Problem in circuit laplace and switch

Discussion in 'Homework Help' started by one(1), Nov 21, 2014.

  1. one(1)

    Thread Starter New Member

    Oct 6, 2014
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    I do this probem by my hand
    plese see my solution true?
     
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  2. one(1)

    Thread Starter New Member

    Oct 6, 2014
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    Hello again
    First thing I want understand the circuit how I can do analysis t<0 and t>0
     
  3. one(1)

    Thread Starter New Member

    Oct 6, 2014
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    if the circuit on t>0 is mean the ( V0 = 0 ) & ( Ix = 0 ) because there is not current flow in circuit
    why in picture 2.jpg there is source?
    from where the source ?
     
  4. one(1)

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    Oct 6, 2014
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    please , can any person told me from where came the source in analysis t>0?
     
    Last edited: Nov 21, 2014
  5. t_n_k

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    Mar 6, 2009
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    Is your concern that you don't know why there is current in the inductor at the instant the switch moves to position B?
     
  6. one(1)

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    Oct 6, 2014
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    yes I not understand how there is current in t>0 , i want understand from where came current source !
    please if you are know from where the source told me from where came current source ?
     
  7. t_n_k

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    Well if you consider firstly what is happening in the circuit when the switch is in position A.
    If the switch has been in that position (A) for a very long time, any transient associated with that situation would have vanished to zero. However one has to account for the current coming from the 15mA source. Given the inductor has zero DC resistance, all of the 15mA source current will eventually flow in the inductor rather than the 2 k ohm parallel resistor.
    So at the moment (t=0) when and the switch instantaneously moves to position B there will still be that 15 mA of current in the inductor - remember the current in an inductor can't change instantaneously. Hence for the analysis with the switch at position B, we must consider the inductor having an initial current of 15mA in the direction indicated.
     
    Last edited: Nov 21, 2014
  8. one(1)

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    Oct 6, 2014
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    please see my draw in multisim :
    I change the switch from A to B but the current in part B it is zero!
     
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  9. t_n_k

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    Multisim possibly can't handle an instantaneous switch transition as an ideal event in which the inductor current condition is preserved - a priori. I am unfamiliar with the use of multisim. Such an operation is theoretically / conceptually possible but impossible in practice, so one has to be careful in applying simulations in such circumstances.
    Besides, would you rather trust me or multisim?:rolleyes:
     
    Last edited: Nov 21, 2014
  10. JoeJester

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    Apr 26, 2005
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    Here are my questions with respect to your two diagrams in your post number 8:

    1. What is the time constants involved in each switch position?

    2. How do you measure current in a circuit?
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello there,

    As others have said, the current for the second part of the solution comes from the final current in the inductor that you get from part 1. The current from part 1 is the final current in the inductor, and that becomes the initial current for the circuit for part 2. Since the current was flowing downward in the inductor in part 1, that means it will be negative for part 2.

    This current is the initial current for the inductor for part 2. It is called the initial current because that is the current that exists in the inductor just before t=0 (for part 2 that is).

    I can not read your writing very well. When you communicate in technical matters you should really learn how to convey your questions a little more clearly. It's better to type whenever possible then to write out unless you learn to write very very neatly. If you took any drafting cless in the past you would learn to write very very neatly, but with practice you can do it on your own too. You will find that people can understand you much faster.

    Since i cant read your writing very well i will post the result of part 2 and you can compare with your own result:
    Vc(t)=4.8e-6*(e^(-2500*t)-e^(-0.0004*t))

    Note that gives us 4.8uv across the capacitor at 't' somewhat large but not very large. At t => +infinity the solution tends to zero but it takes a long time to get there, so the solution may look like it settles at 4.8uv but it really very very gradually dies down to zero. So if you plot it you have to plot for a short time like 5ms, then again for a long time like 10000 seconds to see the whole response plot. This kind of response would be called "stiff" because there are two time constants that are very far apart: one that happens very fast and the other appears almost to not move at all.

    Also, rather than use a switch to turn the circuit to the topology of part 2, instead use an initial condition generator for the inductor current and dont use a switch just start the simulation at t=0.
     
  12. one(1)

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    Oct 6, 2014
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    obs ! sorry for my wrong
    thank you for explain
    there is problem in t>0
    in t>0 there are inductor and capacitor and resister
    the inductor change to current source
    I can use method V0 = (R + Zc ) * ( Ix)
    if I use method V0 = ( Zc ) * ( Ix) the answer is :
    http://www.wolframalpha.com/input/?i=partial fraction (30)/(s*(0.8s+2000))((800000)/(s))
    and If use another method V0 = (R + Zc ) * ( Ix) the answer is :
    http://www.wolframalpha.com/input/?i=partial fraction (30)/(s*(0.8s+2000))(((800000)/(s))+2000)
    I think use first method ,but my answer not same your answer
    please If can you see my solution :
     
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    Last edited: Nov 22, 2014
  13. MrAl

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    Jun 17, 2014
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    Hi,

    From what you wrote it looks like you are going about it in a way that ignores the inductor, which is not good. That's what it looks like and from your statements quoted here:
    "the inductor change to current source"
    "I can use method V0 = (R + Zc ) * ( Ix)"

    it looks like you want to change the inductor to just a current source, and sorry to say that's not correct.

    What you have to do is connect the current source in parallel to the inductor. That leaves the inductor there along with the current source. That also means that to calculate the voltage at the top of the inductor you must put the capacitor and resistor R+zC in parallel to the inductor, then you can apply the current Ix.

    For example:
    Z=zL in parallel with (R+zC)

    Then voltage at top of inductor to ground:
    vL=Ix*Z

    and then voltage at top of cap to ground:
    Vc=vL*zC/(R+zC)

    So if you first put inductor in parallel with R+zC and also in parallel to the current source, you'll get the right result.

    Note that you never loose the inductor. The inductor is what takes the voltage at the left side of the resistor to zero after a certain time by passing all of the initial current after that time, that's where all the initial current has to go (for the circuit theory to work anyway)
     
    Last edited: Nov 22, 2014
  14. one(1)

    Thread Starter New Member

    Oct 6, 2014
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    hi
    I do the problem but the answer is different !
    please see this :
    Vo= Ix * Z
    http://www.wolframalpha.com/input/?...00+2000s+(0.8)(s^2)+500000+2000s++2000000000)

    and laplace transform for V0:
    http://www.wolframalpha.com/input/?i=laplace transform (((24s+9600)/(800000+2000s+(0.8)(s^2))+(((24s+9600)/(800000+2000s+(0.8)(s^2)+500000+2000s +2000000000)
     
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  15. one(1)

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    Oct 6, 2014
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    how got : Vc(t)=4.8e-6*(e^(-2500*t)-e^(-0.0004*t)) ??:(
     
  16. t_n_k

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    Mar 6, 2009
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    Yes - it looks like MrAl may have made a mistake.

    With the circuit values given in the Laplace form:

    Ls=0.8s ohm, Hence L=0.8H
    1/(Cs)=0.8/s M ohm, Hence C=1.25uF
    R=2k ohm

    The characteristic equation would then be

    s^2+\frac{R}{L}s + \frac{1}{LC}= s^2+2500s+100000

    for which the real roots are 500 & 2000

    From which the solution for the capacitor voltage would be more like

    V_c(t)=-8\(e^{-500t}-e^{-2000t}\) \text{\ Volts}
     
    Last edited: Nov 22, 2014
  17. one(1)

    Thread Starter New Member

    Oct 6, 2014
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    nice,
    see what I did in circuit :
    parallel Z=(R + Zc ) || Zl
    Vo = Z * Ix
    Ix= 15m( 1/s + 1/( s+2.5K)
    Z= ( ( 1.6Ks + 640K) /(( 2K +0.8M/s) + (0.8s))
    I got the answer is:
    V0 =
    http://www.wolframalpha.com/input/?i=laplace transform ((0.015/s+0.015/s+2500))*((1600s+640000)/((2000+800000/s)+(0.8s)))
     
    Last edited: Nov 22, 2014
  18. t_n_k

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    Firstly, I'd rather see you solving the inverse transform yourself rather than relying on the Wolfram web site. In my opinion, relying solely on such resources is effectively encouraging a lack of academic rigor. The problem at hand is readily solvable with a modest level of mathematical skill when you correctly approach the problem.

    While I have also applied the Laplace Transform method to the solution, I note that the original question attached in post #1 specifically calls for a "Time Domain Transient" analysis. Presumably, this involves the solution of a second order differential equation in the time domain - perhaps a lack of rigor in the problem statement. I simply note this out of curiosity, notwithstanding the fact that the included schematics have elements designated by their Laplace form values. What do you think?

    I'm also puzzled by a couple of things
    • your duplicate nomenclature for the source current Ix and the inductor current Ix in the included image 8.jpg in post #14.
    • your Laplace formulation for the current Ix shown in your last post, which I believe is possibly incorrect depending (at least) upon the resolution of the previous dot point.
    • your reason for multiplying the current Ix by the impedance Z to find the voltage Vo which (with reference to the original circuit) is in fact just the capacitor voltage - not the voltage across the impedance Z.
     
    Last edited: Nov 22, 2014
  19. one(1)

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    Oct 6, 2014
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    hi
    first I should do second order differential equation in the time domain?
    please see my schematic what I will do :
     
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  20. MrAl

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    Jun 17, 2014
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    Hello again,

    I see what happened here. The chicken scratch was too hard to read so i read the capacitor value as 1/0.8 Farads, not 1.25uf. The big "M" apparently was intended to show a multiplier by a factor of 1e6. This of course changes the solution. It's better to use notation like 1e6 or 1e-6 rather than "m" and "M". This happened with the current amplitude also, which i had mistaken for 15 amps not 15 ma in the beginning.

    I'll repeat the values so we are all on the same page:
    R=2000 ohms
    L=0.8 Henries
    C=1.25 microfarads
    I=0.015 amps

    The solutions then should matche the others:
    Vc(s)=12000/((s+500)*(s+2000))
    Vc(t)=8*(e^(-2000*t)-e^(-500*t))

    So you can verify with this if you like.


    The solution i was doing previously was for C=1.25 Farads, which comes out to:
    Vc=4.8e-6*(e^(-0.0004*t)-e^(-2500*t))

    which is actually a more interesting solution because the time constants are so far apart.
     
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