Recent content by SaShiMe

ohh, so Vc=-ib(-j) =-5.785<-91.46 =-5.785sin(5t-91.46) (sin because current lead voltage in capacitor)

hmm, Vc=ib*(-j) Vc=(5.785<-1.46)(1<-90) did i get the sign wrong?

so Vc=ib*(-j) Vc(t)=5.785<-91.46 =5.785sin(5t-91.46)

ok I've got ib=5.785<-1.46 =(5.783-j0.147) ia=(14.217+j0.147) =14.29<-0.5924 i know i need ib and ic to caculate, but i don't know how...

OHH, ia+ib=20<0 ia-loop ia(3+j5)=0 ib-loop ib(15+j25-j)+(ib-10<0)(12+15j)=0 so instead of ia loop=0 and ib loop=0. i have ia(3+j5)=ib(15+j25-j)+(ib-10<0)(12+15j) is that correct?

ok, but i have no way to know the voltage drop across 20<0 and 10<0 right? should i change my loop?

oh, so i have missed out that there's a voltage drop across 20<0 and 10<0 current source? is that what you meant?

voltage drop across current source in Ic loop? is it (10<0-ib)(12+j15)?

hmm that's what i am confused about, initially, i tried it this way, but i am not sure if its correct. denote phase angle as < notation. ia+ib=20<0 ic-loop (10<0*3)+(10<0-ib)(12+j5)=0 ia-loop ia(3+j5)=0 ib-loop ib(15+j25-j)+(ib-10<0)(12+15j)=0 am i correct?

Homework Statement Find the steady-state voltage Vc(t) across the capacitor. Homework Equations jwL,-j/wC The Attempt at a Solution thats all i can do, i have no idea how to even start... T.T its depressing...