/tex]

    Loop 2:
    [tex]40.0I_3 + 30.0 I_4 = 15.0 V[/tex]

    Loop 3:
    [tex]20.0I_2 + 30.0 I_4 = 10.0 V[/tex]

    Current Law:
    [tex]I_2 + I_3 - I_4 = 0 A[/tex]

    Then I can use Cramer's Rule (heheh) to find each individual current right? I'm afraid I don't know how to make matrices in LaTeX yet. :redface:
     

    Attached Files:

  2. jcsd
  3. Mar 15, 2006 #2
    Eek, why not just ref the matrix with your calculator??

    [tex]\left(\begin{array}{ccccc}10&20&0&30&15\\0&0&40&30&15\\0&20&0&30&10\\0&1&1&-1&0\end{array}\right)[/tex]

    That column of mostly zeros looks interesting if one were to do it by hand.....
     
  4. Mar 16, 2006 #3
    I don't really know how to use Maple either. :tongue:

    It'll ceratinly make my life easier finding the determinant manually.

    But is this system of equation compliant with the attached circuit drawing?
     
  5. Mar 16, 2006 #4
    I can't see your drawing until it gets approved. But you seem to know what your doing.

    Ref it on your Ti-83. Thats what I do.
     
  6. Mar 16, 2006 #5
  7. Mar 16, 2006 #6
    Did you use a supernode? Are you using mesh current analysis, or node voltage?
     
  8. Mar 16, 2006 #7
    Supernode? Sorry I'm not familiar with that term. :(
     
  9. Mar 16, 2006 #8
    Your equations seem good to me, but your picture has your resistors labeled I2 and I4 with the +,- backwards. Based on how you have them drawn, they would be negative voltages.

    Tell me what answers you get and I will compare it to what the computer simulation tells me. (im too lazy to do it by hand :smile:)
     
    Last edited: Mar 16, 2006
  10. Mar 16, 2006 #9
    Whoops, I4 was backwards, but not I2 though. So there's no I5 and I6 for each of the voltage sources?
     
  11. Mar 16, 2006 #10
    Your right, sorry I2 is fine. There is an I5 and I6, but you do not know them because you do not know the resistances. You can find the values of I5 and I6 by going back and applying KCL at each of the nodes after you find all the currents (if you want to.)
     
  12. Mar 16, 2006 #11
    That would make sense since I would know the current comining in and out after. Ok then, thank you for helping me out. :)
     
  13. Mar 16, 2006 #12
    Tell me what values you get and I can tell you if its right or wrong.
     
  14. Mar 16, 2006 #13
    for I1. I get... 13000? That can't be right...
     
  15. Mar 16, 2006 #14
    Whoops again, I have to divide that number by the determinant. :redface:
    I get:
    I1 = 0.5 A

    This is hard doing it manually. :s
     
  16. Mar 16, 2006 #15
    Whoa! I think this problem has been made complicated. I think I1 is the easiest current to find, as it has a potential difference of 5V across it.

    I would reccomend using nodal analysis to find the other currents, as it is a linear equation with 1 unknown variable. The only node where you don't know the voltage is the dot at the bottom of your diagram.

    Thoughts?

    Regards,
    Sam
     
  17. Mar 16, 2006 #16
    I'm at a loss. How is it that the top resistor at I1 has only 5.0V? Can you explain? So I only need to use Cramer's Rule for I2, I3 and I4? (I4's polarity is wrong in the diagram)

    Thanx
     
  18. Mar 16, 2006 #17
    You know that the potential at the right side of the resistor is +15V and the left side is at +10V, because the two cells are either side of it and they have their 0V at the same node (they are connected together).

    Sorry, I have to go, I'll try to log on again in 15 minutes...

    Sam
     
  19. Mar 16, 2006 #18
    He does not know about node voltage analysis yet. Just let him do it his way for now.

    I would urge caution in saying that they have their 0V at the same node. You do not know that the - is zero volts. All you know is the potential difference across the sources is 15V, you have no information what it is on each end relative to ground. For all you know it could be 10V-0v and -15v and 0 V. Then it would 0V and -15V adjacent to eachother, and not both zero.
     
  20. Mar 16, 2006 #19
    All potential is relative, I didn't say it was 0 V relative to ground. Ground is the usual reference for zero, but I used it as the lowest point of potential, so there are no negative voltages. I always try to simplify things before using maths, it makes life easier.

    As for Kirchoff's current loops, now there only need to be 2 loops, so that is a simulaneous equation with only 2 unknowns. I am surprized though that Loops are taught before nodal analysis.

    Sam
     
  21. Mar 16, 2006 #20
    Why, loops are the most basic forms. Node analysis is derived from KVL and KCL. You can't do it without first knowing loops.
     
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