RL Diff Eq Exercise

Here is a problem sent by Neeraj

Let the current in the left inductor be IL1 I_{L1} and let the current in the right inductor be IL2 I_{L2} .

Recall that the voltage across an inductor is equal to its inductance multiplied by the time derivative of its current. The equations governing the circuit are:

(IL1+IL2)R=LI˙L1(IL1+IL2)R=2LI˙L2 -(I_{L1} + I_{L2})R = L \dot{I}_{L1} \\ -(I_{L1} + I_{L2})R = 2L \dot{I}_{L2}

The system matrix is:

[RLRLR2LR2L]\begin{bmatrix} - \frac{R}{L} & - \frac{R}{L} \\ - \frac{R}{2 L} & - \frac{R}{2 L} \end{bmatrix}

The eigenvalues for this system are (3R2L,0) (-\frac{3R}{2L}, 0) , and the associated eigenvectors are [21]\begin{bmatrix} 2 \\ 1 \end{bmatrix} and [11]\begin{bmatrix} -1 \\ 1 \end{bmatrix} . I used Wolfram alpha to find these.

The currents are therefore:

[IL1IL2]=c1[21]e3Rt/2L+c2[11]e0t\begin{bmatrix} I_{L1} \\ I_{L2} \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{-3R t / 2L} + c_2 \begin{bmatrix} -1 \\ 1 \end{bmatrix} e^{0 t }

To solve for the constants, apply the initial conditions from time t=0t = 0 :

[I0I0]=c1[21]+c2[11]\begin{bmatrix} I_0 \\ I_0 \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} -1 \\ 1 \end{bmatrix}

This results in (c1,c2)=(23I0,13I0) (c_1, c_2) = ( \frac{2}{3} I_0, \frac{1}{3} I_0 ).

The equations for the currents are then:

IL1=43I0e3Rt/2L13I0IL2=23I0e3Rt/2L+13I0 I_{L1} = \frac{4}{3} I_0 e^{-3R t / 2L} - \frac{1}{3} I_0 \\ I_{L2} = \frac{2}{3} I_0 e^{-3R t / 2L} + \frac{1}{3} I_0 \\

Interestingly, the inductors don't spend all of their energy as heat lost to the resistor. At t= t = \infty , there is a circulating current of magnitude 13I0 \frac{1}{3} I_0 that only flows in a loop through the inductors.

The resistor current is:

IR=(IL1+IL2)=2I0e3Rt/2LI_R = -(I_{L1} + I_{L2}) = - 2 I_0 e^{-3R t / 2L}

From here, one can calculate the heat and charge as follows:

Eh=0IR2RdtQ=0IRdt E_h = \int_0^\infty I^2_R R \, dt \\ Q = \int_0^\infty I_R \, dt

Note by Steven Chase
2 weeks, 6 days ago

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@Lil Doug Greetings. Here is problem 30

Steven Chase - 2 weeks, 6 days ago

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@Steven Chase Thank you so much.
BTW, what is the meaning of eigen and how to find eigenvalues?

Lil Doug - 2 weeks, 5 days ago

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"Eigen" is the German word for "characteristic" or "own". Here is an article with more details

https://brilliant.org/wiki/eigenvalues-and-eigenvectors/

Steven Chase - 2 weeks, 5 days ago

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You'll have to study Linear Algebra formally to know what this is. I assume you're familiar with matrices, at the most basic level. So, an eigenvalue is found basically when you multiply a matrix by a vector, and you get a scaled version of the vector.

You can find it using the equation below:

Av=λvA \vec{v} = \lambda \vec{v}

Where AA is a matrix, and λ\lambda is the Eigenvalue.

What you're doing is you're going to be solving the equation above, to find the eigenvalue, and the eigenvector (Usually there'll be more than one) that satisfies the equation.

You'll find it really useful in the future.

Krishna Karthik - 2 weeks, 5 days ago

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@Krishna Karthik bro how to type matrix in Brilliant?

Lil Doug - 2 weeks, 4 days ago

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@Lil Doug Like this:

"\begin{bmatrix} 1 & 2 & 3 \ a & b & c \end{bmatrix}"

Renders as: [123abc]\begin{bmatrix} 1 & 2 & 3\\ a & b & c \end{bmatrix}

Krishna Karthik - 2 weeks, 4 days ago

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@Krishna Karthik @Krishna Karthik \[123 abc]\begin{bmatrix} 1 & 2 & 3 \ a & b & c \end{bmatrix}

Lil Doug - 2 weeks, 4 days ago

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@Lil Doug Use two back slashes between the rows. On brilliant, it only comes as one for a weird reason.

Krishna Karthik - 2 weeks, 4 days ago

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@Steven Chase it is written at an instant, the current is flowing I0I_{0} ,
Why we are integrating it from 0 to inftyinfty ?
Maybe it can 3 to \infty??? 4 to \infty ?

Lil Doug - 2 weeks, 5 days ago

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The beginning instant has been arbitrarily chosen as t=0 t = 0

Steven Chase - 2 weeks, 5 days ago

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@Steven Chase isn't there any pure Anayltical way to solve this??

Lil Doug - 2 weeks, 5 days ago

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Which part don't you like?

Steven Chase - 2 weeks, 5 days ago

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@Steven Chase I like all part
But I want a Anayltical way of that solution of differential equation.

Lil Doug - 2 weeks, 5 days ago

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@Lil Doug You mean the calculation of the eigen-quantities? You could do that by hand if you wanted to.

Steven Chase - 2 weeks, 5 days ago

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@Steven Chase @Steven Chase yeah can you guide me

Lil Doug - 2 weeks, 5 days ago

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@Lil Doug I can post a note on the general process tomorrow

Steven Chase - 2 weeks, 5 days ago

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@Steven Chase Hello.

Lil Doug - 2 weeks, 5 days ago

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@Krishna Karthik Bro this is the whole matrix. Now guide me to eigen

Lil Doug - 2 weeks, 4 days ago

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To find the eigenvalues, solve the equation:

det(AλI)=0\det(A-\lambda I) = 0

I want you to solve the equation resulting yourself to find out how to do this. The answer is already above. You need to learn how to solve for eigenvectors and eigenvalues.

We get a quadratic equation; solving this quadratic equation, we get the associated eigenvectors and eigenvalues. And since it's quadratic, there will be two eigenvectors.

Krishna Karthik - 2 weeks, 4 days ago

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@Krishna Karthik bro what is det?

Lil Doug - 2 weeks, 4 days ago

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@Lil Doug Determinant of a matrix. It's almost like finding the volume from the paralleled of the matrix vectors. Or the area, in this case.

Krishna Karthik - 2 weeks, 4 days ago

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@Krishna Karthik and what is II

Lil Doug - 2 weeks, 4 days ago

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@Lil Doug Identity matrix in 2 dimensions: [1001]\begin{bmatrix}1 & 0 \\ 0 &1 \end{bmatrix}

This is not to be confused with current.

Krishna Karthik - 2 weeks, 4 days ago

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@Krishna Karthik @Krishna Karthik Ok thanks bro. I am solving it right now.
I will show my solution within 10 min.

Lil Doug - 2 weeks, 4 days ago

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@Lil Doug Nice. Keep it going bro :)

Krishna Karthik - 2 weeks, 4 days ago

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@Krishna Karthik @Krishna Karthik here it is
Thanks

Lil Doug - 2 weeks, 4 days ago

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@Lil Doug Perfect bro. Well done :)

Krishna Karthik - 2 weeks, 4 days ago

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@Lil Doug Btw, I just want to add, an eigenvalue of 0 is never counted, because 0 is always a solution. You learnt this pretty quickly, I think. Good job.

Krishna Karthik - 2 weeks, 4 days ago

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Also, watch this video guiding how to solve a system of differential equations. You need Linear Algebra knowledge for it:

https://www.youtube.com/watch?v=iVlHPDER0FA

Krishna Karthik - 2 weeks, 4 days ago

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@Steven Chase Help me in the 35th problem.

Lil Doug - 2 weeks, 4 days ago

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That's kind of a nice problem. I wouldn't mind posting that as a problem in the E and M section. And then we could have people post solutions to it (my own included). Or if you know the answer, you could post it.

Steven Chase - 2 weeks, 4 days ago

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@Steven Chase I didn't understand what you want to say.
You can do anything what you want to do, at the end I want Anayltical solution.
By the see my new note please.

Lil Doug - 2 weeks, 4 days ago

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@Steven Chase ok I am. Posting that problem right now.

Lil Doug - 2 weeks, 4 days ago

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@Steven Chase it is up now.

Lil Doug - 2 weeks, 4 days ago

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