Recent content by AdamP

Well it is a little half-hearted to say i like ME more since I have never been exposed to an actual engineering class. I say I like ME more since, I would rather work on a hydrolic arm or an engine than a circuit board.

I am a transfer student so all I have is 2 years :)

Hi guys i am transferring out of a community college in los angeles to either a UC or a cal-state campus. Prospects are ucla, UCsandiego, Ucirvine, UC santabarbara, calstate northridge, cal state long beach... I have to submit my application in 2 days and I still can not decide on what to do...

Before I go by the way tested the above method on one more problem and it works, so its good.

No the above statement is also wrong, for some reason that I do not know, phase angle is always subtracted. No matter if current leads or lags. I think the sign is decided by the sign of the phase angle anyways, which gets it from the phase angle = inversetan((Xl-Xc)/R). So no need to compensate...

Dang it its the exact opposite, so if current leads you add the phase angle if current lags you subtract the phase angle. VL=152 Vc=-68 Vr=122 total is 206volts emf at that time 212 volts, close enough for me. So I guess this is solved.

Hmm, I think if Xc>XL then its a capacitive circuit, current leads voltage, you subtract the phase angle, if XL>Xc than its an inductive circuit, current lags voltage so you add the phase angle, above question XL>Xc so let's add the phase angle. and try it...

Ok I found my own mistake, I am omitting the phase angle at the last calculations. It needs to be added or subtracted inside the sin parenthesis. I still don't know when to add and when to subtract though, so if someone can clear that up that would be great.

Homework Statement R=250 L=6 C=15microf e(t)=300sin(50pi t) Find Z, phase angle, lead/lag, emf(215ms) (got all of this right) Voltage across inductor, capacitor and resistor, also show that the sum of the voltage across the elements at 215ms is equal to total emf at 215ms. Homework Equations...

Ok the assignment states for partial credit you can assume Bi=B(i) for all i, Bx=B(x) By=(y) I guess that is what you are talking about since supposedly this makes the problem much easier, however does not provide full credit.

Well, they are given as vectors with (x,y,z) coordinateness but no function of (t) is implied. However velocity and position r, are supposed to be found as functions of time v(t), r (t). I am not sure if that forces the E B and v to be as functions of time also. How did 1/2dv^2/dt become 1/2...

Oh excellent tnx tim. So dv/dt is a, and we dotted that with v, which gave us vdv/dt and got rid of the -(q/m)vXB. Then vdv/dt integrated to 1/2dv^2 /dt and so then it is equal to intg(qEv). 1) Now by doing all this are we closing in on the E? Like single out E on the right hand side...

I was away due to other exams and did not do any additional work on this problem, but I still need to solve it. It looks like we are making progress all though you are doing most of the work. :) Anyhow, I flat out don't get that last part. I don't see how you got 1/2 d(v.v)/dt . after...

Ok ok I am slowly getting it. I think i struggled because I didint know about the (AxB).C = (CxA).B. So ok (VxB).V =0 from VxV=0 . If I did that I am getting vdv/dt=qEv . Then I played around a little v=dx/dt and dv/dt=d2x/dt=a . Probably vdv/dt equals something else..hmmm I mean can I now...

kinetic energy formula has another variable than the velocity and the problem does not set it as constant so...