3.1 Name and explain TWO types of losses that occur in transformers - NSC Electrical Technology Power Systems - Question 3 - 2017 - Paper 1

To calculate the total kVA, we can use the formula:

$S = \frac{P}{\cos \theta}$

Substituting the given values:

$S = \frac{10,000}{0.8} = 12.5 \text{ kVA}$

The secondary line current can be calculated using:

$I_{L_S} = \frac{P}{\sqrt{3} V_S \cos \theta}$

Substituting the known values:

$I_{L_S} = \frac{10,000}{\sqrt{3} \times 400 \times 0.8} \approx 18.04 \text{ A}$

The secondary phase current is the same as the line current in a star connection:

$I_{phase} = I_{L_S} \approx 18.04 \text{ A}$

The secondary winding must be connected in a star configuration to create a three-phase four-wire system, which allows the transformer to provide both single-phase and three-phase supplies. This arrangement ensures that the transformer can efficiently distribute power to both domestic and industrial loads.

1. Overloading: Operating a transformer above its rated capacity can lead to excessive heat generation.

2. Poor cooling: Inadequate cooling mechanisms can prevent efficient heat dissipation, causing the transformer to overheat.

3. Poor connections: Loose or faulty electrical connections can create resistance, leading to localized heating and potential damage.

If the load is doubled, the primary current of an ideal transformer will also double. This is because the current is directly proportional to the load. As the electrical demand increases, the primary side must provide more current to maintain the power balance.