# Autotransformer Problem



## applepieordie (Jun 26, 2017)

Hi,

Can someone answer these questions:

1) Am I correct in thinking that Vbc = 480V and Vab = 120V?

2) Why is only I2 multiplied by 600V?

Thanks

View attachment 9796


View attachment 9797


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## rg1 (Jun 26, 2017)

applepieordie said:


> Hi,
> 
> Can someone answer these questions:
> 
> ...


a) Your voltages are correct.

b) The answer is right but he has messed up with I1, I2, I3. The output I of auto transformer is output current (which is the rated current of 120V winding in this case= 125A) multiplied by output Voltage which is 600V in this case. So the answer is 600*120=75 KVA. 

Similarly I/P of Xmer is I/P current (which is 31.25 A as reflection or say balancing current) of 125A and 125A going out= 125+31..25A= 156.25A) multiplied by the I/P Voltage 480V which makes again 75KVA and proves we are right.


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## applepieordie (Jun 27, 2017)

rg1 said:


> a) Your voltages are correct.
> 
> b) The answer is right but he has messed up with I1, I2, I3. The output I of auto transformer is output current (which is the rated current of 120V winding in this case= 125A) multiplied by output Voltage which is 600V in this case. So the answer is 600*120=75 KVA.
> 
> Similarly I/P of Xmer is I/P current (which is 31.25 A as reflection or say balancing current) of 125A and 125A going out= 125+31..25A= 156.25A) multiplied by the I/P Voltage 480V which makes again 75KVA and proves we are right.


Thank you! This was very helpful and explained the solution very well.

I wasted so much time trying to understand the author's solution but just got more confused.


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## rg1 (Jun 27, 2017)

applepieordie said:


> Thank you! This was very helpful and explained the solution very well.
> 
> I wasted so much time trying to understand the author's solution but just got more confused.


You are most welcome and it is my pleasure.


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## TNPE (Jun 28, 2017)

Or you can look at it this way, which is much easier and saves a butt load of time, thought and work...

(VH + VL)/VL= 5, where this is V high (480V) and V low (120V)

5 x (15 KVA) = 75 KVA

The math in the diagram is correct, but the labeled currents are wrong.  Swopp I2 and I3.  I2 being 125 amps through the 120V winding, I3 being 31.25 amps through the 480V winding and I1 being the sum of these two currents.


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## applepieordie (Jun 28, 2017)

Hi guys,

I would like a better understand of the numbers in this problem.

does this mean that the transformer in this example is overloaded (75KVA on 15KVA rate transformer)?


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## BigWheel (Jun 28, 2017)

applepieordie said:


> Hi guys,
> 
> I would like a better understand of the numbers in this problem.
> 
> does this mean that the transformer in this example is overloaded (75KVA on 15KVA rate transformer)?


No, not overloaded...the single phase transformer used as an Autotransformer allows the total capacity to be fully realized. You started out with a single-phase transformer rated for 15kVA, but through a clever use of the law conservation of energy, you unlocked the true and total capacity of a 75kVA power source.

A single-phase XFMR has a primary voltage and secondary voltage winding ratio and results in apparent power equality that defines the primary and secondary currents (i.e., 15 kVA in = 15 kVA out).

On the other hand, an Autotransformer kind of works in reverse (i.e., primary maximum apparent power in = secondary maximum apparent power out) primary current available is related to the primary voltage, while secondary current available is related to the secondary voltage.

Once the voltages and currents are known, one can calculate the "true" apparent power on the primary and secondary sides of the Autotransformer, which will be equal (per the conservation of energy law).


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## BigWheel (Jun 29, 2017)

Probably a better way to say this is that a standard transformer couples voltage from the primary side to the secondary side by electromagnetic induction. This electromagnetically coupled voltage induces a current on the secondary. Since power in must equal power out, the amount of current induced is inversely proportional to the voltage on the secondary (i.e., big voltage/small current on primary = small voltage/big current on secondary). All of this happens based on transformer impedances.

An autotransformer doesn't rely on emag coupling; the primary and secondary windings are physically wired together, so the current produced in your secondary follows KCL while the voltage follows KVL, so the power "adds up."


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## TNPE (Jun 29, 2017)

The entire coil is NOT used in transferring power from primary to secondary; hence, the KVA of an autotransformer, depending on design, can be sunstantially higher than a two-winding, single phase XFMR.  

Just as with the example you provided, 125A is delivered to the load through the 120V winding (i.e. the entire coil is not being used).  It is a genius design that allows for small changes in voltage to be accomplished economically, in comparison to a two-winding  XFMR.  Also, autotransformers allow for flexibility across a range of voltages with varying loads, like the ones used in electric distribution.

It is hard to find good material related to autotransformers, but Chapman, if you have his book or access to it, has some good material and explanations for evaluating/understanding their operation.


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## rg1 (Jun 29, 2017)

I will put it in this way. Overloading of a device means bearing of a larger than rated current by device or a part of device. We see in the example above that the rated currents or voltages of none of the windings (125 and 31.5A) are violated. So they are not overloaded. The reason for the magic of high power transfer is the power Transfer in a conventional transformer is through mutual induction (only) of two coils and in auto transformer it is partly by mutual induction only(15KVA in our case) and partly by potentiometer action (60KVA in our case).


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## applepieordie (Jul 5, 2017)

Thank you all for the detailed explanations.


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