Autotransformer - Actual Load vs Equivalent Load: what is the difference? (Shorebrook PE Exam Question 65)

[akyip]

Hi guys,

I have a few autotransformer questions from Shorebrook PE Exam question 65.

1. What is the difference between actual load and equivalent load? From what I see, it seems like actual load is the load that is actually connected to the transformer's secondary... but then what is the equivalent load?

Does the equivalent load have to do with the autotransformer ratings? Or is the equivalent load the load of a 1-phase standard transformer that is somehow equivalent to the autotransformer?

2. Where does this equation (Actual Load VA) / (Equivalent Load VA) = (V high - V low) / (V high) come from anyway? I have not seen this in any other autotransformer materials or PE exam questions so far other than the Shorebrook PE exam...

Thanks for any input on this!

 

[Orchid PE]

Here is the diagram I'll be using:



One big thing Shorebrook did not mention in the problem is that the autotransformer was constructed from a standard 2-winding transformer. That bit of information would've been nice to have.

The typical power advantage of an autotransformer constructed from a 2-winding transformer is:



None of these contain the voltage values that Shorebrook uses, but we can figure that out. He uses V_high and V_low (which is V_H and V_L in the diagram).

So we have:



Substituting we get:



With V_H=2200, V_L=2000, and S2-w=20kVA, the equivalent rating of the autotransformer is 20*2200/(2200-2000) = 220.

We can do the same thing if we use the V1 and V2 values:

Sauto = S2-w * (V1 + V2) / V1 = 20 * (200 + 2000) / 200 = 220.

The 200V value of V1 comes from subtracting V_H - V_L = 2200 - 2000 = 200. This is the voltage across the "secondary" coil of the 2-w transformer or the low voltage side.

 

[akyip]

Here is the diagram I'll be using:

View attachment 18860

One big thing Shorebrook did not mention in the problem is that the autotransformer was constructed from a standard 2-winding transformer. That bit of information would've been nice to have.

The typical power advantage of an autotransformer constructed from a 2-winding transformer is:

View attachment 18861

None of these contain the voltage values that Shorebrook uses, but we can figure that out. He uses V_high and V_low (which is V_H and V_L in the diagram).

So we have:

View attachment 18862

Substituting we get:

View attachment 18863

With V_H=2200, V_L=2000, and S2-w=20kVA, the equivalent rating of the autotransformer is 20*2200/(2200-2000) = 220.

We can do the same thing if we use the V1 and V2 values:

Sauto = S2-w * (V1 + V2) / V1 = 20 * (200 + 2000) / 200 = 220.

The 200V value of V1 comes from subtracting V_H - V_L = 2200 - 2000 = 200. This is the voltage across the "secondary" coil of the 2-w transformer or the low voltage side.

Chattaneer, thanks for the explanation!

I'm still a bit thrown off by how Shorebrook uses the terms actual load and equivalent load without really clarifying. But the best way I thought about this problem after your explanation is simply:

S auto / S 2-winding = V H,auto / (V H,auto - V L,auto)