Auto Transformer Question

In summary, the conversation is about solving for the correct equations for autotransformers and understanding the dot convention. Participants discuss the correct equations for V1 and V2, with one suggesting that V1 = jωL1I1 + jωMI2 and V2 = jωL2(I2 - I1) + jωMI1, taking into account the direction of current flow and the number of turns in each coil. Another participant shares that the gain for V2 is V1/3, based on the ratio of turns in the two coils. The conversation also includes a link to a similar example for reference.
  • #1
Hello, I have the following question regarding autotransformers and dot convention (please see attached).

Can anyone give any guidance? I think I have an answer for part (a) but not completely sure if this is correct as I haven't come across a similar example before, sorry about the lack of subscripts on the numbers.

(a) V1 = jωLI1 + jωM(I2 + I1)

V2 = jωL2 (I2- I1) + jωMI1

If the above is correct I think I can do (b) and (c) but if it's incorrect then I would really appreciate some pointers.

Thanks
 

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  • #2
Your eq. for V2 is correct but that for V1 is not.

Remember fundamentally: v = jwL1i1 +/- jwMi2
where v = voltage across L1, i1 = current thru L1 and i2 = current thru L2. So how does that change your first equation?

Hint: the answer to part (c) does not include R.
 
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  • #3
Thanks for your pointer. So should V1 = jωL1I1 + jωMI2 ? because the current is
entering at the dotted end of L1 and therefore the mutual inductance is added to the self inductance of L1?
 
  • #4
alex.daciz said:
Thanks for your pointer. So should V1 = jωL1I1 + jωMI2 ? because the current is
entering at the dotted end of L1 and therefore the mutual inductance is added to the self inductance of L1?

No, because V1 is not the voltage across L1, and i2 is not the current thru L2!
 
  • #5
Woops! Ok another attempt: V1 = [ jωI1 (L1 + M) ] - [ (I2-I1) (jωL2 + M) ]

My thinking here is that L1 has self and mutually inductance emf components that are positive with respect to the dot and direction of I1. While L2 has current I2-I1 across it but itself and mutual inductances are opposing L1 and therefore subtracted from L1? Am I close!?
 
  • #6
Oh dear.

I'm going to give you the equation and you should think about why it's the correct one:

V1 - V2 = jwL1*i1 + jwM(i1 - i2).

Maybe there is some confusion here, though. I am assumimng L1 is the self-inductance of the top part of the transformer only, not the entire winding. So, with no load,

V2/V1 = L2/(L1 + L2), not L2/L1. Did you see it that way? I do believe my interpretation is the intended one.

Next of course you need to eliminate M. V2/V1 should be a function of L1 and L2 only, since coeff. of coupling = 1.0.
 
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  • #7
BTW I was wrong when I said your other equation was correct. I didn't notice your sign change. It's actually V2 = jwL2(i1 - i2) + jwMi1.
 
  • #8
rude man said:
Maybe there is some confusion here, though. I am assumimng L1 is the self-inductance of the top part of the transformer only, not the entire winding. So, with no load,

V2/V1 = L2/(L1 + L2), not L2/L1. Did you see it that way? I do believe my interpretation is the intended one.

The schematic plainly indicates that L1 is the top inductor. Also, since L1 = 4L2, there are twice as many turns on L1 as on L2. If there are N turns on L2, then there will be 2N turns on L1 and 3N turns for the whole L1+L2 winding. With k=1, the whole thing is wound on a single core, with very tight coupling. Calling it an autotransformer is an apt description.

Given these facts, I can tell you what the ratio V2/V1 is just from knowing how variacs work.

But, if L1=4L2, then the ratio V2/V1 = L2/(L1+L2) = 1/5 doesn't seem right.
 
  • #9
The Electrician said:
The schematic plainly indicates that L1 is the top inductor. Also, since L1 = 4L2, there are twice as many turns on L1 as on L2. If there are N turns on L2, then there will be 2N turns on L1 and 3N turns for the whole L1+L2 winding. With k=1, the whole thing is wound on a single core, with very tight coupling. Calling it an autotransformer is an apt description.

Given these facts, I can tell you what the ratio V2/V1 is just from knowing how variacs work.

But, if L1=4L2, then the ratio V2/V1 = L2/(L1+L2) = 1/5 doesn't seem right.

I goofed when I said the gain was L2/(L1 + L2). That's definitely wrong. That would have been right if coeff. of coupling k = 0. However, it didn't affect my calculations since I computed the gain in terms of L1, L2 and k.

Plugging in L1 = 4L2 gets me V2/V1 = 1/3, based on my solving the equations under the assumption that, as you say, L1 is the self-inductance of the top coil. I think that's right. You didn'd state what you believe the gain is - do you agree with 1/3?
 
  • #10
Thank you both for your input. I have found a similar example which may be useful at the bottom of page 77: http://nptel.iitm.ac.in/courses/IIT-MADRAS/Electrical_Machines_I/pdfs/1_11.pdf [Broken]

It seems to suggest that the current across L2 is I2-I1. In which case the 2 equations would be:

V1 = jωL1I1 + jωM(I2 - I1) + V2

V2 = jωL2(I2 - I1) + jωMI1
 
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  • #11
rude man said:
You didn'd state what you believe the gain is - do you agree with 1/3?

Absolutely. V1 is applied to 3N turns and V2 is taken from a winding with N turns, so V2 = V1/3.
 
  • #12
OK Alex. The ball is in your court now - finish the assignment?
 
  • #13
alex.daciz said:
Thank you both for your input. I have found a similar example which may be useful at the bottom of page 77: http://nptel.iitm.ac.in/courses/IIT-MADRAS/Electrical_Machines_I/pdfs/1_11.pdf [Broken]

It seems to suggest that the current across L2 is I2-I1.

HOW CAN YOU SAY THAT? LOOK AT YOUR SCHEMATIC, IT'S OBVIOUS THAT THE CURRENT FLOWING INTO L2 IS i1 - i2, not i2 - i1.

And since the B fields obviously enhance each other, since there is in reality only one coil, it's equally obvious that i1 and (i1-i2) both enter the dotted ends of L1 and L2. Not to mention the schematic redundantly dots the ends anyway.

If you think your eq's are right, work with them. But you should also work with the equations I gave you, and compare results. It should then be pretty obvious which was right.
 
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  • #14
Alex,

I think you would do well to set up the equations for a mesh solution to the network using KVL.

You almost seemed to be heading in that direction in your first post. What you want to end up with is two equations that look like this:

Z11*I1 + Z12*I2 = V1
Z21*I1 + Z22*I2 = V2

The Z11, Z12, Z21 and Z22 are coefficient expressions such as jω(L2+RL). When RL is connected, V2 = 0. If you will do this, all the parts of the problem fall out immediately.
 
  • #15
electrician - I don't think you want terms like that, do you?

The Z11, Z12, Z21 and Z22 are coefficient expressions such as jω(L2+RL). :uhh:

"When RL is connected, V2 = 0. " ? :uhh:
 
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  • #16
To sum up:

V1 - V2 = jwL1*i1 + jwM(i1 - i2).
V2 = jwL2(i1 - i2) + jwMi1

You have 2 equations and 2 unknowns: V1 and i1.

Remember k=1 so M = √(L1*L2)
 
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  • #17
rude man said:
electrician - I don't think you want terms like that, do you?

The Z11, Z12, Z21 and Z22 are coefficient expressions such as jω(L2+RL). :uhh:

Well, maybe not quite like that; just sort of like that. The right paren got a little too far to the right.

rude man said:
You have 3 equations and 3 unknowns: V2, i1 and i2.
For part (c), I2 isn't an unknown; it's a given. We don't need 3 equations.
 
  • #18
The Electrician said:
Well, maybe not quite like that; just sort of like that. The right paren got a little too far to the right.

Check.

But why V2 = 0 when RL is connected? I'm guessing you meant V2 = 1000.

You're right about needing just the two equations, of course. I hadn't noticed that V2 and i2 are given in (c).
 
  • #19
rude man said:
But why V2 = 0 when RL is connected? I'm guessing you meant V2 = 1000.

I had suggested to Alex that he form the mesh equations for the system:

"What you want to end up with is two equations that look like this:

Z11*I1 + Z12*I2 = V1
Z21*I1 + Z22*I2 = V2

The Z11, Z12, Z21 and Z22 are coefficient expressions such as jω(L2)+RL. When RL is connected, V2 = 0. If you will do this, all the parts of the problem fall out immediately."

I didn't mean that the value of the voltage across RL would be zero. Since the quantities on the right sides of each equation are the voltage sources in each mesh, the absence of any source in the mesh involving RL means that the right hand side of the second equation would be set to zero (where V2 was shown). It would have been less confusing if I had shown the equations as:

Z11*I1 + Z12*I2 = E1
Z21*I1 + Z22*I2 = E2

and said that E2 would be set to zero for part (c) of the problem. I was trying to steer him toward a formal "Z matrix" solution. The solution will give an expression for I2 and multiplying by RL gives the voltage labeled V2 in the problem.
 
  • #20
I'm a big fan of two-port methods myself but if he hasn't had that yet he's going to be more confused than ever.

He should run the Kirchhoff equations (nodal or current, I never remember which is which, I always sum currents = 0 at every dependent node, is that KVL or KCL? As in (v2-v1)/R1 = (v2-v3)/r2 + v3/R3?
 
  • #21
He seems to have departed, so it doesn't matter much anyway.
 
  • #22
Oh dear - our Dearly Departed? And just when we almost got him on the right track! :cry:
 

1. What is an auto transformer?

An auto transformer is an electrical transformer that has a single winding and is used to step up or step down the voltage in a circuit. It differs from a regular transformer in that it uses only one winding instead of two.

2. How does an auto transformer work?

An auto transformer works by using the principle of electromagnetic induction. The voltage is stepped up or down based on the ratio between the number of turns on the primary and secondary windings. The voltage change is achieved by tapping into different points on the single winding.

3. What are the advantages of using an auto transformer?

One advantage of using an auto transformer is that it is more efficient and compact compared to a regular transformer. It also allows for a wider range of voltage adjustments and can handle higher power loads. Additionally, it is less expensive to manufacture.

4. What are the potential risks associated with using an auto transformer?

The main risk associated with using an auto transformer is that it does not provide electrical isolation between the primary and secondary circuits. This can be a safety hazard, especially in high voltage applications. There is also a risk of voltage fluctuation or failure if the single winding becomes damaged.

5. How is an auto transformer different from a regular transformer?

An auto transformer differs from a regular transformer in that it has a single winding, while a regular transformer has two separate windings. It also uses tapping points to adjust the voltage, whereas a regular transformer uses separate windings for different voltage levels. Additionally, an auto transformer does not provide electrical isolation between the primary and secondary circuits, unlike a regular transformer.

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