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3.1 What is the purpose of a transformer? 3.2 Name the type of loss that is dissipated in a transformer due to the internal resistance in the windings - NSC Electrical Technology Power Systems - Question 3 - 2017 - Paper 1

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3.1 What is the purpose of a transformer? 3.2 Name the type of loss that is dissipated in a transformer due to the internal resistance in the windings. 3.3 State T... show full transcript

Worked Solution & Example Answer:3.1 What is the purpose of a transformer? 3.2 Name the type of loss that is dissipated in a transformer due to the internal resistance in the windings - NSC Electrical Technology Power Systems - Question 3 - 2017 - Paper 1

Step 1

What is the purpose of a transformer?

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Answer

The purpose of a transformer is to step down or step up an alternating voltage. It is used to transfer power from one AC circuit to another, changing the voltage and corresponding current flow, while isolating one circuit from another electrically.

Step 2

Name the type of loss that is dissipated in a transformer due to the internal resistance in the windings.

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Answer

The type of loss that occurs due to internal resistance in the windings is known as copper losses, often represented mathematically as I2RI^2R losses.

Step 3

State TWO methods used to cool transformers.

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Answer

  1. Air cooling
  2. Oil filled self-cooling or oil filled water cooling.

Step 4

Describe what could happen if any one of the cooling methods used to cool large transformers failed to perform its function.

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Answer

If a cooling method fails, the transformer may overheat. This overheating could damage the insulation between the windings, depending on the heat degree. In severe cases, this could lead to an internal short circuit, and gas may form, which is highly combustible.

Step 5

Name TWO applications of a three-phase delta-star transformer.

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Answer

  1. Used to distribute power to consumers at substations.
  2. To step down the voltage to a 380 V three-phase and 220 V single-phase supply.

Step 6

Secondary line current

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Answer

Given: S=120kVAS = 120 \, kVA and VL=380VV_{L}= 380 \, V

To find the secondary line current ILI_L:

IL=S3VLI_L = \frac{S}{\sqrt{3}V_{L}}

Calculating: IL=1200003380182.32 AI_L = \frac{120000}{\sqrt{3} \cdot 380} \approx 182.32 \text{ A}

Step 7

Secondary phase current

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Answer

To find the secondary phase current IphI_{ph}:

Since Iph=ILI_{ph} = I_L, we have: Iph=182.32 AI_{ph} = 182.32 \text{ A}

Step 8

Input power to the clinic

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Answer

To determine the input power PP to the clinic:

P=3VLILp.f.P = \sqrt{3} \cdot V_{L} \cdot I_L \cdot p.f.

Calculating: P=3380182.320.9108.47kWP = \sqrt{3} \cdot 380 \cdot 182.32 \cdot 0.9 \approx 108.47 \, kW

Step 9

Explain why the secondary winding of a step-down transformer has a thicker wire.

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Answer

The secondary winding of a step-down transformer has a thicker wire because it carries a higher current compared to the primary winding, which helps to prevent overheating and ensures efficient operation.

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