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The values of the elements of an RLC circuit are given. Solve the initial value problems. L=1, R=0, C=10^-4; e(t)=100 if 0<=t<2pi: e(t)=0 if t>=2pi

Mathematics
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So i was working on this problem about 10 min ago. I am having a hard time determining e(t). Can you please help me find e(t). Once I obtain e(t) I can solve the equations for f(t)
Do we use this piecewise function to determine e(t) |dw:1401498254051:dw|
I'm at a bit of a disadvantage here with the physics concepts, but I'll give it a shot. As I see it, the diff. eq. is \[\frac{dI}{dt}+10^4\int_0^t I(\tau)~d\tau=E(t)\] and \(E(t)\) is defined as \[E(t)=\begin{cases}100&\text{for }0\le t<2\pi\\0&\text{for }t\ge2\pi\end{cases}\] and with initial value \(I(0)=0\). Have I interpreted this correctly?
This looks very familiar. We did one of these RLC circuits in class together. Looking back at my notes I am a bit lost...
But that is what i did, not sure how to proceed after defining E(t) as the piecewise function
Taking the Laplace transform of both sides is the way to go here, but let's focus on the RHS first: \[\large\mathcal{L}\left\{E(t)\right\}=\int_0^\infty e^{-st}E(t)~dt\] \(E(t)=0\) for \(t\ge2\pi\) and \(E(t)=100\) for \(t<2\pi\), so we can reduce to \[\large \mathcal{L}\left\{E(t)\right\}=100\int_0^{2\pi} e^{-st}~dt=100\cdot\frac{1-e^{-2\pi s}}{s}\]
You were on the right track with the step function idea - you can definitely express \(E(t)\) in terms of \(u(t-c)\): \[E(t)=100u(2\pi-t)\]
we have to change that 2pi-t because we can't take the LT
so would it be 100-u(t-2pi)
But if I was to use the unit step function how to identify the equation you have there E(t)=100u(2pi-t)?
Actually, if you want to write it in terms of \(t-2\pi\), the change would be \[E(t)=100-100u(t-2\pi)\] But that's not that important. Using the definition, you can find the same answer without having to memorize any formulas (other than the integral definition).
I was looking at my notes... my instructor went over the definition of the LT so I understand what you did up top with the integral sign. It's just a way I am not use to when finding the laplace transform.
Can we solve this using the definition of the Laplace transform? What you showed me fist?
I think it would help me understand this LT better by using the definiton first because in all these mass spring system or RLC circuits we been using the unit step function
Yeah I'll continue with my method, but if you still want to know how to extract the unit step function expression, I can help you figure that out too.
yeah we can do that later.
I think I got a different result.
I based my result off these notes: http://calculus7.com/sitebuildercontent/sitebuilderfiles/rlc.rtf
After I found Q Then I used Q'=i to use my condition i(0)=0
i also did it without laplace
which you can
Okay, so we have \[\frac{dI}{dt}+10^4\int_0^t I(\tau)~d\tau=E(t)\] with \[E(t)=\begin{cases}100&\text{for }0\le t<2\pi\\0&\text{for }t\ge2\pi\end{cases}\] and with initial value \(I(0)=0\). Taking the LT of both sides: \[\begin{align*} \mathcal{L}\left\{\frac{dI}{dt}+10^4\int_0^t I(\tau)~d\tau\right\}&=\frac{100(1-e^{-2\pi s})}{s}\\ \left(sF(s)-I(0)\right)+10^4\frac{F(s)}{s}&=\cdots\\ \left(s+\frac{10^4}{s}\right)F(s)&=\frac{100(1-e^{-2\pi s})}{s}\\ \left(s^2+10^4\right)F(s)&=100(1-e^{-2\pi s})\\ F(s)&=\frac{100(1-e^{-2\pi s})}{s^2+10^4} \end{align*}\]
We have Q''+10^4Q=100 first find homogenous solution to Q''+10^4Q=0 then find the particular solution I chose to use general form Q=At^2+Bt+C to get there
^^ yup that method works too. I assumed this was a LT question based on previous questions.
okay thanks @SithsAndGiggles that makes a lot of sense.
If I am not able to identify the equation using the unit step function I can always use the definition of the laplace transform?
Yea there looks like a ton of laplace going on in those notes I posted
Yes, the integral from the definition would be reduced to a simple calculation by changing the limits of integration based on the the step function you use.
okay thank you. Please show me how to extract the unit step function from the info given
Here's a plot of \(E(t)\): |dw:1401500411049:dw| So the regular step function is defined as \[u(t-c)=\begin{cases}1&\text{for }t-c\ge 0\\0&\text{for }t-c<0\end{cases}~~~~~~~~\left[\iff~~u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }tc\end{cases}\] which gives the plot |dw:1401500884228:dw| So, multiplying by the scale factor of 100 and letting \(c=2\pi\), we get \(E(t)=100u(2\pi -t)\). However, like you said, you want this in terms of \(t-2\pi\), and to do that you would negate this function and its argument and add 100: |dw:1401501104472:dw|
I got to leave here soon. Let me read this over some more and I will tag you later, k? I just need to study this a bit. If I have any questions I will let you know. For the most part I understand what is going on mathematically.
Sure, take your time.
So what piecewise function do we use to obtain the unit step function? @SithsAndGiggles
Do we use |dw:1401581129762:dw| or |dw:1401581158916:dw|
The second one. The first is the unit step function; the second is the unit step function scaled by a factor of 100, which is what you want.
Okay thanks.
1 Attachment
How does my work look @SithsAndGiggles ?
I think you skipped a step in the last two lines. Partial fractions first, then take the inverse transform. For example, \[F(s)=\frac{1}{s(s^2+4)}=\frac{A}{s}+\frac{Bs+C}{s^2+4}\] Rearranging factors, you have \[\begin{align*}1&=A(s^2+4)+s(Bs+C)\\ 1&=(A+B)s^2+Cs+4A\end{align*}\] so you have \[\begin{cases}A+B=0\\C=0\\4A=1\end{cases}~~\Rightarrow~~A=\frac{1}{4},~B=-\frac{1}{4},~C=0\] Then \[F(s)=\frac{1}{4}\cdot\frac{1}{s}-\frac{1}{4}\cdot\frac{s}{s^2+4}\] Taking the inverse transform yields \[F(s)=\frac{1}{4}-\frac{1}{4}\cos2t\] The same applies for the \(\dfrac{e^{-\pi s}}{s(s^2+4)}\) term.
*That last \(F(s)\) should be \(f(t)\).
img 102 has my work and img 103 has the "advance" LT's. How would I continue my work without the cheat sheet? If I was only given the basic LT?
nvm I got it figured out. Sorry I have not been here to discuss that one problem. I have an exam tomorrow on LTs so I am reviewing material instead of working on hw.

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