Hello once again i am having a hard time understanding why the signs are the way they are, In the pictures bellow i have the solution to a problem where i am suposed to use the star equivalent to solve it . The first thing he does is to change the current I2 in order to be able to determine the values of the new Ls. But after that i dont understand why I2 is always positive when he is doing the mesh analysis in the equivalent circuit



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Once again you don't show us the problem, just the solution. So how should I know why he does what he is doing?

He uses terms I2 and i2, but I don't see i2 defined anywhere except at the end where he claims I2 = -i2, so I assume its direction was shown in the problem statement? The answer to your question is to pay attention to what he is doing with the direction of current in the right hand loop. The sign is arbitrary, it merely depends on how the problem statement defined the direction of I2 or i2 or whatever. For whatever reason he may have chosen to solve with the current in the opposite direction of how it was defined and then just negate it at the end to get the correct polarity.

His mesh equation is wrong, (the signs you noted), if referenced to b). It was bad form for him to define I2 in two different directions, maybe he confused himself. Or maybe if I saw the problem statement it would make some sense to me.

 

Thanks for your answer DG here is the original circuit and the problem is
Solve the problem in Example 13.1 (see Fig. 13.9) using the T-equivalent model for the magnetically coupled coils.


Mod edit: shown image full size again

 

Thanks for your answer DG here is the original circuit and the problem is
Solve the problem in Example 13.1 (see Fig. 13.9) using the T-equivalent model for the magnetically coupled coils.
View attachment 110702

Mod edit: shown image full size again

I still get the feeling I am not getting the context of this problem. Anyway, his equation is correct, apparently, because in eq 1 and eq 2 he is not solving for the I2 you see in figure b) or the original problem statement, he is solving for i2 which he later defines as -I2. A backassward approach if you ask me. No wonder you are confused.

Do his signs make sense to you now?

 

OK, I think I see what's going on. If the currents shown in figure a) had been notated as i1 and i2, then his equations have the correct sign. Though still miss-leading since he suggests that they follow from fig b). It was probably a mistake in the printing of the book.

 

I still get the feeling I am not getting the context of this problem. Anyway, his equation is correct, apparently, because in eq 1 and eq 2 he is not solving for the I2 you see in figure b) or the original problem statement, he is solving for i2 which he later defines as -I2. A backassward approach if you ask me. No wonder you are confused.

Do his signs make sense to you now?

OK, I think I see what's going on. If the currents shown in figure a) had been notated as i1 and i2, then his equations have the correct sign. Though still miss-leading since he suggests that they follow from fig b). It was probably a mistake in the printing of the book.

Thanks for your answers DG
so what he basically does is this ?

 

Forget his garbled presentation. Why don't you solve the problem - show all your work. Then I will tell you if you got it right.

By the way he shows the wrong primary inductance in fig. a).

 

Hi,

Two port theory is sometimes based on the two external currents going INTO the network, so to comply with a given theory you may have to reverse one or both currents.

 

here is what i got in two different approaches

 

While you didn't solve the problem, you did set it up correctly with the dependent sources.

With the T equivalent you set it up wrong. The second loop current (I2) shown in the problem statement and (I3) shown in your diagram are going in the same direction. So why would you claim I3 = -I2 ?

After you establish the T-model, you should simply use the current directions shown
in the problem statement, i.e. two clockwise loop currents. Of course you can choose
whatever direction you want to solve for the mesh currents as long as you take care of the sign reversal so I2 agrees
with the problem statements direction, but there is no reason to complicate the problem
by introducing another variable I3. And in this case your I3 = I2 anyway.

By the way, the authors solution for I2 is correct. Probably I1 is as well but I didn't check it.

 

While you didn't solve the problem, you did set it up correctly with the dependent sources.

With the T equivalent you set it up wrong. The second loop current shown in the problem (I2) and shown in your diagram (I3) are going in the same direction. So why would you claim I3 = -I2 ? It clearly does not.

You and the author are over complicating the problem with these sign reversals.
After you establish the T-model, you should simply use the current directions shown
in the problem statement, i.e. two clockwise loop currents. Of course you can choose
whatever direction you want as long as you take care of the sign reversals so I2 agrees
with the problem statements direction, but there is no reason to complicate the problem
by introducing another variable I3. And in this case your I3 = I2 anyway.

By the way, the authors solution for I2 is correct. Probably I1 is as well but I didn't check it.

Hello both of you,

JoyAm's solution in the first page is correct. I didnt check the second page there. The first page shows I3=-I2 at some point.

I2 must be the negative of I3 because I2 is right for the model and I3 is reversed.

The model theory is based on BOTH currents going INTO the network. That is, with both currents going downward through M. When we apply the clockwise mesh current we reverse this so I3=-I2 or I2=-I3. So if we write the equations with both currents clockwise then we have to set I3=-I2 before we are done.
Note we can do this while writing or just after solving for I1 and I3.

Here is a set of equations with the input capacitor shorted out. To prove how the theory works that is ok because we are just concerned with the transformer and load and input voltage.

I1*Las+I1*Ms-I3*Ms=V1
I3*Ms+I3*Lbs-I1*Ms=I3*R12

Now we could either replace I3 with -I2 or wait until we solve for I3, then just make I3=-1*i3 and then go from there with the numericals.

The values shown are impedances in symbolic form and remember this is for the transformer alone with only the load resistor of 12 ohms.

The most importance aspect of I3 and I2 is the effect on the output PHASE. If I3 is not made equal to -I2 then we get the wrong output phase shift.
Note that for the transformer the phase shift should be roughly close to zero degrees. If we dont use I3=-I2 then we would get something close to 180 degrees which is wrong.

So with this in mind the first solution you found after changing I3 to -I2 should be correct. You can verify the amplitude and phase shift with a simple circuit simulation.

Yes one of the drawings is wrong because they show j3 instead of j5.

 

Hello both of you,

JoyAm's solution in the first page is correct. I didnt check the second page there. The first page shows I3=-I2 at some point.

I2 must be the negative of I3 because I2 is right for the model and I3 is reversed.

The model theory is based on BOTH currents going INTO the network. That is, with both currents going downward through M. When we apply the clockwise mesh current we reverse this so I3=-I2 or I2=-I3. So if we write the equations with both currents clockwise then we have to set I3=-I2 before we are done.
Note we can do this while writing or just after solving for I1 and I3.

Here is a set of equations with the input capacitor shorted out. To prove how the theory works that is ok because we are just concerned with the transformer and load and input voltage.

I1*Las+I1*Ms-I3*Ms=V1
I3*Ms+I3*Lbs-I1*Ms=I3*R12

Now we could either replace I3 with -I2 or wait until we solve for I3, then just make I3=-1*i3 and then go from there with the numericals.

The values shown are impedances in symbolic form and remember this is for the transformer alone with only the load resistor of 12 ohms.

The most importance aspect of I3 and I2 is the effect on the output PHASE. If I3 is not made equal to -I2 then we get the wrong output phase shift.
Note that for the transformer the phase shift should be roughly close to zero degrees. If we dont use I3=-I2 then we would get something close to 180 degrees which is wrong.

So with this in mind the first solution you found after changing I3 to -I2 should be correct. You can verify the amplitude and phase shift with a simple circuit simulation.

Yes one of the drawings is wrong because they show j3 instead of j5.

If you show two diagrams for the same circuit with currents going in the same direction then they must have the same sign. Right? She shows I2 and I3 for the same loop both going clockwise, but writes I2=-I3. They can't both be right.

Check her second solution setup, why do you think it is wrong?

Yes, the standard formula for the two port T-model assumes the currents both go into the top terminals.
But after you have created the T-model, you can chose whatever direction you want for your currents to solve the meshes - it will only change the sign. The problem statement wants to know the value of the two loop currents going clockwise.

jI1 - j3*I2 = 12, not jI1 + j3*I2 = 12

 

Hi there DG,


I think what might be happening here is there is some ambiguity lurking around which we need to clear up. I think i might agree with you once we do this

The only way to be sure is to solve for a NUMERICAL value of Vout with the reference point at the bottom of Vout. That way we will be comparing the same things. We should both come out with the same voltage, which i think we will.

What i would suggest to make the ideas more clear is to short out the capacitor and solve it like that. This way we wont be giving the answer right out either
This also makes it a little easier to solve, and the phase of the output is more apparent then too because we just have a transformer and resistor and no cap that could change the phase to a more unusual value. With just the transformer, we should get either close to 0 degrees or close to 180 degrees with respect to the input voltage which we take to be at 0 degrees.

So if you can render an output voltage using JUST the transformer and load resistor of 12 ohms, then we can both come up with some numerical values and compare directly.
Otherwise one of us might be assuming a negative current and the other a positive current or something like that. With a numerical value this cant happen.
If the phase angle is very small then i wont worry about it, can call it zero, but if it is more like plus or minus 179 degrees then i would have to call it at least estimated at 180 degrees. This also goes for the amplitude which should be always positive, but if the amplitude is shown as negative then i have to assume 180 degree phase shift in addition to the phase shift if also given, or just 180 degrees if no phase shift is given.
So for a few examples:
1.2, 0 degrees
-1.2 (assume 180 degrees then)
4.5, 179 degrees
It would be more preferable to show the amplitude as positive and the phase shift or approximate phase shift with it such as:
1.2, 10 degrees
1.2, -12 degrees
like that.

Also, the input voltage can be taken to be 1 volt peak or rms just for the comparison test.

So if you want to go ahead and do that and quote your finding here, i will do the same. If you dont like removing the cap that's ok too, but it just complicates things and does not help prove the ambiguity with the current I2 or I3 is one way or the other.

I am betting that we get the same result, and that we are just explaining it differently.
I was assuming that I2 was the current in the model, and I3 was JoyAm's assumption then, but a numerical result for Vout will show the right way and the wrong way, and Vout is the voltage across R12 with positive on top and reference 0v on bottom as usual.

Note that if we 'short out' the transformer too then we get Vout=Vin with no phase shift. By short out i mean short the top two terminals together and the bottom two terminals together which are then both at the same ground 0v potential. That's just a quick test set but of course does not provide the correct answer.

We'll get this straightened out

 

Hello again,

We did not hear back from DG yet so i'll post my results now.

These results can not be confused because they are direct numerical values.

With values L1=5 and L2=6 and R12=12 ohms, and with a resistor of 0.001 ohms in series with the first inductor L1 the output across R12 with top positive and bottom ground 0v reference, and with a 1v peak AC sine wave input of frequency 0.02 Hz, i get close to the following values:
Vout=0.6v peak
Phase=0 degrees

I say "close" to the values because these are not exact but are close enough if we were to inspect the output waveforms by eye on a scope. The values above were also confirmed using LT Spice using the transformer model they recommend, and the transformed T network output was confirmed using a different simulator and is in agreement with the LT Spice simulation using the coupled inductors transformer model.

What this tells us is that if we calculate the output voltage and phase we should get close to those values. When the input voltage gets near the positive peak the output voltage gets near the positive peak, and that's really good enough because if we were to really reverse I2 or I3 from what they should be we would see the output peak near the negative most peak when the input is near it's positive most peak. That's a phase reversal of 180 degrees or close to that. So being off by a little would be acceptable but being off by 180 degrees can not be right at all. This is how we confirm the right polarity of I2 or I3 or whatever we want to call the secondary current.

For reference, the circuits in post #1 (a) and (b) are assumed to have the lower ends of each component connected to ground which is 0 volts. This means the circuit in (a) also has the two lower inductor ends connected together.
Also, the capacitor is shorted for simplicity in demonstrating the action of the two coupled inductors and later the three isolated inductors in the T model. We can include the cap later if desired but that wont help us discover the ambiguity in the transformer model currents.

The ambiguity exists because we might want to draw I2 or I3 in one direction or another but then either call it positive or negative, but with the numerical value for the output (referenced to ground 0v) can only be of one phase angle regardless how we do it. We can do it anyway we want to, but if we come up with something close to 180 degrees and not something close to a zero degree phase shift then we did it wrong