# Homework Help: Find thevenin circuit/ phasor domain

1. Apr 28, 2013

### asdf12312

1. The problem statement, all variables and given/known data
obtain a thevenin equivalent circuit given that iS(t)=3cos(4x10^4)t A. to that end:
a) transform the circuit to the phasor domain.
b) apply source-transformation technique to obtain the thevenin equivalent circuit at terminals (a,b).
c) transform the phasor-domain thevenin circuit back to the time domain.

2. Relevant equations
Z(C) = -j/wC
Z(L) = jwL

3. The attempt at a solution
realizing that the phasor counterpart to the current was 3A, and given w=4x10^4, was able to calculate 1st the impedences of capacitor/inductor, Z(C) = -50j, and Z(L) = 40j.

combined Z(R1) + Z(L) to obtain Z1 = 25+40j
ignoring the 350/37ohm, was able to combine R2+Z(C) to obtain Z2= 35-50j

now tried node analysis, by designating V1 as the node just above the current source.
V1/Z1 + V1/Z2 - iS(t) = 0
V1/Z1 + V1/Z2 = iS(t)
V1/25+40j + V1/35-50j = 3

i could then find V(th) as V1*Z(C)/(Z(C)+R2). i haven't worked this out yet but can anyone tell me if this is the right approach?

2. Apr 28, 2013

### Staff: Mentor

It looks like it will lead you to Vth.

3. Apr 28, 2013

### asdf12312

(V1/25 + V1/35) + j(V1/40 - V1/50)= 3
0.0686(V1)+j0.005(V1)=3
V1= 3/(0.0686+j0.005) = 3<0 / 0.0688<4.17= 43.6<-4.17deg

V(th)=(V1*Z(C))/Z2 = [43.6<-4.17 * 50<-90]/61.033<-55
=2180<-94.17/61.033<-55= 35.72<-39.17deg = 35.72 cos(wt-39.17)

i do this right?

Last edited: Apr 28, 2013
4. Apr 28, 2013

### asdf12312

prety sure the math is right, not sure if the answer is though. the problem also asked to use source trasnformation but how to turn the current into voltage source. i could bring out the capacitor but then the voltage source would have an imaginry component only. how does this work?

5. Apr 28, 2013

### Staff: Mentor

Something's gone awry. V1 should be closer to 150V in magnitude. Vth will be higher, too.

edit: Check how you've dealt with your node equation. The math you've used looks suspect.

6. Apr 28, 2013

### Staff: Mentor

Find an impedance in parallel with the current source. Convert the current source and impedance to a Thevenin equivalent.

7. Apr 28, 2013

### asdf12312

my initial issue in applying source transorfmation was that this left me with a voltage source with two eq. impedences on each side. however, i realize that of course they would be in series then...? i'm sorry i just learn these things kinda slow. so what i did was convert the current into voltage source by bringing out the capacitor impedence, since thats the only thing tha seems to be in parallel with it. multiplying magnitudes, 3<0*50<-90 i do indeed get 150cos(wt-90) as Vs(t)=V(th). Z1 would still be 25+40j. for the new value of Z2 on the right of the new voltage source i would add the 350/37ohm, so Z2 would instead be 44.46-50j. adding together Z1 and Z2 i would get Z(eq)=69.46-10j. i still am not sure this is the right way because it seems too easy.

8. Apr 28, 2013

### Staff: Mentor

The capacitor is not in parallel with the current source; there's a junction in the way where two other components join. And, it's the voltage at that junction that you need to find, so best leave it alone.

Instead, look at the 25Ω + 1mH series branch. It parallels the current source.

9. Apr 28, 2013

### asdf12312

that is what i suspected, only for some reason i assumed Z1 was in series with it. 47.17<58*3<0 = 141.51<58deg for Vs(t). also calculated Z(eq) to be 60+40j since i should leave Z(C) alone for now, and the 350/37ohm as well. then V(th)=(Vs(t)*Z(C))/(Z(eq)+Z(C))= (141.51<58*50<-90)/(60-10j)=7075.5<-32 / 60.83<-9.46= 116.32<-22.54deg

Last edited: Apr 28, 2013
10. Apr 28, 2013

### Staff: Mentor

173.98 doesn't look write for 3x47.

11. Apr 28, 2013

### asdf12312

sorry that was a mistake on my part. edited my post.

12. Apr 28, 2013

### Staff: Mentor

Well, it looks much better now