Ideal transformer problem with possible solution

Thread Starter

JoyAm

Joined Aug 21, 2014
126
Hello everyone on the problem bellow i have come up with a possible solution thats needs evaluation, thanks for helping Screenshot_7.png
What we need to find is the value of Zx that will make Vo=0
So first i took the ideal transformer equations and had V2/v1 = a ( assuming that v2 and v1 are the voltages on the secondary and primary respectively
we also have i2/i1 = 1/a => i1=a*i2 or in mesh currents we have I1-I3=a*(I2-I3)
But since vo=0 and ΖL isnt =0 that means I2 = 0 so the last equation becomes I1= (1-a) *I3
Also Vo=v2=0 so V1= 0 too

lets go to the mesh current equations now

Vin=jωL * I1 - jωL * I3
and
0= (jωL+Zx)*I3 - jωL*I1

From these two equation what first came to my mind was this solution Zx=vin/I3 but this doesnt look acceptable

So i make the assumption that the secondary and primari of the transformer are actually shortcircuits and that makes Zx and jωL two parallel impedances so we have the following equation
Vin=jωL*I1=Zx*I3 =>
Zx= jωL*I1 / I3 and because I1=(1-a)I3 we have

Zx=(1-a)*jωL.


I have the feeling that i am missing something, maybe the assumption i made was wrong?
Your help will be much appreciated.
 

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MrAl

Joined Jun 17, 2014
7,849
Hello everyone on the problem bellow i have come up with a possible solution thats needs evaluation, thanks for helping View attachment 138120
What we need to find is the value of Zx that will make Vo=0
So first i took the ideal transformer equations and had V2/v1 = a ( assuming that v2 and v1 are the voltages on the secondary and primary respectively
we also have i2/i1 = 1/a => i1=a*i2 or in mesh currents we have I1-I3=a*(I2-I3)
But since vo=0 and ΖL isnt =0 that means I2 = 0 so the last equation becomes I1= (1-a) *I3
Also Vo=v2=0 so V1= 0 too

lets go to the mesh current equations now

Vin=jωL * I1 - jωL * I3
and
0= (jωL+Zx)*I3 - jωL*I1

From these two equation what first came to my mind was this solution Zx=vin/I3 but this doesnt look acceptable

So i make the assumption that the secondary and primari of the transformer are actually shortcircuits and that makes Zx and jωL two parallel impedances so we have the following equation
Vin=jωL*I1=Zx*I3 =>
Zx= jωL*I1 / I3 and because I1=(1-a)I3 we have

Zx=(1-a)*jωL.


I have the feeling that i am missing something, maybe the assumption i made was wrong?
Your help will be much appreciated.
Hi,

I got a slightly different result and checked it to see that i get zero out.
I did not use mesh though i used Nodal because that is more straightforward.
Is there any reason you must use mesh or can you use Nodal?

Did you assume short circuits because of the zero output voltage?
I am not sure if that is valid but i will check that next. Offhand i would say that it is valid because the output load is variable and since it must work with any load, zero Ohms should be valid as well. Of course 0v also means shorting to ground does not change any current in the circuit too. This will of course change the time domain response but not the steady state AC response which is what you are dealing with here.
 
Last edited:

Thread Starter

JoyAm

Joined Aug 21, 2014
126
Hi,

I got a slightly different result and checked it to see that i get zero out.
I did not use mesh though i used Nodal because that is more straightforward.
Is there any reason you must use mesh or can you use Nodal?

Did you assume short circuits because of the zero output voltage?
I am not sure if that is valid but i will check that next.
Hello and thanks for your time again :)
I assumed that because i had zero voltage on the secondary/primary
Nodal would indeed be more straightforward but i dont like doing it so i always try to use mesh analysis against all odds :p

Did you get Zx= -a*jωL using the nodes ? i think thats what i got.
So if my assumption is wrong do you have any idea what i should do to solve it with meshes ?
Maybe consider a 3rd loop like the one you showed on my previous post ?
 

MrAl

Joined Jun 17, 2014
7,849
Hello and thanks for your time again :)
I assumed that because i had zero voltage on the secondary/primary
Nodal would indeed be more straightforward but i dont like doing it so i always try to use mesh analysis against all odds :p

Did you get Zx= -a*jωL using the nodes ? i think thats what i got.
So if my assumption is wrong do you have any idea what i should do to solve it with meshes ?
Maybe consider a 3rd loop like the one you showed on my previous post ?
Hello again,

Well in your previous reply you stated that you got:
"Zx=(1-a)*jωL"

and i did not get that so i made a comment about that.

Yes, i got -a*j*w*L as your last reply stated. That is correct i believe because when we try that in the transfer function it always leads to zero in the steady state solution.

You might recognize this as a capacitor?

A neat way to do a sanity check is to eliminate the transformer and just use L and Z and see if you can get that value of Z to get zero output, but since you solved it already you may not want ot do that.

No i think your assumption is right.
 

Thread Starter

JoyAm

Joined Aug 21, 2014
126
Hello again,

Well in your previous reply you stated that you got:
"Zx=(1-a)*jωL"

and i did not get that so i made a comment about that.

Yes, i got -a*j*w*L as your last reply stated. That is correct i believe because when we try that in the transfer function it always leads to zero in the steady state solution.

You might recognize this as a capacitor?

A neat way to do a sanity check is to eliminate the transformer and just use L and Z and see if you can get that value of Z to get zero output, but since you solved it already you may not want ot do that.

No i think your assumption is right.
When i solved with meshes i got Zx=(1-a)*jωL but when i did with nodes i got -a*j*w*L which is most likelly the correct answer, but i dont know why i didnt get the right one using meshes :/
 

The Electrician

Joined Oct 9, 2007
2,801
Interesting so by using the initial equations i had and doing only maths ( and no asumptions or whatsoever ) came to the correct result, right ?
You can see the 3 mesh equations and the two constraint equations that I used in the image. I get the same result if I solve for Vo = 0
 

Thread Starter

JoyAm

Joined Aug 21, 2014
126
Yes they look the same to me but i dont understand what i did wrong to conclude to this slightly different result. What will you get if you make I2,Vo,V1 zero from your initial equations ?
 
Yes they look the same to me but i dont understand what i did wrong to conclude to this slightly different result. What will you get if you make I2,Vo,V1 zero from your initial equations ?
I get the same result when I set any of those 3 to zero.

You can see the solution I got for I1, I2, I3, V1, Vo. Compare those to what you get. Maybe that will help you track down the problem.
 

Thread Starter

JoyAm

Joined Aug 21, 2014
126
You can see the 3 mesh equations and the two constraint equations that I used in the image. I get the same result if I solve for Vo = 0
I get the same result when I set any of those 3 to zero.

You can see the solution I got for I1, I2, I3, V1, Vo. Compare those to what you get. Maybe that will help you track down the problem.
Thanks a lot, i see my mistake now. I thought i have reached a dead-end by having 2 equations for 3 unknowns so i devised this shortcircuit thing but looking at it again thats not the case. From my initial post i have
(1) Vin= jωL *I1 - jωL*I3
(2)0= (jωL+Zx)*I3 - jωL*I1
What caused me to feel i am on a dead-end was that i utilized only these two equations and ended up with an expression of Vin that couldnt get rid of even when i used my 3rd equation (3) I1= (1-a) *I3.
But if i just replace I1 from (3) in (2) i get the desired Zx=-ajωL
Thanks a lot both of you for your interest and help:)
 

MrAl

Joined Jun 17, 2014
7,849
Thanks a lot, i see my mistake now. I thought i have reached a dead-end by having 2 equations for 3 unknowns so i devised this shortcircuit thing but looking at it again thats not the case. From my initial post i have
(1) Vin= jωL *I1 - jωL*I3
(2)0= (jωL+Zx)*I3 - jωL*I1
What caused me to feel i am on a dead-end was that i utilized only these two equations and ended up with an expression of Vin that couldnt get rid of even when i used my 3rd equation (3) I1= (1-a) *I3.
But if i just replace I1 from (3) in (2) i get the desired Zx=-ajωL
Thanks a lot both of you for your interest and help:)
Hi,

You're welcome and it was nice of Electrician to work this out too.

I have to ask why do you seem to prefer mesh equations? They are in my opinion the worst way to do this even though they may result in one less equation. You see the problems that come up now with mesh and i think that is because there is more global mechanical logic to it than with Nodal where we only have to concentrate on the mechanics of writing the equations locally. In other words, with mesh we have more to worry about and get correct than with Nodal when we write each equation.
Dont get me wrong though, i am not saying you cant use mesh except of course for a non planar circuit, but Nodal just seems so much more clear.
So i would be interested to hear why you seem to like mesh so much, especially since it is not as general as Nodal.
 

Thread Starter

JoyAm

Joined Aug 21, 2014
126
Hi,

You're welcome and it was nice of Electrician to work this out too.

I have to ask why do you seem to prefer mesh equations? They are in my opinion the worst way to do this even though they may result in one less equation. You see the problems that come up now with mesh and i think that is because there is more global mechanical logic to it than with Nodal where we only have to concentrate on the mechanics of writing the equations locally. In other words, with mesh we have more to worry about and get correct than with Nodal when we write each equation.
Dont get me wrong though, i am not saying you cant use mesh except of course for a non planar circuit, but Nodal just seems so much more clear.
So i would be interested to hear why you seem to like mesh so much, especially since it is not as general as Nodal.
Hmm thats a big question, maybe because having currents is something that i can actually see in the picture apart from that i agree that in this problem a nodal analysis would have been easier, same thing on the last one.
There are many problems that from the begining look like thet " are meant to" be solved with nodes but i still go for meshes :p
All in all maybe its just because i havent used it much and i dont feel comfortable with it but even that its irrational since its a pretty straight forward method :p
 

RBR1317

Joined Nov 13, 2010
573
Hmm thats a big question, maybe because having currents is something that i can actually see in the picture...
Nodal analysis deals with real currents that obey Kirchoff's Current Law, whereas mesh analysis utilizes fictitious currents circulating in loops. If you need something to actually see in the picture, then you can draw the branch currents flowing at each node, although few people using nodal analysis see any need to do this. As an example I have done this for the given circuit and color-coded the currents in the circuit to the corresponding terms in the node equation. But once you have done this a few times there should be no need to actually see it in the picture anymore.
 

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Thread Starter

JoyAm

Joined Aug 21, 2014
126
Nodal analysis deals with real currents that obey Kirchoff's Current Law, whereas mesh analysis utilizes fictitious currents circulating in loops. If you need something to actually see in the picture, then you can draw the branch currents flowing at each node, although few people using nodal analysis see any need to do this. As an example I have done this for the given circuit and color-coded the currents in the circuit to the corresponding terms in the node equation. But once you have done this a few times there should be no need to actually see it in the picture anymore.
Thanks a lot :)
 
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