I then assume put this back into the equation for VL, and derive it, I get:
V0*L/(R2+L2*ω2*cos(2*φ) times ωcos (ωt + φ)
or should I say times -ωcos (ωt) considering the value of φ.
Is this right?
This is just so many kinds of wrong... :(
This is what I meant to write:
I_{0}sin(\omega t + \varphi) + \frac{RI_{0}}{L}(-\frac{cos{\omega t+\varphi}{\omega}= \frac{1}{L} V_{peak} sin (\omega t)
Is this comment meant for the last version of the response?
I edited my response extensively before I noticed your reply, I now realized I should have just made a new post, since this ultimately confused me and makes communication difficult. Sorry, won't do it again.
I've heard of them...
Integrate the whole thing? Like this:
I_{0}\omega cos(\omega t + \varphi)+I_{0}\frac{R}{L}\frac{sin(\omega t + \varphi)-\omega t cos(\omega t + \varphi)}{\omega^2}=\frac{1}{L}V_{peak}\frac{R}{L}\frac{sin(\omega t)-\omega t cos(\omega t)}{\omega^2}
Is this right?
Homework Statement
An alternating current is running through a serially connected inductor(L) and resistor(R). The alternating voltage causing it is:
http://upload.wikimedia.org/math/5/a/0/5a0ecaa1432c6cdce653a943b4962a21.png
How does the voltage on the inductor change over time?
Homework...
Ok, that seems blindingly obvious in retrospect, have no idea how I missed it.
Thank you for the help! :)
This is then the correct answer:
P(X ≥ 4) = 1 - P(X < 4) = 1 - P(X = 0) - P(X=1) - P(X=2) - P(X=3)
P(X ≥ 4) = 0,73497
A probability of 73,5%.
I apologize if I'm posting this in the wrong subforum, I wasn't quite sure where statistics and distribution questions should go. I have major problems understanding the concepts involved with this kind of stuff.
Homework Statement
On a country road an average of 30 cars pass per hour. Four...