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3.1 State how eddy currents may be limited in the iron core of a transformer - NSC Electrical Technology Power Systems - Question 3 - 2016 - Paper 1

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3.1 State how eddy currents may be limited in the iron core of a transformer. 3.2 Name TWO similarities between a single-phase transformer and a three-phase transfo... show full transcript

Worked Solution & Example Answer:3.1 State how eddy currents may be limited in the iron core of a transformer - NSC Electrical Technology Power Systems - Question 3 - 2016 - Paper 1

Step 1

State how eddy currents may be limited in the iron core of a transformer.

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Answer

Eddy currents in the iron core of a transformer can be limited by laminating and insulating the core. This construction reduces the eddy current paths, effectively lowering their magnitude and minimizing energy losses.

Step 2

Name TWO similarities between a single-phase transformer and a three-phase transformer.

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Answer

Both single-phase and three-phase transformers serve the same functional operations by stepping voltage up or down. Additionally, both can be designed as either closed core or shell core types.

Step 3

State TWO factors that may cause excessive heating in a transformer.

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Answer

  1. Insufficient ventilation can hinder heat dissipation, leading to higher temperatures.
  2. Constant overloading causes excessive current flow, which generates more heat than the transformer can handle.

Step 4

Name TWO applications of a delta-star connected distribution network transformer.

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Answer

  1. Supply of three-phase and single-phase electricity to commercial sites.
  2. Providing power to industrial areas requiring robust energy supply systems.

Step 5

Explain the function of a transformer in a distribution network.

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Answer

A transformer in a distribution network serves to adjust voltage levels to suitable values for transmission and utilization. It steps down high voltage from transmission lines to lower voltage suitable for end-users, ensuring efficiency and safety in electrical distribution.

Step 6

Primary phase voltage

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Answer

Using the formula VL=VPH3V_{L} = V_{PH} \sqrt{3}, the primary phase voltage can be calculated as follows:

VPH=VL3=660033810.66 VV_{PH} = \frac{V_{L}}{\sqrt{3}} = \frac{6600}{\sqrt{3}} \approx 3810.66 \text{ V}

Step 7

Secondary phase current

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Answer

The secondary phase current IPHI_{PH} is the same as the line current in this case, so:

IPH=IP=30extAI_{PH} = I_P = 30 ext{ A}

Step 8

Turns ratio

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Answer

The turns ratio (Np:NsN_p:N_s) is calculated using:

NpNs=VPHVSH=660023028.7\frac{N_p}{N_s} = \frac{V_{P_{H}}}{V_{S_{H}}} = \frac{6600}{230} \approx 28.7

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