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3.1 State how eddy currents may be limited in the iron core of a transformer - NSC Electrical Technology Electronics - 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 and a three-phase transformer. 3.3 S... 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 Electronics - 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 can be limited in the iron core of a transformer by laminating and insulating the laminations of the core. This reduces the flow of eddy currents, as the lamination creates high resistance paths that restrict the circulation of these currents within the core.

Step 2

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

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Answer

  1. Both transformers serve the same functional operations, such as transforming voltage levels.
  2. Both can be designed to step down or step up the supply voltage, depending on the application.

Step 3

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

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Answer

  1. Insufficient ventilation within the transformer can lead to heat accumulation.
  2. Constant overloading of the transformer results in excessive current flow, which increases heat generation.

Step 4

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

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Answer

  1. Providing single and three-phase supply to commercial sites.
  2. Supplying three-phase power to industrial areas.

Step 5

Explain the function of a transformer in a distribution network.

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Answer

A transformer in a distribution network functions to adjust voltage levels to suitable values for efficient power delivery. It steps down high transmission voltages to lower voltages for consumer use, ensuring safety and minimizing energy loss during transmission. Additionally, transformers maintain the balance of electric power across the network, allowing for reliable distribution to end users.

Step 6

Calculate the: 3.6.1 Primary phase voltage.

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Answer

To calculate the primary phase voltage, we use the formula: V_{ip} = V_{sp} imes rac{V_{ip}}{V_{sp}} Given that Vip=6.6kVV_{ip} = 6.6 kV, the primary phase voltage is 6.6kV6.6 kV.

Step 7

3.6.2 Secondary phase current.

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Answer

The secondary current can be calculated using the power formula: P=VspimesIsimesextCoshetaP = V_{sp} imes I_{s} imes ext{Cos} heta Rearranging gives: I_{s} = rac{P}{V_{sp} imes ext{Cos} heta} Substituting given values:

ightarrow I_{s} = 30 A$$

Step 8

3.6.3 Primary phase power.

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Answer

The primary phase power can also be calculated using: Pip=VipimesIpimesextCoshetaP_{ip} = V_{ip} imes I_{p} imes ext{Cos} heta With Vip=6.6kVV_{ip} = 6.6 kV and using the relationship of power in primary and secondary, we can assume:

ightarrow P_{ip} = 15 kW$$

Step 9

3.6.4 Turns ratio.

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Answer

The turns ratio can be calculated as follows: ext{Turns Ratio} = rac{V_{ip}}{V_{sp}} = rac{6.6 kV}{230 V} Calculating gives: extTurnsRatio=28.7 ext{Turns Ratio} = 28.7

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