RL Diff Eq Exercise

Here is a problem sent by Neeraj

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

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:

$-(I_{L1} + I_{L2})R = L \dot{I}_{L1} \\ -(I_{L1} + I_{L2})R = 2L \dot{I}_{L2}$

The system matrix is:

$\begin{bmatrix} - \frac{R}{L} & - \frac{R}{L} \\ - \frac{R}{2 L} & - \frac{R}{2 L} \end{bmatrix}$

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

The currents are therefore:

$\begin{bmatrix} I_{L1} \\ I_{L2} \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} e^{-3R t / 2L} + c_2 \begin{bmatrix} -1 \\ 1 \end{bmatrix} e^{0 t }$

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

$\begin{bmatrix} I_0 \\ I_0 \end{bmatrix} = c_1 \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} -1 \\ 1 \end{bmatrix}$

This results in $(c_1, c_2) = ( \frac{2}{3} I_0, \frac{1}{3} I_0 )$ .

The equations for the currents are then:

$I_{L1} = \frac{4}{3} I_0 e^{-3R t / 2L} - \frac{1}{3} I_0 \\ I_{L2} = \frac{2}{3} I_0 e^{-3R t / 2L} + \frac{1}{3} I_0 \\$

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

The resistor current is:

$I_R = -(I_{L1} + I_{L2}) = - 2 I_0 e^{-3R t / 2L}$

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

$E_h = \int_0^\infty I^2_R R \, dt \\ Q = \int_0^\infty I_R \, dt$

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Lil Doug Greetings. Here is problem 30

Steven Chase - 9 months, 1 week ago

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

Talulah Riley - 9 months, 1 week ago

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

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

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:

$A \vec{v} = \lambda \vec{v}$

Where $A$ is a matrix, and $\lambda$ 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 - 9 months, 1 week ago

@Krishna Karthik bro how to type matrix in Brilliant?

@Talulah Riley – Like this:

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

Renders as: $\begin{bmatrix} 1 & 2 & 3\\ a & b & c \end{bmatrix}$

@Krishna Karthik – @Krishna Karthik \ $\begin{bmatrix} 1 & 2 & 3 \ a & b & c \end{bmatrix}$

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

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

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

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

Which part don't you like?

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

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

@Steven Chase – @Steven Chase yeah can you guide me

@Talulah Riley – I can post a note on the general process tomorrow

@Steven Chase Hello.

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

To find the eigenvalues, solve the equation:

$\det(A-\lambda I) = 0$

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 bro what is det?

@Talulah Riley – 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 and what is $I$

@Talulah Riley – Identity matrix in 2 dimensions: $\begin{bmatrix}1 & 0 \\ 0 &1 \end{bmatrix}$

This is not to be confused with current.

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

@Talulah Riley – Nice. Keep it going bro :)

@Krishna Karthik – @Krishna Karthik here it is
Thanks

@Talulah Riley – Perfect bro. Well done :)

@Talulah Riley – 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.

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

@Steven Chase Help me in the 35th problem.

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 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.

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

@Steven Chase it is up now.

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$