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Euler's Equation and KVL

by iRaid
Tags: equation, euler
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iRaid
#1
Feb13-13, 03:04 PM
P: 550
In one of my engineering classes we discussed these two topics and I have two questions about this stuff.

First question is how does euler's equation work exactly..
[tex]e^{j\varphi}=cos\varphi+jsin\varphi[/tex]

Second question is how do you solve this:
[tex]V_{M}cos\omega t=Ri(t)+L\frac{di(t)}{dt}[/tex]

Thanks for the help..
(These aren't homework problems, just notes I do not understand)
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Whovian
#2
Feb13-13, 03:15 PM
P: 642
Quote Quote by iRaid View Post
First question is how does euler's equation work exactly..
[tex]e^{j\varphi}=\cos\varphi+i\cdot\sin\varphi[/tex]
I, unfortunately, don't understand the question. What do you mean by "how?" (Note that I think you mean i, not j.)

And, to your second, what are Ri(t) and i(t)? Or [itex]V_M[/itex]? I might be being an idiot and missing something.
mathman
#3
Feb13-13, 03:22 PM
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Euler's equation is an identity. What do have in mind with the question "How does it work?"?

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Feb13-13, 04:10 PM
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Euler's Equation and KVL

Quote Quote by iRaid View Post
In one of my engineering classes we discussed these two topics and I have two questions about this stuff.

First question is how does euler's equation work exactly..
[tex]e^{j\varphi}=cos\varphi+jsin\varphi[/tex]

Second question is how do you solve this:
[tex]V_{M}cos\omega t=Ri(t)+L\frac{di(t)}{dt}[/tex]

Thanks for the help..
(These aren't homework problems, just notes I do not understand)
Hi iRaid!

Euler's equation is more than an equation - it's an identitiy.
It is always true.
It's not so easy to explain why it is true, but there are proofs lying around.
For instance here.


For your 2nd question, it's an inhomogeneous first order ODE.

The way to solve it is one of:
- method of undetermined coefficients
- with a standard formula for it
- variation of parameters


@Whovian: in electrical engineering, the function ##i(t)## is often used to define an input signal, usually a current.
That makes the use of the symbol ##i## for the imaginary constant ambiguous.
That's why the symbol ##j## is used for the imaginary constant.
The expression ##R i(t)## is Ohm's law (##\Delta V = I \times R##).
iRaid
#5
Feb13-13, 04:17 PM
P: 550
Quote Quote by I like Serena View Post
Hi iRaid!

Euler's equation is more than an equation - it's an identitiy.
It is always true.
It's not so easy to explain why it is true, but there are proofs lying around.
For instance here.


For your 2nd question, it's an inhomogeneous first order ODE.

The way to solve it is one of:
- method of undetermined coefficients
- with a standard formula for it
- variation of parameters


@Whovian: in electrical engineering, the function ##i(t)## is often used to define an input signal, usually a current.
That makes the use of the symbol ##i## for the imaginary constant ambiguous.
That's why they use ##j## for the imaginary constant.
The expression ##R i(t)## is Ohm's law (##\Delta V = I \times R##).
Ah I remember euler's formula now, the j is what got me..
Well I don't know ODE's so this sucks lol.
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Feb13-13, 04:20 PM
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Quote Quote by iRaid View Post
Ah I remember euler's formula now, the j is what got me..
Well I don't know ODE's so this sucks lol.
Ah well, let's just say that in your case the general solution is of the form (method of undetermined coefficients):
$$i(t) = C_0 e^{-\frac R L t} + K \cos ωt + M \sin ωt$$

Can you find the constants ##C_0, K, M##?
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#7
Feb13-13, 04:27 PM
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Oh, and since we're not in electrical engineering here, let me add that KVL is the abbreviation for Kirchhoff's Voltage Law, which states that the sum of voltages in any electrical loop must be zero. The 2nd equation sort of represents this for a specific situation.
SteamKing
#8
Feb13-13, 04:31 PM
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Remember, electrical engineers use j = sqrt(-1) in place of i to avoid confusion with current.
iRaid
#9
Feb13-13, 04:41 PM
P: 550
Quote Quote by I like Serena View Post
Ah well, let's just say that in your case the general solution is of the form (method of undetermined coefficients):
$$i(t) = C_0 e^{-\frac R L t} + K \cos ωt + M \sin ωt$$

Can you find the constants ##C_0, K, M##?
If they are given I can lol. I'm not sure what you mean.
joeblow
#10
Feb13-13, 05:08 PM
P: 71
The imaginary unit is also symbolized as j in physics. I believe that they denote it that way because it is consistent with the quaternions.

You can take the expedient yet historically dishonest path of defining exp(j*x)= cos x + j sin x. If you want to go the more honest path, consider the movement about the unit circle with a constant arc length speed and do some diff eq. with the IC that exp(0)=1.
iRaid
#11
Feb13-13, 05:11 PM
P: 550
Yeah I see what you're saying. I knew that j was the imaginary unit, but when I saw that equation, I forgot that or something, I'm not sure lol it just screwed me up. And as stated above the j is used as i in EE so that it isn't confused with current as current is used im many equations.
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Feb13-13, 05:24 PM
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Quote Quote by iRaid View Post
If they are given I can lol. I'm not sure what you mean.
When solving a differential equation, we're usually left with a general solution that has 1 or more as yet unknown constants.
When you fill in the general solution in the differential equation, you should find that some of those constants will have to have a specific value to satisfy the equation (for all values of t).
The question would then become: what do you get if you fill in this general solution in your equation?
(Only if you're interested of course.)
iRaid
#13
Feb13-13, 05:59 PM
P: 550
Doing di(t)/dt it would be:
[tex]C_{0}=\frac{-R}{L}\\K=-\omega\\M=\omega[/tex]

I believe?
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Feb13-13, 06:34 PM
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Quote Quote by iRaid View Post
Doing di(t)/dt it would be:
[tex]C_{0}=\frac{-R}{L}\\K=-\omega\\M=\omega[/tex]

I believe?
Let's see...
$${di(t) \over dt} = - C_0 \frac R L e^{- \frac R L t} - Kω \cos ωt + Mω \sin ωt$$

So
$$V_M\cos ωt = Ri(t)+L{di(t) \over dt}$$
$$ \qquad = R(C_0 e^{- \frac R L t} +K \cos ωt + M \sin ωt) + L(- C_0 \frac R L e^{- \frac R L t} - Kω \sin ωt + Mω \cos ωt)$$
$$\qquad = (RK +LMω) \cos ωt + (RM - KLω) \sin ωt$$

Since it has to be true for any t, we find:
$$\left\{ \begin{array}{l}
RK +LMω = V_M \\
RM - KLω = 0
\end{array}\right.$$

From this you can solve K and M.
You still have a free choice for ##C_0##, but that choice follows if you know ##i(0)=i_0##.
iRaid
#15
Feb13-13, 06:57 PM
P: 550
I see thank you for the clarification.


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