Laplace circuit

Discussion in 'Homework Help' started by full, Nov 4, 2014.

  1. full

    Thread Starter Member

    May 3, 2014
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    hello everyone

    I do this problem on Laplace circuit

    please see my solution true?

    thanks
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    What have you done to check your answer?

    For instance, what do you expect the voltage to be once steady state is achieved? Does that agree with your answer?

    Always, always, ALWAYS ask if the answer makes sense.

    If you want me to look over your work, then you will need to track your units properly.
     
  3. full

    Thread Starter Member

    May 3, 2014
    225
    2
    I do 2 loops in this circuit but I think there is wrong in inverse Laplace in last
    because I new in Laplace
    please see the inverse Laplace in last
    thanks sir
     
  4. full

    Thread Starter Member

    May 3, 2014
    225
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    there is any person help my ;):rolleyes:
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Your problem is not in your inverse transform, it is a bit before that. But, as I said, I'm not going to assist with it unless you properly track your units. If you don't care enough about getting your work correct to avail yourself of arguably the single most effective error detection tool available to the engineer, why should I?

    Besides, you need to start relying on yourself to review your work and find your own mistakes -- there usually won't be someone to find them for you as a practicing engineer, since people are paying you to solve their problems in the first place.
     
    Last edited: Nov 4, 2014
  6. full

    Thread Starter Member

    May 3, 2014
    225
    2
    yes , I want check my solution with myself but I don't know how!
    from where I can start to check my solution?

    sir ,and how I can check my answer true?

    thanks sir
     
    Last edited: Nov 4, 2014
  7. WBahn

    Moderator

    Mar 31, 2012
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    Start by answering the questions I have asked you several times already:

    1) What do you expect the answer to be at t=0?

    2) What do you expect the answer to be at t=∞?

    3) Does your solution agree with both of those expectations?
     
  8. full

    Thread Starter Member

    May 3, 2014
    225
    2
    what is mean t=0?
    is mean capacitor =open circuit , inductors =short circuit?
    and get to V0= ,I0= ?

    thanks sir
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    You are asked to find vo(t), right?

    What is 't' in that expression? Time!

    What do you expect the result to be at time equals zero? What do you expect the result to be at time equals infinity (in other words, after a sufficiently long time such that all of the transients have died out and the system is in steady state)?

    You are not asked for any current, you are asked for vo(t), so what is vo(t=0) and what is vo(t=∞)?
     
  10. full

    Thread Starter Member

    May 3, 2014
    225
    2
    I don't know what happen in t=0 and t=infinity
    how I'm expect?

    thanks sir
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    You are expected to apply what you are supposed to have already learned.

    What are the constraint equations regarding instantaneous changes in voltage and current for a capacitor and an inductor?

    What are the voltages and currents in the capacitor and inductor just before t=0 (i.e., at t=0-)?

    Given the constraints, what do you know about the voltages and currents in the capacitor and inductor just after t=0 (i.e., at t=0+)?

    What does this then require about he current through and the the voltage across the resistor be at t=0+?

    How do capacitors and inductors behave in DC steady state?

    What does this then require about he current through and the the voltage across the resistor be at t=∞?
     
  12. full

    Thread Starter Member

    May 3, 2014
    225
    2
    update my solution in post #1:
     
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  13. WBahn

    Moderator

    Mar 31, 2012
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    Does this satisfy the conditions you expect the voltage across the resistor to conform to at t=0 and t=∞?

    What about for t<0?
     
  14. full

    Thread Starter Member

    May 3, 2014
    225
    2
    What are the constraint equations regarding instantaneous changes in voltage and current for a capacitor and an inductor?
    not a lot.

    What are the voltages and currents in the capacitor and inductor just before t=0 (i.e., at t=0-)?

    capacitor =open circuit (current =0)
    inductor = short circuit

    Given the constraints, what do you know about the voltages and currents in the capacitor and inductor just after t=0 (i.e., at t=0+)?

    I don't know

    What does this then require about he current through and the the voltage across the resistor be at t=0+?

    is mean 't' gather than zero

    How do capacitors and inductors behave in DC steady state?


    capacitor open circuit

    inductor short circuit

    What does this then require about he current through and the the voltage across the resistor be at t=∞?

    zero
     
  15. full

    Thread Starter Member

    May 3, 2014
    225
    2
    t<0
    capacitor and inductor =assuming zero
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    Well, thanks for at least trying to answer the questions, finally.

    You have a lot to learn about basic electronics and you really need to get a handle on them before you go much further since, as you can see, it is already crippling your ability to work problems and check the reasonableness of your own work.

    What are the constraint equations regarding instantaneous changes in voltage and current for a capacitor and an inductor?
    not a lot.

    WRONG! Because energy is stored in a capacitor in terms of the voltage across it and in an inductor in terms of the current through it, and because energy is a conserved quantity and therefore can't change instantaneously, neither the voltage across a capacitor nor the current through an inductor can change instantaneously. This constraint does NOT apply to either the current through a capacitor or the voltage across an inductor, which can change radically in an instant. These constraints are the basis for how many common electrical circuits work, ranging from automotive ignition systems, switch-mode power supplies, and many others. It is also why ignoring these principles and doing something such as open-circuiting a high-inductance magnet can kill you instantly.

    What are the voltages and currents in the capacitor and inductor just before t=0 (i.e., at t=0-)?

    capacitor =open circuit (current =0)
    inductor = short circuit

    WRONG! Even more to the point, it is a complete non-answer to the question. Before t=0 there are no sources that are active in the circuit, thus there are no currents or voltages present anywhere in the circuit.

    Given the constraints, what do you know about the voltages and currents in the capacitor and inductor just after t=0 (i.e., at t=0+)?

    I don't know

    CORRECT -- albeit that, in and of itself, is a problem.

    Because the voltage across the capacitor was zero just before t=0, it remains zero just after t=0. Similarly, because the current through the inductor was zero just before t=0, it remains zero just after t=0. Because the inductor is in series with the resistor, that means that it, too, has no current flowing through it at t=0+ and, therefore, the initial voltage across it must also be zero.

    We don't know anything about the current through the capacitor or the voltage across the inductor, at least not based on the constraints directly, but the circuit is simple enough for us to figure them out by inspection. Since the current in the inductor is zero just after t=0, all of the current must be going though capacitor. Because the capacitor, which starts out with zero volts across is, is in parallel with the inductor-resistor branch, and because we've already established that the resistor has zero volts across it initially, the voltage across the inductor is also zero volts. This fact can help us determine yet another sanity check for our result that hasn't been hinted at yet and we'll come back to that later.

    What does this then require about he current through and the the voltage across the resistor be at t=0+?

    is mean 't' gather than zero

    No. Look at the prior question. "just after t=0 (i.e., at t=0+). The notation t=0- means just a tiny, immeasurable amount of time before t=0 and the notation t=0+ means just a tiny, immeasurable amount of time after t=0. Not one hour after t=0, not one second after t=0, not one nanosecond after t=0, but just immediately after t=0.

    This question is answered as part of the question above: "Because the inductor is in series with the resistor, that means that it, too, has no current flowing through it at t=0+ and, therefore, the initial voltage across it must also be zero."

    How do capacitors and inductors behave in DC steady state?

    I don't know

    CORRECT -- albeit that, in and of itself, is a problem.

    What does "steady state" mean? It means unchanging. If something is unchanging, that means that it's time derivatives are zero, right? After all, the time derivative tells you the rate at which something is changing with time and if it is unchanging, this rate must be zero. For a capacitor, the current is proportional to the time derivative of the voltage which means that, in DC steady state, there is no current flowing in it. Thus, it behaves like an open circuit. For an inductor, the voltage is proportional to the time derivative of the current which means that, in DC steady state, there is no voltage across it. Thus, it behaves like a short circuit.

    What does this then require about he current through and the the voltage across the resistor be at t=∞?

    zero

    WRONG! If this were correct, then that would mean that your latest answer is wrong because it has a voltage of 20V (I'm assuming it's "volts" since you can't be bothered to even indicate units on your answers -- inexcusably sloppy).

    Since the capacitor looks like an open circuit in DC steady state, that means that all of the current from the source must eventually go through the resistor, producing a 20V drop across it. We also know that the inductor has no voltage drop across it in DC steady state, so we know that the capacitor is eventually charged to 20V as well.

    Getting back to an earlier point, because we know that the initial voltage across the inductor is zero (due to the clamping behavior of the capacitor), and because we know that for an inductor di/dt is equal to v(t)/L, we know that di/dt for both the inductor and the resistor is zero at t=0+ which means that dv/dt for the resistor is also zero. That is another check we can make.

    So let's consider your proposed solution:

    vo(t) = 20V [1 - e^(-t/0.5s) - (1/0.5s)·t·e^(-t/0.5s)] u(t)

    At t=0+, this reduces to vo(0+) = 20V [1 - 1 - 0] = 0V (good)
    At t=∞, this reduces to vo(∞) = 20V [1 - 0 - 0] = 20 V (good)

    What about dvo/dt at t=0+?

    dvo(t)/dt = 20V [ (1/0.5s)·e^(-t/0.5s) - (1/0.5s)·e^(-t/0.5s) + (1/0.5s)²·t·e^(-t/0.5s)] u(t)
    dvo(t)/dt = 80V/s² · t · e^(-t/0.5s) u(t)

    which, at t=0+, is zero.

    And notice how the units work all work out properly.
     
    full likes this.
  17. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    You are give the initial conditions: "assuming zero initial conditions."
    The capacitor is not charged, the inductor is not charged.

    At t=0 the current source is conected into the circuit. The capacitor and inductor do not change instantly.

    At t=infinity, the capacitor is fully charged and is seen as open circuit, inductor is also fully charged and is seen as short circuit. So all the current will be going through 4 Ohm resistor. Vo(infitity)=4*5u(infinity)
     
    full likes this.
  18. full

    Thread Starter Member

    May 3, 2014
    225
    2
    thanks sir ,you are great man ,I will learn a lot from this
     
  19. full

    Thread Starter Member

    May 3, 2014
    225
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    capacitor : 1 / (1/4)s = 4/s

    thanks sir
     
  20. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    You are right. My mistake.
     
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