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Homework Help: Circuit Analysis: Simple Problem

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data
    We are given the following equations in my circuit analysis class.
    [itex]q(t) = 2 - 2e^{-100 \cdot t}[/itex]
    Where I am told to find [itex] q(0) = ?, q(\infty) = ?, I(0) = ?, I(\infty) = ?[/itex]
    q represents the number of coulombs in a charge of a circuit I believe?
    I represent the amount of current any a given time, where time t is defined in milliseconds



    2. Relevant equations



    3. The attempt at a solution
    I know charge is defined as number of coulombs per second, 1 coulomb per 1 second.
    My main problem is I am getting these terms and concepts confused, now my first problem I would like to address is now that because time is defined in milliseconds, using the appropriate unit conversion so to speak, would my solutions come to millicoulombs per milliseconds and milliamperes per milliseconds ? I know to evaluate the current it is a simple integral which I know how to evaluate with a limit of integrating containing infinity using the method of improper integrals using limits, How would I evaluate such a thing as q(t) at "infinity"? I know that there is some sort of simple trickery my professor tried implying here. All help is appreciated. I just need some clarifications not exact solutions.
     
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  3. Jan 16, 2010 #2
    Simple Circuit Analysis Concepts

    This is not an homework question, I posted there just in case as well, this was a class given, I took on my own initiative to do instead of waiting for the next class

    Given the following equation in my circuit analysis class.
    [itex]q(t) = 2 - 2e^{-100 \cdot t}[/itex]
    Where I am told to find LaTeX Code: [itex]q(0) = ?, q(\\infty) = ?, I(0) = ?, I(\\infty) = ?[/itex]
    q represents the number of coulombs in a charge of a circuit I believe?
    I represent the amount of current any a given time, where time t is defined in milliseconds


    I know charge is defined as number of coulombs per second, 1 coulomb per 1 second.
    My main problem is I am getting these terms and concepts confused, now my first problem I would like to address is now that because time is defined in milliseconds, using the appropriate unit conversion so to speak, would my solutions come to millicoulombs per milliseconds and milliamperes per milliseconds ? I know to evaluate the current it is a simple integral which I know how to evaluate with a limit of integrating containing infinity using the method of improper integrals using limits, How would I evaluate such a thing as q(t) at "infinity"? I know that there is some sort of simple trickery my professor tried implying here. All help is appreciated. I just need some clarifications not exact solutions.
     
  4. Jan 16, 2010 #3
    Re: Simple Circuit Analysis Concepts

    Actually, no integration is necessary here.

    An important concept to keep in mind is that the current is the rate of change of charge with respect to time. In other words to find an equation for the current, the only thing that you need to do is take the derivative of q(t).

    Since they are only asking for charge and current at t=0 and t=infinity, you do not need to worry about units, as 0 milliseconds is exactly the same as 0 seconds. For t=infinity, just take the limits of both equations.

    Hope this clarified things.
     
  5. Jan 16, 2010 #4
    Re: Simple Circuit Analysis Concepts

    Current is the first derivative of charge:

    I(t) = d q(t)/dt

    Bob S
     
  6. Jan 16, 2010 #5
    Re: Simple Circuit Analysis Concepts

    It seems that the equation is designed taking into account the fact that when you substitute time in milliseconds the solution for charge comes in coulomb. You can be sure of the solution to be in coulomb unless and otherwise specified in the question.
     
  7. Jan 17, 2010 #6
    Re: Simple Circuit Analysis Concepts

    I listed all possible data that was given, he stated t was in milliseconds and the equation for charge was [itex]2-2e^{-100t}[/itex]

    So I am wondering as to writ the solution for charge in terms of millicoulombs and milliamperes or using the fact every 1000 milliseconds equates to 1 ampere and 1 coulomb, only part I am confused on, Thanks for the clarification fellas, I just have to plot this graph but I just took the limit to infinity of the equation for charge so far and it obviously equates 2 coulombs constant for any time greater than zero or 2000 millicoulombs
     
  8. Jan 17, 2010 #7

    tiny-tim

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    Hi IronBrain! :smile:

    (have an infinity: ∞ and try using the X2 tag just above the Reply box :wink:)
    There's probably some simple rule, but it's safest (and I prefer) just to convert to seconds … it minimises the chance of mistakes! :rolleyes:

    (and millicoulombs per millisecond and milliamperes per millisecond (no "s" :wink:) are just coulombs per second and amperes per second)
    Simple rules: e+∞ = ∞, e-∞ = 0 (look at the graph!). :smile:
     
  9. Jan 17, 2010 #8

    vela

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    If you write q(t) (I assume q is in coulombs) with the units explicitly written in, it would be

    [tex]q(t) = 2~\mbox{C} - (2~\mbox{C}) e^{-(100~\mbox{ms}^{-1}) t}[/tex]

    When you differentiate that, the -100 ms^-1 comes down from the exponent, so the units would work out to be C/ms.
     
  10. Jan 17, 2010 #9
    Re: Simple Circuit Analysis Concepts

    Another question as I differentiate q(t) which equates to:
    [itex]I(t) = \frac{dq(t)}{dt} = 200e^{-100t}[/itex]

    Now when I take the limit as t goes to infinity I evaluate it to:
    [itex]\lim_{t\to\infty}200e^{-100t}=0[/itex]

    Correct? So what does this mean? Initial the current at time t=0 is 200 amperes or 200,000 milliamperes, at t > 0, the flow of current begins to dissipate as time increases? I do not think I accessed this correctly

    Next...

    I evaluate q(t) limit to infinity as:
    [itex]\lim_{t\to\infty}2-2e^{-100t}=2[/itex]

    Meaning that at any time of t > 0, the charge in coulombs are 2C or 2000 milliCoulmbs ever second or every 1000 milliseconds? To me this does not seem to work in conjunction of the evaluation of current at any time t > 0, Just need a bit of clarifiaction
     
  11. Jan 17, 2010 #10

    tiny-tim

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    I really dislike that …

    the "something" in esomething should always be dimensionless.
     
  12. Jan 17, 2010 #11

    berkeman

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    Two threads merged. Please do not multiple-post.
     
  13. Jan 17, 2010 #12

    vela

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    It is dimensionless. The constant in front of the t has to have units of 1/time to make the argument of the exponential dimensionless.
     
  14. Jan 17, 2010 #13

    vela

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    Re: Simple Circuit Analysis Concepts

    The initial current is 200,000 A, actually. The coefficient has units of

    [tex]\frac{\mbox{C}}{\mbox{ms}} = \frac{\mbox{C}}{\mbox{ms}} \times \frac{1000 \mbox{ms}}{1 \mbox{s}} = 1000 \frac{\mbox{C}}{\mbox{s}} = 1000~\mbox{A}[/tex].

    Your description of the current's behavior is correct. It starts off at its maximum value and decays to 0 as [itex]t \rightarrow \infty[/itex].

    I'm not sure what you're trying to say here. You've calculated the limit at infinity, so no, it's not for "any time t>0." And charge is measured in coulombs, but you're talking about coulombs per second which is units of current. Are you are asking how can the current be so large initially yet only a relatively small amount of charge accumulates?
     
  15. Jan 17, 2010 #14
    Re: Simple Circuit Analysis Concepts

    Yes.
    The few posts above some what confused me, I'll analyse and repost my thoughts.
     
  16. Jan 17, 2010 #15

    vela

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    The current is that large only for an instant, and it falls off very quickly. After just one millisecond, the current has already dropped off to [itex]i=7.44\times10^{-39}~\mbox{A}[/itex].
     
  17. Jan 17, 2010 #16
    What about the charge since the limit of the equation q(t) is 2 Coulombs constant after t = 0?
    I was just trying to find a way to form an articulate explanation of the solution
     
  18. Jan 17, 2010 #17

    vela

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    I'm not sure what you mean by this.
     
  19. Jan 17, 2010 #18

    berkeman

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    meters per second multiplied by seconds is dimensionless? That's what tiny-tim is pointing out, I believe.
     
  20. Jan 17, 2010 #19

    berkeman

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    Oh wait, was that meant to be milliseconds? Maybe that's where the confusion is. milliseconds isn't mks units, so that's maybe what confused tiny-tim and me.
     
  21. Jan 17, 2010 #20
    Ok well for my equation for charge I evaluate the charge, with t in millisecond to be:

    [itex]q(\infty)\lim_{t\to\infty} 2 - 2e^{-100t}= 2[/itex]
    [itex]q(0) = 0[/itex]

    As for my equation for current, which is said to be the time rate of change of charge to be:

    [itex]I(t)=\lim_{t\to\infty}200e^{-100t}=0[/itex]

    [itex]I(0) = 200[/itex]

    My finally question being, how would I write out the solutions for these equations evaluate at what I have provided above being this is an analysis class and the professor expects clear, 100% correct, explanations and analysis from us, I want to prepare myself properly.
     
  22. Jan 17, 2010 #21

    vela

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    Yeah, I meant milliseconds. :) The idea of m/s didn't even occur to me until you said that, but I can see why it was confusing now.
     
  23. Jan 17, 2010 #22
    Yes, because when given the equation for q(t) he did not state the charge was either measured in coulombs or millicoulomb, only time, t, is measured in milliseconds. Then precede to ask how to find I(t), and q(t) and 0, and + infinity

    So my last step or asking from the posters in this thread is that, are my evaluations correct?
    and that what would be the best way to write a solution to this simple analysis properly? Sorry, to be a pain in the a**, just trying to learn things the best way possible
     
  24. Jan 19, 2010 #23
    Bump...
     
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