Question

A single-phase two-winding transformer is rated at 15 kVA, 1100/220 V. It is reconnected as an autotransformer with a voltage rating of 1320/1100 V. Find the kVA rating of the autotransformer.

Hint:

When a two-winding transformer is reconfigured as an autotransformer, its kVA rating increases by a factor of \((k+1)\), where \(k\) is the turns ratio of the original high-voltage winding to the low-voltage winding. This is a quick way to find the new rating.

Answer 1

Step 1: Analyze the Original Transformer
High Voltage (\(V_{HV}\)) = 1100 V, Low Voltage (\(V_{LV}\)) = 220 V. Rated current for the 220V winding (\(I_{LV}\)) = \(15000 / 220 \approx 68.18\) A. Rated current for the 1100V winding (\(I_{HV}\)) = \(15000 / 1100 \approx 13.64\) A.
Step 2: Reconnection as Autotransformer
The autotransformer rating is 1320/1100 V. The high voltage (1320V) is the sum of the two original voltages: \(1100 + 220 = 1320\) V. This indicates that the two windings are connected in series-additive polarity.
Step 3: Calculate the New kVA Rating
In an autotransformer, the power rating is determined by the current capacity of the common winding or the series winding. The kVA rating of an autotransformer (\(S_{auto}\)) is given by: \[ S_{auto} = \left( \frac{V_H}{V_H - V_L} \right) \times S_{2-winding} \] Where \(V_H = 1320\) V and \(V_L = 1100\) V. \[ S_{auto} = \left( \frac{1320}{1320 - 1100} \right) \times 15 = \frac{1320}{220} \times 15 = 6 \times 15 = 90 \text{ kVA.} \]