Question about a book

Discussion in 'Homework Help' started by Piggins, Mar 19, 2015.

  1. Piggins

    Thread Starter New Member

    May 5, 2014
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    In the book Practical electronics for inventors there is a part about capacitors that got me puzzled. It states the following about RC circuit behaviour:

    "By applying Kirchhoff's coltage law to the circuit, you can set up the following expression to figure out how the current will behave after the switch is closed:

    V0=IR + 1/C *⌠Idt

    To get rid of the integral, differentiate each term:

    0=R* dI/dt + 1/C*I or dI/dt + 1/RC*I = 0

    This expression is a linear, first-order, homogenous differential equation that has the following solution:

    I=I0e^-t/RC"

    I cant understand how it arrived to this conclusion, can someone help me?

    Thanks.

    Edit: I mean how it solved the differential equation that is.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Have you had differential equations?

    If not, then it isn't too obvious.

    Let's consider the equation we are trying to solve:

    <br />
\text{\frac{dI}{dt} \; + \; \frac{1}{RC}I \; = \; 0}<br />

    Note that when you said, "1/RC*I" what you were technically saying was "((1/R)C)I". Be careful about order of operations. It will bite you when you start entering equations into programs and simulators and such if you don't.

    Getting back to solving the differential equation -- this expression tells us that the first term and the second term have to be equal and opposite at all moments in time. This means that I and dI/dt have to have the same form. We have two obvious candidates for this, namely

    <br />
I(t)=A \cdot sin(\omega t)<br />

    and

    <br />
I(t)=A \cdot e^{\omega t}<br />

    Try both and see if you get a solution.

    In this particular case, being homogeneous, we can use methods typically taught in calc (Calc I?) by simply rearranging it as

    <br />
\text{\frac{dI}{dt} \; = \; - \frac{1}{RC}I}<br />

    Now multiply both sides by dt.

    For further assistance you need to show your best attempt to take it from here.
     
  3. Piggins

    Thread Starter New Member

    May 5, 2014
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    Thanks for answering.

    Yes I dont know much about differential equations but it bothers me how books about basic electronics just solve them without telling what is happening.
    I am not sure what you want me to do with those candidates since they are already in the form of the final answer.

    However, I found http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx and http://circuitscan.homestead.com/files/ancircp/rcdiff1.htm

    the second site says: "
    [​IMG]
    The above equation can now be integrated directly to give give the following general solution.
    [​IMG]"
    What I dont understand is where the alpha(t) went from the left side after it was integrated.

    In the first link, it had a solution process for differential equations so I tried that too.
    So we start with dI/dt + I/RC = 0 integrating factor is then u(t) = e^⌠1/RCdt = e^t/RC
    then everything is multiplied by u(t) and we get dI/dt * e^t/RC + 1/RC*I * e^t/RC = 0 and the left side is the product rule now.

    So we get d(I*e^t/RC)dt = 0
    after integrating this becomes I*e^t/RC + c = 0 the books answer was I = I0*e^-t/RC so I must have gotten something wrong. Can you give me a hint? Thanks for your help.
     
  4. WBahn

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    Mar 31, 2012
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    It's annoying, but all too common. There are two possible reasons -- first, the book is assuming a certain level of prior understanding and is not going to spend time teaching you things that you should already know. But it is also frequently the case that material in a science or engineering curriculum outpaces the typical math curriculum. Part of that is just because each field needs somewhat different math at different times and it is not feasible to construct a math curriculum that can serve that many masters. Usually, the particular curricula that is getting in front of the math takes on the responsibility of at least introducing the students to the basics they need.

    What about if I had offered up

    <br />
I(t) \; = \; 5A<br />

    That is also in the form of the final answer, but it is not correct. You determine this by plugging the candidate solution into the differential equation and seeing if it is a solution to the differential equation.

    <br />
\text{\frac{dI}{dt} \; + \; \frac{1}{RC}I \; = \; 0}<br />

    Plugging in our candidate solution above we have

    <br />
\text{\frac{5A}{dt} \; + \; \frac{1}{RC}5A \; = \; 0}<br />
\;<br />
\text{0 \; + \; \frac{5A}{RC} \; = \; 0}<br />

    which is not a solution, thus our candidate is not a solution.

    Now you try it with the two original candidates I offered up.

    The stuff you found from those other sites is making things unnecessarily complicated. Let's go back to what I recommended at the end of my earlier post:

    <br />
\text{\frac{dI}{dt} \, = \, - \frac{1}{RC}I}<br />

    Separate the variables so that I is on one side and t is on the other (it doesn't matter where you put the constants):

    <br />
\text{\frac{dI}{I} \, = \, - \frac{1}{RC}dt}<br />

    Now integrate both sides

    <br />
\text{\int{\frac{dI}{I}} \, = \, - \frac{1}{RC}\int{dt}}<br />

    Carry out the integration

    <br />
\text{ln(I) + C_0 \, = \, -\frac{1}{RC}\(t+T_0\)}<br />

    Keeping in mind that C0 and T0 are arbitrary constants, we can rearrange things so as to combine them into a single constant.

    <br />
\text{ln(I) \, = \, K-\frac{t}{RC}}<br />


    Since we want I and not ln(I), we can exponentiate both sides to get:

    <br />
I \, = \, \(e^K\)\(e^{\frac{-t}{RC}}\)<br />

    If K is a constant, then e^K is a constant and we can replace it with a simpler arbitrary constant, I0.

    <br />
I \, = \, I_0 e^{\frac{-t}{RC}}<br />
     
  5. Piggins

    Thread Starter New Member

    May 5, 2014
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    Ok, thanks for clearing it out for me. I tried solving the equation with the candidates from the earlier post and the results were:
    equation: dI/dt + I / RC = 0
    w + 1/RC * tan(wt) = 0 for I(t) = A*sin(wt) and
    t = - 1/RC for I(t) = A*e^wt

    Not sure if I got them right. A is a constant and you need to use chain rule for sin(wt) and e^fx, right?

    Also, I didnt know you could break dI/dt like that. I can understand that dI/dt is like 5 amps/second or change in current but if its only dI then its just current changing regard to nothing?
    When you integrate dI/I and use the rule for integrating 1/x, does that mean that dI is basically a constant?
     
  6. WBahn

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    Mar 31, 2012
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    dt, dx, and dI are differentials, which just means that that are infinitesimally small changes in t, x, and I respectively.

    If you know that

    v(t) = dx(t)/dt

    where x(t) is the position, v(t) is the velocity, and t is time, then you can rearrange this and get

    dx(t) = v(t)·dt

    Let's step back from differentials a bit is use incrementals.

    v = Δx/Δt

    If I want to get an estimate of my velocity over some short period of time, then I can measure how far I move during that short period of time and divide by that short period of time. But I can manipulate this and get

    Δx = v·Δt

    which means that if I take my velocity and multiply it by a time interval, I get how far I move during that time interval.

    If I am moving at a constant velocity, then it won't matter if that "short period of time" is one microsecond or one century. But if my velocity isn't constant, then this approach will just give me the average velocity over that time interval. If I choose an interval that is short enough such that my velocity doesn't change noticeably over that interval, then the average velocity is a really good estimate of the instantaneous velocity throughout that interval. A differential is simply the next logical step in which we make the time interval so small that the velocity can't change at all over that interval -- this means making it infinitesimally small, which merely means looking at the behavior of the math in the limit that the time interval goes to zero.
     
  7. WBahn

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    Let's take the first candidate:

    <br />
\text{I(t) = A\cdot\sin(\omega t)<br />

    Take the derivative of this with respect to t:

    <br />
\text{\frac{dI(t)}{dt} = A\cdot\frac{d\sin(\omega t)}{dt}}<br />
\.<br />
\text{\frac{dI(t)}{dt} = A\omega\cdot\cos(\omega t)}<br />

    Substituting the candidate expressions for I(t) and dI(t)/dt, you have

    <br />
\text{\frac{dI}{dt} + \frac{1}{RC}I = 0}<br />
\,<br />
\text{A\omega\cdot\cos(\omega t) + \frac{A}{RC}\sin(\omega t) = 0}<br />

    The question then becomes where there is any value for the constant A that will satisfy this equation (and for all values of ω). If the answer is, "No," then this candidate is not a solution.

    Now you try the same approach using the other candidate, namely

    <br />
\text{I(t) = A\cdot e^{\omega t}<br />

    EDIT: Fixed type in equation pointed out by TS.
     
    Last edited: Mar 22, 2015
  8. Piggins

    Thread Starter New Member

    May 5, 2014
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    So Δx = v·Δt and dx(t) = v(t)·dt are the same thing? Trying to get that last part. if its about the slope of the tangent that can be drawn on the graph of a differentiated thing. If the curve goes up and down the two points on the curve need to be real close to get a good estimate what the change will be at that point?

    But then again you said that the differentials are infinitesimally small changes so

    dx(t) = v(t)·dt
    Small change in the position at time t equals velocity at time t times small change in time?
    But what does this tell us if both change in place and time are small? How do I use this? But it is how much the object moved during the measurement of the speed right?



    How come the A does not get substituted to the I / RC term? Ill try to do the second substitution again but ill post this now so I dont forget.
     
  9. Piggins

    Thread Starter New Member

    May 5, 2014
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    equation: dI/dt + I / RC = 0
    and I(t) = A * e^wt dI(t)/dt = A * e^wt * w
    so substituting this I get

    A*e^wt * w + (A*e^wt)/RC = 0 So now I need to figure out if A can be something that makes ends meet? 0 could do it but it dont sound too smart.
    if I divide by e^wt*A I get
    w + 1/RC = 0
    w = -1/RC
     
  10. WBahn

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    Mar 31, 2012
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    Very close. In the first case the quantities are finite, while in the second they are infinitesimal. For most purposes you can reasonably treat the two as interchangeable concepts (there are those that would vehemently disagree and for those I would just ask that we allow a bit of sloppiness to slide by).

    I'm having a hard time telling exactly what you are saying, but I think you've got the right idea. If we want the slope of the curve at a point then we can get an estimate by taking the slope between two points located on either side of the desired point. But this will only be an estimate. As we move the two points closer together we get better and better estimates. The actual slope is what we get with the two points are infinitesimally close together.

    It's a mathematical construct to answer questions about how changes in one thing are related to changes in another. If you drop an object off a roof its speed is constantly changing, right? But it still makes sense to talk about its speed at a particular moment in time, right? We don't have to talk about average speeds over some period of time, we can talk about instantaneous speeds at specific moments. But think about it -- at a specific moment, the object has a specific position and if we just focus on that one moment in time then it would seem like it is not moving; yet we can talk about it also having a speed at that specific moment and that doesn't bother us. So even though dx and dt are infinitesimally small, we can still use them to describe the system and, in this case, we know that they are related to each other by the velocity of the object at that moment in time.

    It should have -- my bad when editing an equation. Thanks for catching it.
     
  11. WBahn

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    Yep.

    Though, note again what I said earlier about order of operations. What you have is really

    w = (-1/R)C = -(C/R)

    which is not what you meant. Write what you mean:

    w = -1/(RC)


    A=0 is known as the "trivial solution" and while it does satisfy the differential equation, it is seldom interesting and would almost never let us meet the boundary (initial) conditions.

    So now you would just use the initial conditions to find the value of A.
     
    Last edited: Mar 22, 2015
  12. Piggins

    Thread Starter New Member

    May 5, 2014
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    So in dx(t) = v(t)·dt the dx(t) is that time the object moves at time t during dx? But its just so small that its 0 basically?

    Edit I mean during dt
     
  13. WBahn

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    Yes. That is a very reasonable way to view it.
     
  14. Piggins

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    May 5, 2014
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    Hi, thanks for helping me. This will be my last post as its getting late. So how do I get the solution to

    A*e^wt * w + (A*e^wt)/RC = 0

    its still far off from what was in the book I = I0*e^(-t/RC)
     
  15. WBahn

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    How is it far off? You've already determined that w = -1/(RC), right?

    So that makes your candidate solution go from

    <br />
\text{I = A e^{-\omega t}}<br />

    to

    <br />
\text{I = A e^{-\frac{t}{RC}}}<br />

    Well, Io is a constant and A is a constant, and since A is an arbitrary constant, can you not simply declare that

    A = Io

    More specifically, what is the value of I at t=0? It's A. So A is equal to the current at t=0, which is presumably Io.
     
  16. Piggins

    Thread Starter New Member

    May 5, 2014
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    ops, forgot brackets

    so with I(t) = A*e^(-t/(RC)) I end up with
    -A*e^(-t/(RC)) + A*e^(-t/(RC)) = 0 this is true with all values of A it would seem.
    Hmm, I might need to learn more about differential equations before going back to that book. Its new to me how the process of solving these equations goes.
     
  17. WBahn

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    Keep in mind that much of this discussion has been Calc I stuff. So you might need to strengthen your basic calc (and possibly algebra) skills, too.
     
  18. Piggins

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    May 5, 2014
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    That is most likely smart, I have never taken maths classes that went this far apart from online guides. also I guess I need to think more what I am doing when I am trying to solve something instead of just spinning numbers around using given rules and such.
     
  19. WBahn

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    That last part is really key. Look for the concepts and the connections and not just the equations and procedures. You've got the right attitude and I'm sure you'll do fine.
     
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