Kirchoffs Current Law, Complex Circuit

Discussion in 'Homework Help' started by Alexj1402, Jan 24, 2015.

  1. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
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    Hi guys. Im new to forums as you can see and im struggling a little breaking down the circuit correctly. I have performed calculations and simulated it and found the exact values using Gauss_Jordans rule with a spreadsheet i made up.

    Here is the simulations with both the actual circuit with the shortened down version. I have simplified from the right to reach the point where i get a value of 362 ohms with a 10v supply

    https://www.dropbox.com/s/lpb574val39pa0g/Screenshot%202015-01-24%2015.14.49.png?dl=

    If it is correct, what would be my next step? Itotal = -i1 --i2 - i3 - i4?

    Your help will be appreciated.

    Thanks
    Alex
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Are you asking how to do it algebraically without simulating it?

    Oh, and would you mind posting your attachments here instead of an external site?
     
  3. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
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    Hi Mike,

    Thanks for the reply. I'm asking to work it out algebraically, i have done it using math to produced a matrix using Gauss Jordan, I would like to have a more visual method like breaking down the circuit as i have sort of got a current just figuring out whats best to work out each loop current.

    I apologise for the link, this is to my dropbox account but i have uploaded it here.

    Thanks
    Alex
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Combine R1//R2 with the familiar Ra =R1*R2/(R1+R2)
    Rb = R6+R7
    Rc=R5//Rb=R5*Rb/(R5+Rb)
    Rd=R4+Rc
    Re=Ra//Rd
    Rf=R3+Re.
     
  5. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
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    Thanks for the reply Mike,

    That was one of the resistance values i had calculate before this and i have 3 different number slighlty out E.g 362 and 355. (Probably my rounding up or down). So now i have the Value 28.25mA (10/354) do i then do branch work so i calculate Voltage drop across each resistor?

    Thanks
     
  6. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Once you know the total effective resistance, you know the current through R3, which leads to finding the voltage at the top of R3. Kirchoff tells us that I(R3) = I(Ra)+I(Rd), so everything else follows from there...
     
  7. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
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    Thanks for you help and support.

    I have now done a quick calculation using Loop theory for the 2 loops using R3, Ra, Rd. I have acheived the value of 5.46mA for the RD value and for Ra (22.7mA). If i wanted to split though values down even more to get to Current for each resistor whats the best approach, Using Ratio or plugging it into a simultaneous equation for the whole circuit?

    These are the loops i have used to mathematically calculate the currents.
    Loop1: 900, -120, 0, -0 = 0
    Loop2: -120, 390, -270, 0 = -10
    Loop3: 0, -270, 730, -130 =10
    Loop4: -0, 0, -130, 670 = 0

    I1 = -3.04mA
    I2 = -22.8mA
    I3 = 5.45mA
    I4 = 1.058mA

    Thanks
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Imagine someone came to you asking for help and gave you the information you have given us. Would you know what I1 and such refer to within the circuit? Would you know what Loop1 refers to within the circuit, let along what a comma separated list of four numbers being equal to zero means? Would you KNOW? Or would you have to guess and/or reverse engineer their work? Engineering is not about guessing. You need to be clear and complete. If you talk about Loop1, then I should be able to look at your diagram and see exactly what you mean by Loop1. If you talk about I3, then I should be able to look at your diagram and know exactly which current (and in which direction) I3 is referring to.
     
  9. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
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    I apologize for the lack of clarity, The way it is laid out in most equations for a KCL is in an order of I1, I2, I3 etcc...
    Loop1: 900i1, -120i2, 0i3, -0i4 = 0
    Loop2: -120i1, 390i2, -270i3, 0i4 = -10
    Loop3: 0i1, -270i2, 730i3, -130i4 =10
    Loop4: -0i1, 0i2, -130i3, 670i4 = 0

    I have attached another image referencing the loop and the current correspondence. If you have any other issue, then please let me know.

    Thanks
    Alex
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    That helps, but there's no indication of what direction each loop current is in. So if i2 comes out to be -10mA and i3 turns out to be +5mA, is the voltage is the voltage at the top of R3 greater than or less than 10V? You don't know, because you haven't defined what direction each current is in. It doesn't take that much to do it right:

    KVL2.png

    Also, a comma separated list of terms is not an equation.

    Code (Text):
    1.  
    2. I1:   (900Ω)I1 - (120Ω)I2                       =   0
    3. I2: - (120Ω)I1 + (390Ω)I2 - (270Ω)I3            =  10V
    4. I3:            - (270Ω)I2 + (730Ω)I3 - (130Ω)I4 = -10V
    5. I4:                       - (130Ω)I3 + (670Ω)I4 =   0
    6.  
    I wrote these down by inspection, which you will be able to do with a bit of practice. Note that mine differ from yours only in the sign of the right-hand side. That is almost certainly because you chose your currents in the other direction, but since you don't indicate the direction I can't tell.

    Note a simple check you can make: sum up all of the equations and you get

    Code (Text):
    1.  
    2. (780Ω)I1 + (0)I2 + (330Ω)I3 + (540Ω)I4 = 0
    3.  
    Now look at the diagram and do KVL around the outside loop. It should match this. This check will catch LOTS of mistakes that you might make in setting up your equations (but not all of them) and this is important because if you don't catch them then the equations you are working with from that point on are for a different circuit and the math doesn't care, so even if you do everything perfect from this point on you will simply end up with results that are correct for a different circuit.

    Note how I carried the units throughout the work (zero does not need units because zero is zero, but it doesn't hurt to put the units on zero as well).

    If you carry your units throughout your work, from beginning to end, then you will catch the vast majority of mistakes that you make (and you WILL make them) almost immediately and it will be easy to track down the ones that you don't. Otherwise you will make a mistake early on, spend hours, days, or weeks working through a design, tack the units you "know" it should have on at the end, not realize that you made that mistake that would have been glaringly obvious had you only tracked your units, and then sometime later wonder why you are having to explain to a jury about to consider what damages are appropriate for the wrongful death of some poor mother's teenage daughter why you didn't catch a mistake that would have been glaringly obvious had you only tracked your units.
     
  11. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
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    Hi Wbahn,

    I really appreciate all of your help, i have managed to solve it with some minor variances but very close to where i need it to be.

    My sole intention was not to get an answer out of you guys but to get some inspiration as to whether i was going in the correct direction which is where i was. Sometimes i lack confidence and double guess my self alot of the time, You are also right Wbahn. Minor mistakes happen all the time and i think because you are set at looking at one method seeing the bigger picture is not always obvious.

    Those numbers i have used with the comma's were from my matrix using Gauss-Jordan's rule which allowed me to achieve my previous results. That method in itself takes hours to get used to and didn't it just.

    Anyways, i do appreciate the help and i will be sure to stick around and help others in similar predicament.

    Thanks Alex
     
  12. WBahn

    Moderator

    Mar 31, 2012
    17,768
    4,802
    Glad you got it. As you learn more techniques and gain experience with them you will start being more confident at choosing the best (or at least a good) approach for a given problem. You won't always pick well, but you should be able to get the correct answer even if you pick the worst approach (it will just be a pain) and you should always be willing to step back and ask if the path you have chosen might not have been a good one and be willing to jump ship in mid stream when appropriate.

    You're doing fine. Keep up the good work.
     
    Last edited: Jan 25, 2015
  13. Alexj1402

    Thread Starter New Member

    Jan 24, 2015
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    Yeah that makes a lot of sense WBahn, im fairly fresh to Mechanical/Electrical engineering as i did ICT before this. Although i have got my head around alot of things i was just stumped. As you said, i need to take a step back and look at what im doing instead of steaming ahead and going completely wrong.

    Thanks for all the help once again!
     
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