Find current with source transforms

[eehelp150]

Homework Statement

Homework Equations

V=IR

The Attempt at a Solution

First I combined the -j3 capacitor and 4ohm resistor to get:
$$\frac{32-j48}{25}$$.
Then I combined that with the j4 ohm inductor to get:
$$\frac{32+j52}{25}$$.
Then I combined that with the -j2 ohm capacitor to get:
$$\frac{104-72j}{36+j2}$$
My circuit now looks something like:

I then transform the Current Source into a voltage source:
3<30° -> $$\frac{3\sqrt{3}}{2} + 1.5i$$
Z total = $$\frac{36-28j}{13}$$
V=I * Z = $$\frac{84+103\sqrt3}{26} + \frac{54-84sqrt3j}{26}$$

I now have something like this:

Have I been doing everything right so far? How do I find I?

 

[gneill]

The Attempt at a Solution

First I combined the -j3 capacitor and 4ohm resistor to get:
$$\frac{32-j48}{25}$$.

You might want to check that result.

As you "absorb" components you must modify the source value, too. Effectively you are converting between current and voltage sources as you progress, doing a Norton → Thevenin → Norton shuffle.

 

[eehelp150]

You might want to check that result.

As you "absorb" components you must modify the source value, too. Effectively you are converting between current and voltage sources as you progress, doing a Norton → Thevenin → Norton shuffle.

What is wrong with the result? Am I not supposed to combine those two parts?

 

[gneill]

What is wrong with the result? Am I not supposed to combine those two parts?

No, combining them is fine. Check your calculated result.

 

[The Electrician]

The Attempt at a Solution

First I combined the -j3 capacitor and 4ohm resistor to get:
$$\frac{32-j48}{25}$$.
Then I combined that with the j4 ohm inductor to get:
$$\frac{32+j52}{25}$$.
Then I combined that with the -j2 ohm capacitor to get:
$$\frac{104-72j}{36+j2}$$

The two steps in red are where you made your mistake. You must combine the j4 and -j2 in series, and then that equivalent combines in parallel with what you have up to that point.

 

[eehelp150]

The two steps in red are where you made your mistake. You must combine the j4 and -j2 in series, and then that equivalent combines in parallel with what you have up to that point.

this is what my circuit looks like now, with Z3 = (96-72j)/(36+2j)
Everything good?
how do I solve for current I

 

[gneill]

Not good. Your current source has to evolve as you absorb the components. Think of it as taking successive Thevenin or Norton equivalents as you go.

You started by reducing the -j3 capacitor and 4 Ohm resistor to a single impedance, then took it and the current source as a Norton model. All good so far. Then you wanted to absorb the j4 inductor. To do that you have to convert the Norton circuit into its Thevenin equivalent first. So the Norton impedance becomes the series impedance for the Thevenin model and you also have to convert the current source to a Thevenin voltage by multiplying it by that impedance. Then you can include the j4 inductor. The same thing goes for absorbing the -j2 capacitor. You need to convert your new Thevenin model back to a Norton model (source current and parallel impedance). The original current source will be "buried" behind a set of multipliers that it accumulates during the switching back and forth between model types.

Your first change of source goes like this:

 

[eehelp150]

Not good. Your current source has to evolve as you absorb the components. Think of it as taking successive Thevenin or Norton equivalents as you go.

You started by reducing the -j3 capacitor and 4 Ohm resistor to a single impedance, then took it and the current source as a Norton model. All good so far. Then you wanted to absorb the j4 inductor. To do that you have to convert the Norton circuit into its Thevenin equivalent first. So the Norton impedance becomes the series impedance for the Thevenin model and you also have to convert the current source to a Thevenin voltage by multiplying it by that impedance. Then you can include the j4 inductor. The same thing goes for absorbing the -j2 capacitor. You need to convert your new Thevenin model back to a Norton model (source current and parallel impedance). The original current source will be "buried" behind a set of multipliers that it accumulates during the switching back and forth between model types.

Your first change of source goes like this:
View attachment 107409

I = 4.47515-2.2127j
Z = (4-72j)/(36-48j)

To get the current, can I use a current divider here or do I need another conversion?
I ended up getting
I = 1.32 - 1.44i

 

[gneill]

Practically speaking you can do it either way. If they are expecting you to use source changes though, you might want to follow through to the end.

You should check your results using another method when you're done. Perhaps using Nodal analysis.

 

[eehelp150]

Practically speaking you can do it either way. If they are expecting you to use source changes though, you might want to follow through to the end.

You should check your results using another method when you're done. Perhaps using Nodal analysis.

I ended up getting
1.32 - 1.44i Amps

 

[gneill]

I ended up getting
1.32 - 1.44i Amps

Not what I'm seeing. I did the whole series of source changes down to a final Thevenin model in series with the 2 Ohm load, and I checked using nodal analysis to find the voltage at the output node then determined the current through the load from that. The results agreed, and they weren't the same as your result above

 

[eehelp150]

Not what I'm seeing. I did the whole series of source changes down to a final Thevenin model in series with the 2 Ohm load, and I checked using nodal analysis to find the voltage at the output node then determined the current through the load from that. The results agreed, and they weren't the same as your result above

My Vth is 3-5.2i
is that off as well?

 

[eehelp150]

Not what I'm seeing. I did the whole series of source changes down to a final Thevenin model in series with the 2 Ohm load, and I checked using nodal analysis to find the voltage at the output node then determined the current through the load from that. The results agreed, and they weren't the same as your result above

Is it right up to here?

I = 4.47515-2.2127j
Z = (4-72j)/(36-48j)

 

[gneill]

I don't believe so, no. In that configuration your Z should match my final Thevenin impedance and it doesn't.

Your Z:

Mine:

 

[eehelp150]

I don't believe so, no. In that configuration your Z should match my final Thevenin impedance and it doesn't.

Your Z:
View attachment 107414

Mine:
View attachment 107415

So your Z is 2.769 - 2.154j at the point in my picture?

 

[gneill]

So your Z is 2.769 - 2.154j at the point in my picture?

That would be my understanding, yes.

Perhaps we need to go through the transformations one at a time from the beginning?

 

[eehelp150]

That would be my understanding, yes.

Perhaps we need to go through the transformations one at a time from the beginning?

I know where I screwed up.
Final I should be:
-0.067 - 1.9i A

 

[gneill]

I know where I screwed up.
Final I should be:
-0.067 - 1.9i A

Yes, that looks good.