3.1 Name and explain TWO types of losses that occur in transformers. 3.2 The delta-connected primary winding of a three-phase transformer is supplied with 11 kV. The secondary winding is star-connected and supplies 400 V to a balanced star-connected load of 10 kW with a power factor of 0,8. Calculate the following at full load: Given: $V_{P} = 11 \text{ kV}$ $V_{S} = 400 \text{ V}$ $P = 10 \text{ kW}$ $\cos \theta = 0,8$ 3.2.1 The total kVA of the load. 3.2.2 The secondary line current. 3.2.3 The secondary phase current. 3.3 Describe why the secondary winding of a transformer must be connected in star if the transformer is to supply both a domestic and an industrial load. 3.4 Name THREE causes of heat build-up in transformers. 3.5 Explain what will happen to the primary current of an ideal transformer if the load is doubled.

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

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3.1 Name and explain TWO types of losses that occur in transformers. 3.2 The delta-connected primary winding of a three-phase transformer is supplied with 11 kV. Th... show full transcript

Worked Solution & Example Answer:3.1 Name and explain TWO types of losses that occur in transformers - NSC Electrical Technology Electronics - Question 3 - 2017 - Paper 1

Step 1

3.1 Name and explain TWO types of losses that occur in transformers.

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Answer

In transformers, two key types of losses are:

Copper Losses: These losses occur due to the resistance of the copper wires used in the windings. When current flows through these wires, a portion of the energy is lost as heat, which is proportional to the square of the current. This can be expressed mathematically as: $P_{loss} = I^2 R$ where $I$ is the current flowing through the winding and $R$ is the resistance of the winding.
Stray Losses: These losses happen when some of the magnetic field generated by the primary winding does not intersect with the secondary winding. As a result, some energy is lost in the form of a magnetic field that does not contribute to the transformer’s output.

Step 2

3.2.1 The total kVA of the load.

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Answer

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

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

Substituting the given values:

$S = \frac{10000}{0.8} = 12.5 \text{ kW}$

Thus, the total kVA of the load is 12.5 kVA.

Step 3

3.2.2 The secondary line current.

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Answer

To find the secondary line current, we use the formula:

$I_{L(S)} = \frac{S}{\sqrt{3} V_S \cos \theta}$

Substituting the values:

$I_{L(S)} = \frac{10000}{\sqrt{3} \times 400 \times 0.8} = 18.04 \text{ A}$

Therefore, the secondary line current is 18.04 A.

Step 4

3.2.3 The secondary phase current.

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Answer

The secondary phase current can be calculated using the line current as:

$|I_{ph}| = I_{L(S)}$

Thus, the secondary phase current is also 18.04 A.

Step 5

3.3 Describe why the secondary winding of a transformer must be connected in star if the transformer is to supply both a domestic and an industrial load.

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Answer

The secondary winding of a transformer must be connected in star when supplying both domestic and industrial loads to allow for:

Neutral Point Availability: A star connection provides a neutral point, which is essential for domestic loads that require a neutral connection for single-phase supply.
Voltage Flexibility: It enables the use of different line-to-neutral voltages for various applications, accommodating the voltage needs of domestic appliances and industrial equipment.

Step 6

3.4 Name THREE causes of heat build-up in transformers.

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Answer

Three causes of heat build-up in transformers include:

Overloading: When the transformer is subjected to loads exceeding its rated capacity, excess heat is generated due to increased copper losses.
Poor Cooling: Inadequate cooling mechanisms can lead to inefficient heat dissipation, causing temperature to rise.
Lack of Ventilation: In a confined space, poor air circulation can prevent heat from escaping, resulting in elevated temperatures.

Step 7

3.5 Explain what will happen to the primary current of an ideal transformer if the load is doubled.

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Answer

In an ideal transformer, if the load is doubled, the primary current will also double, as it is directly proportional to the load. This is because the transformer’s primary current is determined by the load connected on the secondary side, following the relationship: $I_P = \frac{P_{out}}{V_P}$ where both $P_{out}$ and $V_P$ remain constant under ideal conditions.

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

Worked Solution & Example Answer:3.1 Name and explain TWO types of losses that occur in transformers - NSC Electrical Technology Electronics - Question 3 - 2017 - Paper 1

3.1 Name and explain TWO types of losses that occur in transformers.

3.2.1 The total kVA of the load.

3.2.2 The secondary line current.

3.2.3 The secondary phase current.

3.3 Describe why the secondary winding of a transformer must be connected in star if the transformer is to supply both a domestic and an industrial load.

3.4 Name THREE causes of heat build-up in transformers.

3.5 Explain what will happen to the primary current of an ideal transformer if the load is doubled.

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