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?
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- 60 kVA
- 75 kVA
- 90 kVA
- 100 kVA
The Correct Option is C
Solution and Explanation
A different way to see this is to track the current in the series winding directly and multiply it by the full input voltage, instead of quoting the ratio formula first.
The 220 V winding of the original two-winding transformer is designed to carry a rated current of:
\[ I_{series}=\frac{15{,}000\text{ VA}}{220\text{ V}}=68.18\text{ A} \]When this same winding becomes the series winding of the autotransformer, it still carries this same rated current, since its own conductor and insulation have not changed. But now this current is drawn from the full 1.32 kV input line, not from a separate 220 V source.
So the autotransformer's throughput rating, seen from its high-voltage line terminal, is:
\[ S_{auto}=V_1\times I_{series}=1320\times68.18=90{,}000\text{ VA}=90\text{ kVA} \]This is exactly six times the original 15 kVA rating, matching the ratio \(V_1/(V_1-V_2)=1320/220=6\) used in the standard formula.
\[ \boxed{90\text{ kVA}} \]