# Solving a second order circuit with phasors

1. May 21, 2012

### Evales

The question requests that I solve the circuit below for v0(t). I'm solving for the voltage over the inductor. I'm getting a result that's close to what I expect, however I think the phase angle of the voltage is slightly off (some friends of mine said they all got 46°, whereas I am consistantly getting 45.02°).

My equations:
ix=(16∠45°)/(R)=(16∠45°)/(4E3) = 4E-3∠45°

KVL of second loop:
0=0.754j(i1+8E-3∠45°)+4E3(4E-3∠45°)-53.05j(i1)
i1=0.306∠-44.98°

v0=0.754j(0.306∠-44.98°) = 0.231∠45.02°
v0(t)= 0.231cos(377t+45.02°)V

Image of the circuit:

2. May 21, 2012

### stef6987

EC203 assignment? ahah i got 46.59 degrees as the final angle, don't use KVL, you know at the 4k resistor the voltage is 16<45, all you need to do is apply a KCL at node v0(t) and just solve. your final answer is very close to mine, try my method and i'm sure your answer will be the same as your friends :)

3. May 21, 2012

### stef6987

Sorry i just realised you can use your method, you made one mistake, the current through the inductor is i1-2ix

4. May 21, 2012

### Evales

Thanks, I knew one day I would bump into someone doing the same course as me! I'm not sure why I added the currents rather than subtracting them. Weird, and I've redone this one like 3 times haha.

5. May 21, 2012

### stef6987

haha :) My final answer was 2.30<46.59 volts, i don't think a little difference would matter much :) how did you go on the rest of the assignment?

but... i'm pretty sure my answer is off a bit, the angle should be 46.59 - 180... due to it being in the 3rd quadrant.. so my angle would be -133.41... what was your final answer?

Last edited: May 21, 2012
6. May 22, 2012

### Evales

How can you tell the answer is in the third quadrant? I'm redoing the question now, I'll edit this comment as I complete it.

EDIT: I now get an even more incorrect result: 0.231∠44.98°
Redoing the same calculations on my calculator now gives: 0.231∠45°. *facepalm.jpg*

Last edited: May 22, 2012
7. May 22, 2012

### stef6987

i re did mine and got 0.231<-133.4 degrees or something around their,

your answer is rectangle form should be -0.158 - j0.167. because both terms are negative it lies in the third quadrant, so you have to subtract Pi radians or 180 degrees, and that is the answer,

8. May 22, 2012

### Evales

I think I might just grab my previous answer and run with it/ switch it to the third quadrant, rather than going over this question again and again.

Just to check results from other questions:
4. RTH=3Ω, Max Power = 3 Watts
5. IO=1.301∠-77.47° (this one I converted the current source to a voltage source to reduce the circuit)

Thanks, don't know anybody in the course really well, it's awesome to have someone to bounce ideas off XD

9. May 22, 2012

### stef6987

just use KCL, it's pretty straight forward if you do it like that

And yeah i got the same results for 4 & 5. have you done 6 yet?

10. May 22, 2012

### Evales

Just completed it, I got VS=0.252∠165.04°

11. May 22, 2012

### stef6987

i got 9.4j - 18.36 volts :S how'd did you do it?

12. May 22, 2012

### Evales

I made the relationships with the transformers first.
Image of the work:

And then did 3 KVL's for the last three loops. I just looked at my work though and I think I made an incorrect relationship with the current of the source loop.

Just to make it less confusing i1 is the current of the loop to the right of the first transformer and the I1 is the current given.

13. May 22, 2012

### stef6987

why are you multiplying the voltages by the currents? :S

14. May 22, 2012

### Evales

Good freaking question, zomg, my mind is empty atm. Lol

15. May 22, 2012

### stef6987

haha :P this one was probably the longest, just write all 4 kvl's, because they are ideal the only purpose of the dots is to show the relationship between the transformer voltages and currents, so if you start from the right mesh you can work your way to the left and solve Vs, takes time but eh, look at the lecture notes he solved a really similar problem :)

16. May 22, 2012

### Evales

I did 4 KVL's and a KCL (because I couldn't find a relationship with the source current).
The KCL is the middle node of the, between the transformers.

Got the same result. Yeh I saw him solve the similar one, just brain farted and started multiplying voltages and currents to get voltages lol.

Thanks for the help! I bet I would have spent ages looking at that thinking there was nothing wrong, lol.

17. May 22, 2012

### stef6987

No need for the KCL, if all currents are going clockwise, you can say I2 = 4<0 + I3, i'll post up my results when i get home if you like?

and have you done Q7, i only have that to do and i'm not sure if my transfer function is right, my numbers are ridiculous haha.