Autotransformer phasor diagram

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I'm studying Auto Transformer phasor diagram, and am having some doubts. I'd be grateful if someone could help me out with them.

Book has equation:
upload_2017-10-31_10-36-34.png

I understand that using Kirchoff's law we get minus sign for (I2 - I1) term, but shouldn't it be positive as per concept?
Negative sign means the mmf of (I2 - I1) aids the source, which can't be.
It is V1 that is eventually providing I2 So I2 has to oppose V1 and not aid it.

I understand that using KVL sign conventions we get the above equation with the signs above, but conceptually shouldn't (I2-I1) oppose V1

Doubt#2: And in the phasor diagram here:
upload_2017-10-31_10-38-46.png

They've added the terms V1 = E1 + I1R1 + I1X1 + (I2-I1)R1 +(I2-I1)X1

Shouldn't the direction of I2-I1 be upwards when added with E1?
SInce we're adding -(I2-I1)(R1)
So if (I2-I1)R1 is facing south then the negative of it must face north.
 

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  • #2
cnh1995
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Negative sign means the mmf of (I2 - I1) aids the source, which can't be.
It is V1 that is eventually providing I2 So I2 has to oppose V1 and not aid it.
Nice catch!

Actually, the mmf balance equation for an autotransformer is I1(N1-N2)=N2(I2-I1).
Edit: I was considering N1 to be what the textbooks call N1-N2. So this totally makes sense.
Just a matter nomenclature!

@jim hardy, could you shed some light?
 
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  • #3
jim hardy
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Rigor is a necessary habit for us mortals.

We have to assign polarities.
Here's how i would assign currents and voltage polarities
autoxfmr4cnh.jpg

and i don't see E1
so i'd write KVL, walking I1's loop
V1 = I1(r1+jx1) + (I1 - I2) (r2 + jx2)

which expands to
V1 = I1r1 + jI1x1 + I1r2 + jI1x2 - I2r2 - jI2x2.
which i think is the same as yours because your -(I2 - I1) equals my +(I1 - I2)

Now I2 is doubtless more than I1, see
http://www.nptel.ac.in/courses/Webcourse-contents/IIT Kharagpur/Basic Electrical Technology/pdf/L-27(TB)(ET) ((EE)NPTEL).pdf
upload_2017-10-31_23-7-5.png


so I1r2 points up and I2r2 down ? And I2 is bigger so their sum points down ?
Just too many minus signs to keep track of in your mind. . I always break these things down into individual terms so my pea-brain can focus on them one at a time......
@cnh1995 Do you see any mistakes in my thinking there ?

The mystery to me is - why does his phasor diagram say (I2-I1)r1 instead of r2 ?

autoxfmr4cnh3.jpg



old jim
 

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The mystery to me is - why does his phasor diagram say (I2-I1)r1 instead of r2 ?

View attachment 214153


old jim
I guess that's a typo. But shouldn't it be pointing upwards.
Since (I2 - I1) is pointing down, like south west, so shouldn't the negative of it point north east, like upwards?
 
  • #5
cnh1995
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The mystery to me is - why does his phasor diagram say (I2-I1)r1 instead of r2 ?
Yes, as jaus said, I also think it's a typo.

Consider the following the two cases (magnetizing current not shown):
1) Primary and secondary conncected in series aiding fashion (same winding sense for both).
Screenshot_20171102-175247.png

Using right hand rule, we can see primary and secondary fluxes cancel each other and core flux remains constant.

Case 2: Primary and secondary are connected in series opposition (winding senses are opposite).
Screenshot_20171102-175147.png

Using right hand rule, we can see both primary and secondary fluxes add up.

What happens to the net core flux in the second case? Did I miss something in the drawing?
 

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  • #6
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I think in the second case the direction of I2 is wrong.
I2 is caused because of I1. It will always flow in a direction so the flux produced by it oppose the flux that caused it.
If you change winding from clockwise to anticlockwise then current direction will change.
So I2 is -6A as in the direction is downward. The load current will follow same direction of I2-I1
I2 is caused by I1.
6I1 = 4I2
And I2 - I1 = 10A
Solving this i got: I1 = 20A, and I2 = 30A. Somewhere I'm wrong but can't figure out where.
The Emf induced will be same as emf is subtraction of both fluxes which is always same.
 
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cnh1995
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Solving this i got: I1 = 20A, and I2 = 30A.
That contradicts the the power balance equation. In absence of losses, V1I1 should always be equal to V2I2. Plus, the currents should satisfy KCL at the junction of load, primary and secondary.
 
  • #8
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The currents satisfy KCL. i1 is flowing downward and is 20A.
I2 is flowing downward and is 30A.
Load current is flowing upward and is 10A.

*had mentioned wrong direction of load current in above post.
 
  • #9
cnh1995
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The currents satisfy KCL. i1 is flowing downward and is 20A.
I2 is flowing downward and is 30A.
Load current is flowing upward and is 10A.
If I1=20A, V1I1=100×20= 2000W.
The load consumes only 400W.

This is a step-down transformer. Source current must be smaller than the load current.
 
  • #10
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Yeah you're right. But then in both pics the orientation of winding is different. So even the current direction must be changed.
I2 produced flux which produced I2
So I1 produced flux (twisting changed) so I2 must also flow opposite, i guess.
 
  • #11
cnh1995
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Yeah you're right. But then in both pics the orientation of winding is different. So even the current direction must be changed.
I2 produced flux which produced I2
So I1 produced flux (twisting changed) so I2 must also flow opposite, i guess.
The values of primary current, load current and secondary current are not decided by the sense of winding. The sense of the windings affects only the magnetizing current. In the second case, magnetizing current will be more than that in the first picture.

Load voltage=40V, load resistance=4 ohm (both irrespective of winding sense).
So, load current=10A.

Using power balance equation (independent of the winding sense), primary current is 4A.
Which means secondary current is 6A. But then, in the second picture, primary and secondary fluxes add up instead of cancelling each other.

In a conventional two-winding transformer, there's perfect mmf balance no matter what the sense of windings is, since primary and secondary are electrically isolated.

In the second picture, I'm afraid I am missing something elementary.
 
  • #12
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I understand that in 2 winding transformer the winding sense doesnt change much as there is no common winding so the problem of kcl doesnt arrive.
the only equation needed to be satisfied is Ampere turns must be same and no KCL.

I'm not sure but I think the sense of winding changes the direction of current. that's what right hand flux rule says. curled fingers of right hand gives direction of current as outstretched thumb.
now if winding sense is reversed then the outstretched thumb points in opposite direction.

and since auto Xmer has to satisfy kcl, then the values of current will also change if the relative polarity changes.
 
  • #13
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The sense of the windings affects only the magnetizing current. In the second case, magnetizing current will be more than that in the first picture.
As the magnetizing current is more, the source voltage will also have to be more.
 
  • #14
cnh1995
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As the magnetizing current is more, the source voltage will also have to be more.
No, the effective magnetzing inductance Lm seen by the source will reduce in the second case (as the windings are in series opposition). But since the applied voltage is fixed ( 100V) , core flux is also fixed in both the circuits. So, to create the same amont of flux, the transformer draws more magnetizing current.
 
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  • #15
jim hardy
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Since (I2 - I1) is pointing down, like south west, so shouldn't the negative of it point north east, like upwards?
The whole is the sum of its parts.
Don't draw (I2 - I1)r2, draw I2r2 and I1r2 separately then add them. The positive one points up, the negative one points down, and their sum is whatever it is.
 
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  • #16
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I got confused in vector subtraction. If vectors point in same direction then their subtraction is difference. Like if both are horizontal and pointing right then their difference is difference of their absolute values.

Here I1 and I2 point in opposite directions, so shouldn't their vector subtraction be a bigger value than both I1 and I2?
 
  • #17
cnh1995
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@jim hardy, what are your thoughts on the second case in #5?
Where do you think is the mistake?
 
  • #18
jim hardy
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@jim hardy, what are your thoughts on the second case in #5?
Where do you think is the mistake?
Sometimes the way to test the mental model of something complex is to subject it to a simpler test.

Take your second case and disconnect the load.

autoxfmr4cnh4.jpg


you have reversed the dots and connected the windings in series across the source.

Looks to me like flux must rise to support 100 volts across 6-4 = 2 turns instead of ten turns. I dont think you could energize it long enough to connect a load, saturation would trip it on overcurrent...
It'd be interesting to try that connection on a core that's oversized enough to support 5X normal flux though. If you have a real transformer with multiple windings and a source of low AC you could try it at low enough voltage to not saturate it.

Remember - in normal operation magnetizing current mmf's add, load current mmf's oppose .


autoxfmr4cnh5.jpg


If you reverse the dots what happens ?
Well, if flux were able to somehow make it up to the 50 volts per turn per your 6-turn 4-turn model requires
With no load you'd have a LOT of voltage across each winding and their difference would be 100 volts..
I'll have to think a couple days on what happens when you connect a load.
 

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  • #19
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Why is the orange arrow facing up and then facing down?
magnetizing mmfs add like add to each other?
load mmfs oppose... oppose what? there is only 1 load current, so only 1 load mmf?


and why will the flux rise in the diagram you have above? There is 100V across an inductance above. The current is 100/reactive impedance. So high current.
 
  • #20
jim hardy
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Why is the orange arrow facing up and then facing down?
Because load current flows up through the bottom winding.
Apply right hand rule.
 
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  • #21
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Got it. Thanks.
But shouldn't secondary mmf also oppose primary mmf?
Primary current causes flux which causes secondary current(which is the ampere turns).
 
  • #22
cnh1995
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If you reverse the dots what happens ?
Well, if flux were able to somehow make it up to the 50 volts per turn per your 6-turn 4-turn model requires
With no load you'd have a LOT of voltage across each winding and their difference would be 100 volts..
Awesome! :smile:This is the part I was missing in my reasoning.
The transformer will act as a 2-turn transformer and there will be 50V/turn. Going from top to bottom, there will be 300V drop across the upper winding and 200V rise across the lower winding.

I'll have to think a couple days on what happens when you connect a load.
This is what it will look like. And here, load amp-turns cancel, as they should.
Screenshot_20171102-224152.png

The transformer now acts as a step-up transformer.

Thank a lot for your help! :smile:
 

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  • #23
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Looks like i wasn't completely wrong. The current direction in secondary winding reverses with change of winding sense.
 
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  • #24
jim hardy
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This is what it will look like.
By golly i think you're right ! That's what i hadn't figured out yet - direction of load current reverses... ...
You beat the old guy to it ! And it makes old guys happy to see young guys' nimble minds race past us.
Think in simple steps.
Express your thoughts one step per line..
Rejoice in the power of the basics.

Don't set that example aside just because it's impractical to run a transformer at 5X intended flux.
Current transformer problems require similar out of the ordinary thinking.

NICE JOB @cnh1995 - :smile::smile::smile:.!!!!!

old jim
 
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