RLC AC circuit can't get right answer

[AdamP]

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

XL=942 Xc=424 XL>Xc therefore current lags voltage, phase angle 1.12 rad emf@215ms=212V
Z=575ohm

The Attempt at a Solution

Up to this point I have everything correct. While solving the VL,Vc,Vr I use
VL=ImaxXL sin(wt+pi/2) = -339
Vc=ImaxXc sin(wt-pi/2) = 155
Vr=ImaxR sin(wt)= 91

I got Imax from Emax/Z = 0.52A.
emf(215ms)=300sin (50pi . 215 . 10^-3)= 212 . VL+Vc+Vr should be equal to emf at that time, mine does not equal to that.
What did I do wrong? Am I supposed to use the phase angle somewhere while calculating the voltage across the capacitor, inductor and resistance at a particular time( 215ms in this case)?

Edit:Also when using the phase angle, do we add or subtract it to (wt) depending on if the current is leading or lagging the voltage? How does that actually work out? Like what is a good way to decide + or - phase angle?

 

[AdamP]

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.

 

[AdamP]

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...

 

[AdamP]

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.

 

[AdamP]

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 for it again.

Now its solved :)

Ps: I just had a conversation with my self for about 30 minutes on a public forum, I think it is time to take a break from studying for the physics final...

 

[AdamP]

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