Answer Posted / tuhin

an autotransformer will be smaller than an equivalent
conventional transformer. The reason is an autotransformer
essentially has one winding on a core, with one tap
including the whole winding, and another tap including only
part of the winding. In a conventional transformer, power
is transferred from one winding to the other by induction.
In an auto, some is by conduction, and some is by induction.

Here's an example comparison:

Say we have a 10MVA 100kV/50kV 3 winding auto transformer.
There are two parts of the winding - the series winding
(the section of winding between the 100kV connection and
50kV connection), and the common winding (the section of
winding between the 50kV connection to the neutral). If
powered at 10MVA (full load), there will be:

10M / (100k * sqrt(3)) = 57.7 Amps flowing into the
primary, and
10M / (50k * sqrt(3)) = 115.5 amps flowing out of the
secondary

So, 57.7A flow into the primary, down through the series
winding and out the secondary. At the same time 57.7A flow
up from the neutral through the common winding and out the
secondary, resulting in a total secondary current of (57.7
+ 57.7) = 115.5 (sorry, rounding is making my math look
wrong!).

Thus, the series winding is handling a total of:
(100k - 50k)* 57.7A*sqrt(3) = 5MVA, and the common winding
is handling a total of:
(50k - 0) *57.7A*sqrt(3) = 5MVA

So the series and common windings only have to be made to
handle 1/2 the total load(for this example - it is 1/2
because the primary / secondary voltage is 2/1). In a
conventional transformer, each winding carries the full
load (10MVA), because it operates on induction (no
conduction).

If the votlage ratio between primary and secondary is too
high, there is no real net savings in size (and thus cost),
so a conventional transformer is often used. The cutoff
point is around 4.

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