QUESTION 3: THREE-PHASE TRANSFORMERS 3.1 State TWO functions of the oil used in oil-filled transformers - NSC Electrical Technology Power Systems - Question 3 - 2016 - Paper 1

The primary line current can be calculated using the formula:

$I_{PL} = \frac{S}{\sqrt{3} V_{L_{P}}}$

Substituting the given values:

$I_{PL} = \frac{20000}{\sqrt{3} \times 6600} = 1.75 A$

The secondary phase voltage can be found using the formula:

$V_{PH} = \frac{V_{L_{S}}}{\sqrt{3}}$

Substituting the values:

$V_{PH} = \frac{380}{\sqrt{3}} = 219.39 V$

The transformation ratio can be calculated with the formula:

$N_{P} : N_{S} = \frac{V_{PH_{P}}}{V_{PH_{S}}}$

Where:

• $V_{PH_{P}} = 6600$ V
• $V_{PH_{S}} = 219.39$ V

Thus,

$TR = \frac{6600}{219.39} \approx 30 : 1$