Question

A 20KVA, 2000/200 V single-phase transformer has name-plate leakage impedance of 8%. The voltage required to be applied on the HV side to circulate full load current with the LV winding short-circuited will be:

• A. 16 V
• B. 56.56 V
• C. 160 V
• D. 68.68 V

Hint:

To calculate the short-circuit voltage, use the formula \( V_{SC} = V_{rated} \times \sqrt{(Z_{%})} \), where \( Z_{%} \) is the percentage of the impedance.

Answer 1

The Correct Option is B

The voltage required to circulate full load current when the LV side is short-circuited can be calculated using the formula:
\[ V_{SC} = V_{rated} \times \sqrt{(Z_{%})} \] Where:
- \( V_{SC} \) is the short-circuit voltage,
- \( V_{rated} = 2000 \, \text{V} \) (rated voltage on the HV side),
- \( Z_{%} = 8% = 0.08 \).
Now, substituting the values:
\[ V_{SC} = 2000 \times \sqrt{0.08} = 2000 \times 0.2828 = 56.56 \, \text{V} \] Thus, the required voltage to be applied on the HV side is 56.56 V.