Electrical Engineering - Steady-State Voltage across Capacitor

[SaShiMe]

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...

 

[vela]

So far so good. With the complex impedances, you can treat everything like a resistor.

Write down the three loop equations for the circuit.

 

[SaShiMe]

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?

 

[vela]

No, you forgot the 3-ohm resistor in the Ic loop. You didn't account for Vc in the Ib loop, and you didn't account for the voltage drop across the current source in both the Ia and Ib loops.

EDIT: Oh, I see you did have the 3-ohm resistor in there. You forgot the voltage drop across the current source in the Ic loop.

 

[SaShiMe]

voltage drop across current source in Ic loop?

is it (10<0-ib)(12+j15)?

 

[vela]

No, that's not quite right. KVL applied to that loop tells you it's going to be (10<0-ib)(12+j15) plus the drop across the 3-ohm resistor. In any case, I don't think the Ic loop equation is too important because you already know Ic=10 A.

The basic mistake you made in your equations is assuming the voltage across a current source is 0. It's not. It will be whatever voltage is needed so that the given amount of current will flow through the source.

 

[SaShiMe]

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?

 

[vela]

Yes, that's what I meant.

 

[SaShiMe]

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

 

[vela]

No, you really only have two unknowns: Ia and Ib, so all you need are two equations. You already have one from KCL: Ia + Ib = 20. The other one you can get by combining the two equations for the Ia and Ib loops and eliminating the unknown voltage drop across the 20-A source.

 

[SaShiMe]

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?

 

[vela]

Close. The righthand side of the Ia and Ib equations shouldn't be 0; it should be equal to the voltage drop across the 20-A source:

ia(3+j5)=v₂₀
ib(15+j25-j)+(ib-10<0)(12+15j)=v₂₀

Then when you combine them to eliminate v₂₀, you get

ia(3+j5)=ib(15+j25-j)+(ib-10<0)(12+15j)

 

[SaShiMe]

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...

 

[vela]

Just use Ohm's law to calculate the voltage across the capacitor since you know the current flowing through it now.

 

[SaShiMe]

so Vc=ib*(-j)

Vc(t)=5.785<-91.46
=5.785sin(5t-91.46)

 

[vela]

Looks good! (Maybe check the overall sign?)

 

[SaShiMe]

hmm,

Vc=ib*(-j)
Vc=(5.785<-1.46)(1<-90)

did i get the sign wrong?

 

[vela]

Consider the direction of Ib and the polarity of Vc given on the circuit diagram.

 

[SaShiMe]

ohh,

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