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Euler's Equation and KVL 
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#1
Feb1313, 03:04 PM

P: 551

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) 


#2
Feb1313, 03:15 PM

P: 642

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. 


#3
Feb1313, 03:22 PM

Sci Advisor
P: 6,059

Euler's equation is an identity. What do have in mind with the question "How does it work?"?



#4
Feb1313, 04:10 PM

HW Helper
P: 6,187

Euler's Equation and KVL
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##). 


#5
Feb1313, 04:17 PM

P: 551

Well I don't know ODE's so this sucks lol. 


#6
Feb1313, 04:20 PM

HW Helper
P: 6,187

$$i(t) = C_0 e^{\frac R L t} + K \cos ωt + M \sin ωt$$ Can you find the constants ##C_0, K, M##? 


#7
Feb1313, 04:27 PM

HW Helper
P: 6,187

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.



#8
Feb1313, 04:31 PM

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Thanks
PF Gold
P: 6,475

Remember, electrical engineers use j = sqrt(1) in place of i to avoid confusion with current.



#9
Feb1313, 04:41 PM

P: 551




#10
Feb1313, 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. 


#11
Feb1313, 05:11 PM

P: 551

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.



#12
Feb1313, 05:24 PM

HW Helper
P: 6,187

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.) 


#13
Feb1313, 05:59 PM

P: 551

Doing di(t)/dt it would be:
[tex]C_{0}=\frac{R}{L}\\K=\omega\\M=\omega[/tex] I believe? 


#14
Feb1313, 06:34 PM

HW Helper
P: 6,187

$${di(t) \over dt} =  C_0 \frac R L e^{ \frac R L t}  Kω \cos ωt + Mω \sin ωt$$ 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##. 


#15
Feb1313, 06:57 PM

P: 551

I see thank you for the clarification.



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