Autotransformer question

[Ampera18 PE]

Hello, i cannot wrap my head around autotransformers. it's just not working. This question is giving me the most trouble.

I have a few specific questions.

1. Can anyone explain how the adding and subtracting works? As far as I can see, any of those are possible, but perhaps the least likely is B.

2. I am struggling with the power rating question below it. What is the trick?

 

[Dothracki PE]

Hello, i cannot wrap my head around autotransformers. it's just not working. This question is giving me the most trouble.

I have a few specific questions.

1. Can anyone explain how the adding and subtracting works? As far as I can see, any of those are possible, but perhaps the least likely is B.

2. I am struggling with the power rating question below it. What is the trick?

View attachment 19919View attachment 19920

Autotransformers aren't my greatest strength so if I am saying anything wrong or someone has a better way to explain, please step in. 

For an autotransformer turns ratio (N1/N2 or V1/V2) made from a single phase transformer, the turns ratio can only be made from a combination of the primary and secondary windings of the single phase transformer. Depending on desired configuration (step-up or step-down) and what side the autotransformation is applied to, the new ratio is often one of the base voltage values over or under a combination of the base voltage values. More commonly they are added, but voltages can be sutracted as well. So in this example, a step-up configuration can be made from the primary with new ratio of Npri/(Npri+Nsec) to make a 480V/600V autotransformer, just to mention one possibility. In this problem 240V cannot be obtained from any configuration of 120V and 480V.

As for the power rating, the currents as shown in the diagram depict how this is calculated. The trick with autotransformers is to draw it out and depict the current flows between the primary, seconary, and common winding. In this example of 600V to 120V, the current at the output is the summation of the primary current winding and the common current winding as determined by KCL. The current ratings of the single phase transformer are still in play when converted to an autotransformer. To get the power (magnitude), all you need to multiply the current going out by the output voltage. 

Typically an autotransformer configured from a single phase transformer will have a larger kva rating by like 15-30%. Although that only helps eliminate A in the second part.

 

[jd5191]

Remember that auto-transformers have a single winding in series. While some auto-transformers are specifically manufactured as auto-transformers, this problem is "transforming" (haha) a normal two-winding transformer to an autotransformer by connecting one high leg and one low leg together. Since its easier to draw these out, below are two ways to wire up the two-winding transformer. (I picked polarities arbitrarily to demonstrate the effect)



So you can have any combination of 120 and 480 on one side and 360 and 600 on the other. (I'm going to leave out the 120/480 option out of this convo ... although you could still use those ratios if you really wanted to)

For the apparent power, you can solve it the way described in the the solution, you'll have to be more specific about what you don't understand ... or use Section 4.3.1.7 of the NCEES handbook to find your formula:





Since Vc = 120V (because this is the winding being used) and Vse = 480V, the winding ratios are Nc=1 and Nse=4.

Set Sw = 45kVA because it is provided

Solve for Sio and you get Sio = 56.25kVA

 

[akyip]

IMO, The best way to deal with autotransformer questions is to visualize the windings directly in series, and seeing which common winding and series winding combination will lead to the desired primary and secondary voltages.

The autotransformer rating will always be: S auto = V in, auto * I in, auto = V out, auto * I out, auto

 

[Mohammed Ahmed]

Remember that auto-transformers have a single winding in series. While some auto-transformers are specifically manufactured as auto-transformers, this problem is "transforming" (haha) a normal two-winding transformer to an autotransformer by connecting one high leg and one low leg together. Since its easier to draw these out, below are two ways to wire up the two-winding transformer. (I picked polarities arbitrarily to demonstrate the effect)

View attachment 19928

So you can have any combination of 120 and 480 on one side and 360 and 600 on the other. (I'm going to leave out the 120/480 option out of this convo ... although you could still use those ratios if you really wanted to)

For the apparent power, you can solve it the way described in the the solution, you'll have to be more specific about what you don't understand ... or use Section 4.3.1.7 of the NCEES handbook to find your formula:

View attachment 19929

View attachment 19930

Since Vc = 120V (because this is the winding being used) and Vse = 480V, the winding ratios are Nc=1 and Nse=4.

Set Sw = 45kVA because it is provided

Solve for Sio and you get Sio = 56.25kVA

When you go to Wildi Book for electric machine under autotransformer section, he states that voltages are added when H1 and X2 or H2 and X1 are connected. The voltages are su tracted when H1 and X1 or H2 and X2 are connected. is it general rule we can apply anywhere? what is the difference between his statement and your sketch for autotransformer connection. 

 

[Ampera18 PE]

Remember that auto-transformers have a single winding in series. While some auto-transformers are specifically manufactured as auto-transformers, this problem is "transforming" (haha) a normal two-winding transformer to an autotransformer by connecting one high leg and one low leg together. Since its easier to draw these out, below are two ways to wire up the two-winding transformer. (I picked polarities arbitrarily to demonstrate the effect)

View attachment 19928

So you can have any combination of 120 and 480 on one side and 360 and 600 on the other. (I'm going to leave out the 120/480 option out of this convo ... although you could still use those ratios if you really wanted to)

For the apparent power, you can solve it the way described in the the solution, you'll have to be more specific about what you don't understand ... or use Section 4.3.1.7 of the NCEES handbook to find your formula:

View attachment 19929

View attachment 19930

Since Vc = 120V (because this is the winding being used) and Vse = 480V, the winding ratios are Nc=1 and Nse=4.

Set Sw = 45kVA because it is provided

Solve for Sio and you get Sio = 56.25kVA

 i think i get it, but it might just be a "practice until comfortable" type of thing. Maybe I will never instinctually know what the common winding is.

 

[Dothracki PE]

As I've said, I still am not 100% on autotransformers after studying for over a year. But it's unlikely you will get a lot of autotransformer problems based on 7-11 device problems including transformers, testing, capacitors, and reactors (per the syllabus). Practice is always good to reinforce problem areas, even if you struggle through it on your own, review the solution, and retry the problem afterwards. 

Anyway, the common winding is the winding that is shared by the autotransformer primary and secondary. 

 

[rburns18 PE]

The autotransformer problems that are in a lot of the practice exams are pretty straightforward when they take a 480V/120V transformer and make something like a 480V/600V autotransformer. The currents through the windings are the primary and secondary rated currents from the original transformer. And depending on if it's a step up or step down autotransformer, the common winding current will be adding with the primary current (step down) or subtracting from the primary current (step up) to get the secondary autotransformer current. Also in problems like these, the series and common winding will have the same power going through them since each winding has the original voltage across it and original rated current.

But Example 11.2 in Wildi is where I start to not understand them. 

 

[Ampera18 PE]

"the common winding current will be adding with the primary current (step down) or subtracting from the primary current (step up) to get the secondary autotransformer current."

I hope that's correct, because that's easy for me to think about.

"But Example 11.2 in Wildi is where I start to not understand them. "

I just can't get how its 480V between the 120 and 600  hahahahahhahah

The autotransformer problems that are in a lot of the practice exams are pretty straightforward when they take a 480V/120V transformer and make something like a 480V/600V autotransformer. The currents through the windings are the primary and secondary rated currents from the original transformer. And depending on if it's a step up or step down autotransformer, the common winding current will be adding with the primary current (step down) or subtracting from the primary current (step up) to get the secondary autotransformer current. Also in problems like these, the series and common winding will have the same power going through them since each winding has the original voltage across it and original rated current.

But Example 11.2 in Wildi is where I start to not understand them. 

 

[DilutedAr18_PE]

"the common winding current will be adding with the primary current (step down) or subtracting from the primary current (step up) to get the secondary autotransformer current."

I hope that's correct, because that's easy for me to think about.

"But Example 11.2 in Wildi is where I start to not understand them. "

I just can't get how its 480V between the 120 and 600  hahahahahhahah

Subtractive polarity 600-120=480. 

 

[Ampera18 PE]

Yes that's what it boils down to. I definitely didn't spend more than 10 minutes imagining actual potential differences -- nope!

 

[rburns18 PE]

@Ampera18 that is how it works for the example I have above, the Wildi examples don't seem as straight forward. To help remember, in a step up auto transformer (and regular step up transformer), the current on the secondary will be less since the voltage will be higher. So the common winding won't add with the primary current in an auto transformer. And the opposite for a step down transformer 

 

[akyip]

Hey all,

I'm revisiting this topic to try to help out on autotransformers. I mentioned earlier in this topic that it helps to visualize the 2 windings to see how you need to connect them.

I drew out illustrations for each of the possible choices from the given problem, starting with breaking down the 1-phase standard transformer.

I learned how to do this from Zach Stone's Electrical PE Review session on transformers and autotransformers.

Choice A is pretty easy for me because I've seen many such examples from other practice problem sets.

For Choices B and C, the only thing I might be off is the direction of the currents. I haven't dealt with autotransformers in which the windings' polarities oppose each other all that much. But I know for sure that I got the voltages right.

@Zach Stone, P.E.: if you see this, can you let me know if I drew Choices B and C right? I'm not sure about the currents for when the windings' polarities oppose each other.

 

[akyip]

One more illustration, to answer the apparent power (KVA) problem for the 600 V/120 V step-down autotransformer.

Also, I included the note in red about the relation between V1, V2, I1, and I2 at the bottom. This will help visualize how the currents will flow, especially the common current in the common-winding coil.

 

[akyip]

Hello, i cannot wrap my head around autotransformers. it's just not working. This question is giving me the most trouble.

I have a few specific questions.

1. Can anyone explain how the adding and subtracting works? As far as I can see, any of those are possible, but perhaps the least likely is B.

2. I am struggling with the power rating question below it. What is the trick?

View attachment 19919View attachment 19920

Ampera,

I hope that my illustrations for this problem help.

Also of course, I'll mention that the reason Choice D cannot be made from the given 1-phase standard transformer is because there is no way to add or subtract between 480 V and 120 V to get 240 V.

 

[Ampera18 PE]

I'm so sorry you spent so long drawing it out for silly ol' me. 

On the power, why is it just 600 * 93.75, but not 375-93.75? I thought, because it's a step up, the currents have to subtract. >_< this is over my head lmao 

Ampera,

I hope that my illustrations for this problem help.

Also of course, I'll mention that the reason Choice D cannot be made from the given 1-phase standard transformer is because there is no way to add or subtract between 480 V and 120 V to get 240 V.

 

[akyip]

I'm so sorry you spent so long drawing it out for silly ol' me. 

On the power, why is it just 600 * 93.75, but not 375-93.75? I thought, because it's a step up, the currents have to subtract. >_< this is over my head lmao 

I'll try to explain as best as I can.

With the 600 V/120 V autotransformer configuration, the primary winding of the original regular 1-ph transformer (which has the 93.75 A current) becomes the series winding on the high side of the 600 V/120 V autotransformer, and the primary current 93.75 A of the 1-phase transformer becomes the input current of this autotransformer.

In any ideal transformer (both autotransformers and regular 2-winding transformers): S = Vin * Iin = Vout * Iout

So here in this 600 V/120 V autotransformer, your Vin = 600 V and your Iin = 93.75 A. So Sauto = 56250 VA = 56.25 KVA.

 

[Ampera18 PE]

I think I get it now. you use the current for the 480V side, because that's what's changing? so current for 480 (I old) * 600 (Vnew)

If it was 360/120 it's a 33.750 KVA transformer, right?

 

[akyip]

I think I get it now. you use the current for the 480V side, because that's what's changing? so current for 480 (I old) * 600 (Vnew)

If it was 360/120 it's a 33.750 KVA transformer, right?

I used the current from the original transformer's 480 V side, because that current corresponded with the 480 V winding. And with how it's drawn and laid out in the autotransformer diagram, that current is the input current for this autotransformer.

If you want to calculate the S auto from the secondary side:

The total output current is: I2 auto = I1 auto + I common = 93.75 A + 375 A = 468.75 A

The I common (375 A) is the I2 of the original transformer, because the autotransformer's common winding corresponds to (is the) secondary winding of the original 1-phase transformer.

Sauto = V2 auto * I2 auto = 120 V * 468.75 A = 56250 VA = 56.25 KVA.

As I said before, you really have to visualize and draw out how the windings from the original 1-phase transformer are connected as an autotransformer.

I can't really answer well for the 360 V/120 V autotransformer, because I'm not 100% sure about how the currents flow on autotransformers whose windings have opposing polarities. But I can answer for autotransformers whose windings have the same polarity.

(Most of the autotransformer practice problems that I have done have used same-polarity windings. I haven't seen a practice problem where the series and common windings have opposing polarities.)

 

[Philipsalt]

Hello, i cannot wrap my head around autotransformers. it's just not working. This question is giving me the most trouble.

I have a few specific questions.

1. Can anyone explain how the adding and subtracting works? As far as I can see, any of those are possible, but perhaps the least likely is B.

2. I am struggling with the power rating question below it. What is the trick?

View attachment 19919View attachment 19920

There are two ways to construct the equivalent of an autotransformer. I will ignore for the m\oment the construction of the equivalent circuit using one isolation transformer, although that is the way most buck/boost configurations are constructed.

An autotransformer has one magnetic core and a single winding around that core which has at least three connection points: the two ends and a point in between them.
As with any transformer with multiple windings or winding sections, the voltage between any two points is proportional to the number of turns between those points. .
So lets take a hypothetical tranformer with 120 turns and a tap at the 100 turn point and apply 120 volts to the two ends, A and B. That sets the configuration as 1 volt per turn. And if the tap point, C is 100 turns from A the voltage A-C will be 100V. And the voltage C-B will be 20V. This is an autotransformer configuration that can be used to produce a lower output voltage.
Next, apply the 120V input to points A and B. That sets the voltage at 1.2V per turn. So the voltage from A to C will now be 144V. This is an autotransformer configuration that produces a higher output voltage.
Now note that there is no reason for the the wire guage to be the same from one end of the winding to the other. In our example, we will make the BC section of the winding 5 times the diameter of the AB section and revisit the second (boost) configuration. Ignoring magnetizing current, the current in AB will have to be 1/5 of the current in BC to satisfy the condition of power in = power out.
If the goal of the example is to provide 144V to a load which draws 1A, the transformer will only need to be rated for 24 Watts (120V times .2A on the input side, AB = 1A times 24V on the output side, BC.)

Instead of using a single winding wrapped continuously on a core, you can get exactly the same effect using an isolation transformer with a 5-1 turn ratio and connecting the two windings together in either a buck or a boost configuration.