\text{A }15\text{ kVA, }1100\text{ V/}220\text{ V, single-phase two-winding transformer is configured as a }1.32\text{ kV/}1.1\text{ kV autotransformer.}

$$\begin{gathered} \text{What will be the rating of the autotransformer?} \\ \end{gathered}$$

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The correct option is 90kVA.

Let us solve the problem step-by-step to determine the kVA rating of the transformer when reconfigured as an autotransformer.

Step 1: Identify the given ratings of the two-winding transformer
Rating of the single-phase two-winding transformer, ${\text{S}}_{2\text{W}}=15\text{kVA}$
Primary voltage rating, V1=1100 V
Secondary voltage rating, V2=220 V

Step 2: Calculate the rated currents of the windings
The rated current for the primary winding (1100 V) is:

${\text{I}}_1=\frac{{\text{S}}_{2\text{W}}}{{\text{V}}_1}=\frac{15000\text{VA}}{1100\text{V}}=13.636\text{A}$

The rated current for the secondary winding (220 V) is:

${\text{I}}_2=\frac{{\text{S}}_{2\text{W}}}{{\text{V}}_2}=\frac{15000\text{VA}}{220\text{V}}=68.182\text{A}$

Step 3: Analyze the autotransformer configuration
The autotransformer is configured to operate at $1.32\text{kV /}1.1\text{kV}$, which corresponds to voltages:
High-voltage side, ${\text{V}}_{\text{H}}=1.32\text{kV}=1320\text{V}$
Low-voltage side, ${\text{V}}_{\text{L}}=1.1\text{kV}=1100\text{V}$

Notice that $1320\text{V}=1100\text{V}+220\text{V}$. This shows that the two windings are connected in series in additive polarity, where the $1100\text{V}$ winding serves as the common winding, and the $220\text{V}$ winding is connected in series with it to carry the input/output current at the high-voltage side.

Step 4: Determine the maximum autotransformer rating
The maximum current that can flow through the series winding ($220\text{V}$ winding) without exceeding its thermal rating is ${\text{I}}_2=68.182\text{A}$.

Since the high-voltage terminal current is equal to the current passing through this series winding:

${\text{I}}_{\text{H}}={\text{I}}_2=68.182\text{A}$

Therefore, the power rating of the autotransformer is given by:

${\text{S}}_{\text{auto}}={\text{V}}_{\text{H}}\times {\text{I}}_{\text{H}}$

${\text{S}}_{\text{auto}}=1320\text{V}\times 68.182\text{A}=90000\text{VA}=90\text{kVA}$

Alternative Method using the transformation ratio formula:
For an autotransformer, the relationship between autotransformer rating and two-winding transformer rating is:

${\text{S}}_{\text{auto}}=\frac{1}{1-k}\times {\text{S}}_{2\text{W}}$

where $k=\frac{{\text{V}}_{\text{L}}}{{\text{V}}_{\text{H}}}=\frac{1.1}{1.32}=\frac{1100}{1320}=\frac{5}{6}$.

Substituting $k$ into the formula:

${\text{S}}_{\text{auto}}=\frac{1}{1-\frac{5}{6}}\times 15\text{kVA}=6\times 15\text{kVA}=90\text{kVA}$

Thus, the rating of the autotransformer is 90kVA.