Question:
A 15 kVA, 1100 V/220 V, single-phase two-winding transformer is configured as a 1.32 kV/1.1 kV autotransformer.What will be the rating of the autotransformer?
A 15 kVA, 1100 V/220 V, single-phase two-winding transformer is configured as a 1.32 kV/1.1 kV autotransformer.
What will be the rating of the autotransformer?
Show Hint
The 1100 V winding becomes the common winding and the 220 V winding becomes the series winding; multiply the two-winding rating by V1/(V1-V2), not V2/(V1-V2).
Updated On: Jul 20, 2026
- 60 kVA
- 75 kVA
- 90 kVA
- 100 kVA
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The Correct Option is C
Solution and Explanation
Step 1: Identify how the two-winding transformer is reconnected.
The original two-winding transformer is rated 1100 V/220 V. As an autotransformer, its two windings are connected in series so that the combined winding spans \(1100+220=1320\ \text{V}=1.32\ \text{kV}\), and the tap between the two windings gives the \(1.1\ \text{kV}\) output. So the \(1100\ \text{V}\) winding acts as the common winding, and the \(220\ \text{V}\) winding acts as the series winding.
Step 2: Recall the standard autotransformer rating formula.
For an autotransformer built from a two-winding transformer of rating \(S_{2w}\), with high-side voltage \(V_1\) and low-side voltage \(V_2\), the autotransformer's throughput rating is
\[ S_{auto}=S_{2w}\times\frac{V_1}{V_1-V_2} \]
This formula reflects that the series winding, unchanged in current rating from its two-winding operation, now carries the full autotransformer's high side line current, and this current flows through a much larger voltage \(V_1\) at the input terminal, not just its own \(220\ \text{V}\) rated voltage.
Step 3: Substitute the given values.
\[ V_1=1.32\ \text{kV},\qquad V_2=1.1\ \text{kV},\qquad V_1-V_2=0.22\ \text{kV} \]
\[ S_{auto}=15\times\frac{1.32}{0.22}=15\times6=90\ \text{kVA} \]
Step 4: Sanity check using the winding current.
The two-winding transformer's \(220\ \text{V}\) winding is rated for a current of \(15000/220=68.18\ \text{A}\). As the series winding of the autotransformer, this same current of \(68.18\ \text{A}\) now flows in from the \(1.32\ \text{kV}\) line terminal, giving an autotransformer throughput of \(1320\times68.18=90{,}000\ \text{VA}=90\ \text{kVA}\), confirming Step 3.
Step 5: Rule out the other options.
Option (A), \(60\ \text{kVA}\), would come from using the ratio \(V_2/(V_1-V_2)=1.1/0.22=5\) instead of \(V_1/(V_1-V_2)\), understating the true throughput. Option (B), \(75\ \text{kVA}\), and option (D), \(100\ \text{kVA}\), do not follow from any correct combination of the given voltages and the \(15\ \text{kVA}\) rating.
Step 6: Final Answer.
\[ \boxed{90\ \text{kVA}} \]
The original two-winding transformer is rated 1100 V/220 V. As an autotransformer, its two windings are connected in series so that the combined winding spans \(1100+220=1320\ \text{V}=1.32\ \text{kV}\), and the tap between the two windings gives the \(1.1\ \text{kV}\) output. So the \(1100\ \text{V}\) winding acts as the common winding, and the \(220\ \text{V}\) winding acts as the series winding.
Step 2: Recall the standard autotransformer rating formula.
For an autotransformer built from a two-winding transformer of rating \(S_{2w}\), with high-side voltage \(V_1\) and low-side voltage \(V_2\), the autotransformer's throughput rating is
\[ S_{auto}=S_{2w}\times\frac{V_1}{V_1-V_2} \]
This formula reflects that the series winding, unchanged in current rating from its two-winding operation, now carries the full autotransformer's high side line current, and this current flows through a much larger voltage \(V_1\) at the input terminal, not just its own \(220\ \text{V}\) rated voltage.
Step 3: Substitute the given values.
\[ V_1=1.32\ \text{kV},\qquad V_2=1.1\ \text{kV},\qquad V_1-V_2=0.22\ \text{kV} \]
\[ S_{auto}=15\times\frac{1.32}{0.22}=15\times6=90\ \text{kVA} \]
Step 4: Sanity check using the winding current.
The two-winding transformer's \(220\ \text{V}\) winding is rated for a current of \(15000/220=68.18\ \text{A}\). As the series winding of the autotransformer, this same current of \(68.18\ \text{A}\) now flows in from the \(1.32\ \text{kV}\) line terminal, giving an autotransformer throughput of \(1320\times68.18=90{,}000\ \text{VA}=90\ \text{kVA}\), confirming Step 3.
Step 5: Rule out the other options.
Option (A), \(60\ \text{kVA}\), would come from using the ratio \(V_2/(V_1-V_2)=1.1/0.22=5\) instead of \(V_1/(V_1-V_2)\), understating the true throughput. Option (B), \(75\ \text{kVA}\), and option (D), \(100\ \text{kVA}\), do not follow from any correct combination of the given voltages and the \(15\ \text{kVA}\) rating.
Step 6: Final Answer.
\[ \boxed{90\ \text{kVA}} \]
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