I am wondering if there is a way to transform figure 1 to figure 2 as in the pictures.

In this case Vs, Rs, L1, L2, Iout and Vout are the same in both pictures.
V3 and I3 change to V3' and I3' respectively.

Are there V3' and I3' that satisfy that conditions?

PS. AB terminal is connected to other part of the circuit.

 

You've posted in the homework help forum, and the rules:

http://forum.allaboutcircuits.com/showthread.php?t=3002

require that you show effort to solve the problem yourself before someone will help you.

If this isn't a homework problem (like your previous thread wasn't), then you probably shouldn't be posting in the homework help forum.

 

OK, I will do so. The problem is that I wanted to know if that is possible before diving into the calculation.
If that is impossible, then all my work seems a waste of time.

 

Because AB is connected to other part of the circuit I get stuck.
My thought is to find Vout and Iout in the first figure as a function of Vs, Rs, L1, L2, V3, I1.
Then replace that results into figure 2 and find V3' and I3'.
Does that method work?

 

Does that method work?

Probably, but the only way to know for sure is to give it a try. So either you have to dive into the calculation, or someone else will have to. I vote for you doing it.

If you have a problem with the calculation, then you can explain just where you're having the problem, and help will be forthcoming.

 

If you have a problem with the calculation, then you can explain just where you're having the problem, and help will be forthcoming.

In this case, all sources are ac signals (3 ac sources). This means that I have to divide it into 6 cases and calculate for all 6 cases?
I mean that we don't know the direction of each source and we have to compute for all cases.

 

I can't get started. The problem is that the AB terminal is connected to other circuit. If it stands alone then I think I can manage to do it. However, the circuit is connected to other part and in this case we also can't use Thevenin theorem.
Is there any suggestion?
I can calculate myself. Therefore, I need some hints.

 

I used the Thevenin theorem with two circuits.

The first one:
Open circuit voltage: Voc1
Short circuit voltage: Isc1
Thevenin impedance: Zth1

The second one:
Open circuit voltage: Voc2
Short circuit voltage: Isc2
Thevenin impedance: Zth2

Here is the calculation for the first one:

 

And here is the result for the second circuit:

 

Now I set up equations:

Voc1 = Voc2
Zth1= Zth2 ( or Isc1 = Isc2)
Here:
Voc1 = Voc2, I got:

Isc1 = Isc2, I got:

And finally I have a set of equations. However, it has no solution.

Please tell me where I am wrong.

 

It may or may not be possible to get those two circuits to be equivalent as seen at the terminals A:B. But consider this -- if they are equivalent and since both topologies are linear, they would both have to have the same Thevenin or Norton equivalent, right?

So you could find the equivalent circuits for both topologies and then set the sources and the impedances equal. That will give you the constraints that have to be met.

 

I can't get started. The problem is that the AB terminal is connected to other circuit. If it stands alone then I think I can manage to do it. However, the circuit is connected to other part and in this case we also can't use Thevenin theorem.
Is there any suggestion?
I can calculate myself. Therefore, I need some hints.

I see you've been busy while I was in bed, and now that I'm up, you're probably in bed!

Looking at the two circuits in your first post, clearly V3 can be either to the left of L1 or to the right of L1 and it makes no difference. Just merge V3 with Vs and forget about it.

The only problem as I see it is what I1' has to be. Try setting V3' = V3 and then solving for I1'.

 

I had two equations.

The first equation, Voc1 = Voc2, and the second one (with a little simplification the equation in my previous post), Isc1 = Isc2:

Now if I set, V3' = V3.
From the first equation, we have:

From the second equation:

Two solutions are different. This means that we have no solution?

 

It may or may not be possible to get those two circuits to be equivalent as seen at the terminals A:B. But consider this -- if they are equivalent and since both topologies are linear, they would both have to have the same Thevenin or Norton equivalent, right?

So you could find the equivalent circuits for both topologies and then set the sources and the impedances equal. That will give you the constraints that have to be met.

Yes, that is what I did in previous posts. However, according to that result, there is no solution.
I am wondering if there is a way to get two circuit equivalent if L1 and L2 in the second picture also can be changed.

 

From my result in post #13, can I transform it as follows?

And then we loose a bit about conditions. Assuming that L2 also can be changed and in the second picture, it is L2' not L2.

That doesn't seem right to me because the expression of L2 also consists of "s".

 

Could anyone confirm if the method works?

 

Yes I1' will be given by the relationship

\(I_1^'=\frac{R_s + (L_1+L_2)s}{R_s}I_1\)

If you go back to the complex frequency domain form it is probably more clearly perceived as:

\(I_1^'=\frac{R_s + j(X_{L1}+X{_{L2}})}{R_s}I_1\)

So there is both an amplitude and phase (lead) adjustment required for the equivalent current source.

In my opinion nothing else needs to change in the equivalent dual circuit other than the current source. V3' and V3 will be the same, Vs will remain the same and all passive component values remain unchanged.

 

Thanks. The problem is that when I set V3' = V3 and solve for I1' there is no I1' satisfies the conditions to make two circuit the same.

I set V3' = V3 and then transformed each circuit into Thevenin model.
Two circuits are equal if:
Voc1 = Voc2
Isc1 = Isc2 (or Rth1 = Rth2)

But as in my post #13, the set of equations from this have no solution.
This means that two circuits are impossible to equal with given conditions.
I am wondering if we can make them equal by loosing a bit conditions, for example, in this case, Ls' or Lg' doesn't need to be the same as Ls or Lg.

 

Thanks. The problem is that when I set V3' = V3 and solve for I1' there is no I1' satisfies the conditions to make two circuit the same.

I respectfully beg to differ.

 

I respectfully beg to differ.

I hope I am wrong somewhere. Then maybe there is an answer.
It would be great if you could tell me where I am wrong.