Recent content by SimpliciusH

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?

Ok I get 2*pi+pi for the value of φ and V0/(R2+L2*ω2*cos(2*φ) for I0)

Ok I've simplified it to: Lcos(t\omega + \varphi)+Lsin(t\omega + \varphi)=sin(t\omega + \varphi)\frac{V_{0}}{I_{0}}

V0=Vpeak So I go from: LI0cos(ωt+φ)+RI0sin(ωt+φ)=V0sin(ωt) To: LI0cos(ωt)cosφ - LI0sin(ωt)sinφ + RI0sin(ωt)cosφ + RI0cos(ωt)sinφ = V0sin(ωt) And collect the terms: RI0sin(ωt)cosφ - LI0sin(ωt)sinφ - V0sin(ωt)+ RI0cos(ωt)sinφ + LI0cos(ωt)cosφ = 0 sin(ωt)(RI0cosφ - LI0sinφ - V0)+...

Did you mean the trig sum rule? Edit: It dawned on me a minute too late. :) Thank you for being so patient!

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...

Will remember that, I'm new to this so please don't refrain from pointing out any other things like that.

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...

Better? v=\frac{d(R_{e}-r)}{dt}=-\frac{dr}{dt}=\sqrt{\frac{2*G*M}{r}}

Did I completely misunderstand you or is this what you are referring to? v=\frac{d(R_{e}-r)}{dt}=\sqrt{\frac{2*G*M}{r}}

Can anyone give me a hint on b)?