loop mesh analysis

Discussion in 'Homework Help' started by lemon, Dec 16, 2011.

  1. lemon

    Thread Starter Member

    Jan 28, 2010
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    Hi:
    I have the attached circuit to analyse using Loop and Mesh analysis.
    I think I am ok with this method when all components are resistors, but in this circuit there are also a capacitor and an inductor. Do I just continue with the analysis as normal, ignoring the non-resitor components, or do I need to account for them in some way?
    thanks
     
    Last edited: Dec 16, 2011
  2. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    You need to account for them.

    You also need to account for time

    Assuming the sources are DC, which they are marked as,
    Call the point at which the power goes on as t=0

    At t=0, the inductor will look like an open circuit, and the capacitor will look like a short circuit.

    at t >> 0 the inductor will appear as a short circuit, and the capacitor will look like an open circuit.
     
  3. steveb

    Senior Member

    Jul 3, 2008
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    The thing is, it's not clear what the exact question is. The problem is vague as stated. The voltage sources seem to be written as batteries, which implies DC. Also, there is no switch to indicate that the battery voltages are switched in.

    If the problem is asking for the DC solution, then the method is to leave all capacitors open and to short all coils. Then solve in the normal way for resistors.
     
  4. lemon

    Thread Starter Member

    Jan 28, 2010
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    I'm totally lost by this. I'm only just learning the standard loop mesh analysis with resistors. Could somebody direct me to the pages where I can study this method with circuits like the one above please?

    @SteveB
    What facts would you need to make this problem clear SteveB?
     
    Last edited: Dec 17, 2011
  5. steveb

    Senior Member

    Jul 3, 2008
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    You haven't clearly defined the problem to us, and perhaps not even to yourself.

    Are you trying to do a DC solution? If so, the method I mentioned is correct and simple, and uses techniques you already know.

    Are you trying to solve a transient solution where the batteries are switched in? if so, thatoneguy is giving you the correct gist of it, and we can direct you to more formal theory on that method.

    Or, maybe your DC voltages are mislabeled and they are supposed to be AC sources with given frequency? If so, that would be a different approach (called phasor analysis).

    Basically, before you can find an answer, you have to know the question. Also, in order for us to help properly, you have to inform us of the constraints of the question.

    My gut feeling is that the first case is the one of interest, but you need to confirm it, as this is just a guess on my part. If that is true, then the coils are capacitors are inserted to give you a new twist to your method of finding DC solutions. You need to know that caps are like opens and coils are like shorts; so, you are basically replacing these components with equivalent resistance.
     
  6. thatoneguy

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    Feb 19, 2009
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    Assuming DC Analysis (t>>0) [time is much greater than zero, and a steady state has been reached]

    The inductor is just more wire with resistance of 0Ω, and the capacitor would be completely erased.

    You could then solve for currents from the various voltage sources, assuming the inductor is a short circuit and there is nothing where the capacitor is shown.

    Assuming Transient analysis:

    It gets fairly difficult, so if you haven't been shown how to calculate transient or AC analysis in class, assume DC analysis.
     
  7. lemon

    Thread Starter Member

    Jan 28, 2010
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    So, is the attached file looking ok?
     
  8. thatoneguy

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    If you erase the capacitor, there won't really be a loop 3
     
  9. lemon

    Thread Starter Member

    Jan 28, 2010
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    I don't see how that would happen. If I remove the capacitor, I can see that maybe loop 2 will disappear. But then there are two voltage sources which would make it impossible to put the results in matrix form.

    So if I do the analysis using the same technique as before and loop 2 goes, I get: Please see attached image
     
    Last edited: Dec 22, 2011
  10. lemon

    Thread Starter Member

    Jan 28, 2010
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    Actually, I should have shown the file which our tutor gave us. He stated that the file 'CY2B9 Electronics Tutorials Autumn 2011 v4' is a guide for the questions in the Assignment CY2B9 Assignment v8. Both these files can be found here: https://public.me.com/defunktlemon

    He said:

    Ignoring the first 4 pages , hand-written tutorial questions/model answers are numbered from 1 to 23.

    Assignment Q1 has Examples 1-4. But to be honest I'm having difficulty seeing the relationships as none of these circuits are RLC.

    If anyone had time and patients please help and advise on this.
    thank you.
     
  11. lemon

    Thread Starter Member

    Jan 28, 2010
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    Hi:
    Actually I may have confused things here. The question didn't ask me to perform loop-mesh analysis. The question only stated:

    For the network shown in Figure Q1-1, perform the loop analysis. But that probably means the same thing right.

    If so, If I assume it is a DC network and leave all capacitors open and to short all coils. Then solve in the normal way for resistors, I will only have two loop currents and the question seems to be asking for 3. Would I just assume, therefore, that i2 and i3 are equal?
     
  12. thatoneguy

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    I1=I2=0

    All the current will be in I3, as there is no path for i1 and i2 as drawn.
     
  13. lemon

    Thread Starter Member

    Jan 28, 2010
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    sorry i can't understand why that is:
    I thought it would look like the image attached
    except without the inductor - put that in by mistake
     
    Last edited: Jan 2, 2012
  14. thatoneguy

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    Feb 19, 2009
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    That is correct, after seeing redrawn, had a brain fart moment.

    It should solve fine if you have the voltages, treat the inductor as a piece of wire, the same way you totally erased the cap.
     
  15. lemon

    Thread Starter Member

    Jan 28, 2010
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    ok- I have a reply from my query to my tutor. He said, all that is required to do is: Please first draw the corresponding current loops and write the corresponding expressions for them assuming voltage drops across the elements
     
  16. thatoneguy

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    Did the tutor state your answer was incorrect?
     
  17. lemon

    Thread Starter Member

    Jan 28, 2010
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    i haven't handed this in yet
     
  18. lemon

    Thread Starter Member

    Jan 28, 2010
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    so - based upon what I have drawn last
    and what my tutor has said
    should i take it that the drawing is correct, except I should delete the inductor, and i should now write the corresponding expressions for them assuming voltage drops across the elements?
     
  19. thatoneguy

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    Yes, just do the calculation with inductors as wire, and capacitors as open circuits.
     
  20. lemon

    Thread Starter Member

    Jan 28, 2010
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    How is my start?
     
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