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QUESTION 3: THREE-PHASE TRANSFORMERS 3.1 State TWO functions of the oil used in oil-filled transformers - NSC Electrical Technology Electronics - Question 3 - 2016 - Paper 1

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QUESTION 3: THREE-PHASE TRANSFORMERS 3.1 State TWO functions of the oil used in oil-filled transformers. 3.2 Name TWO losses that occur in transformers. 3.3 Stat... show full transcript

Worked Solution & Example Answer:QUESTION 3: THREE-PHASE TRANSFORMERS 3.1 State TWO functions of the oil used in oil-filled transformers - NSC Electrical Technology Electronics - Question 3 - 2016 - Paper 1

Step 1

3.6.1 Primary line current

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Answer

To calculate the primary line current (IPLI_{PL}), we use the formula:

IPL=S3VLPI_{PL} = \frac{S}{\sqrt{3} V_{LP}}

Substituting the values:

  • S = 20000 VA (20 kVA)
  • VLP=6.6kV=6600VV_{LP} = 6.6 kV = 6600 V

We have: IPL=200003×6600=2000011425.511.75AI_{PL} = \frac{20000}{\sqrt{3} \times 6600} = \frac{20000}{11425.51} \approx 1.75 A

Step 2

3.6.2 Secondary phase voltage

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Answer

To calculate the secondary phase voltage (VPHV_{PH}), we use the relationship between line voltage (VLV_{L}) and phase voltage in a star-connected system:

VPH=VL3V_{PH} = \frac{V_{L}}{\sqrt{3}}

Given:

  • VL=380VV_{L} = 380 V, we calculate:

VPH=3803219.39VV_{PH} = \frac{380}{\sqrt{3}} \approx 219.39 V

Step 3

3.6.3 Transformer ratio

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Answer

The transformer ratio (TRTR) can be calculated using the formula:

TR=VPH(primary)VPH(secondary)TR = \frac{V_{PH (primary)}}{V_{PH (secondary)}}

We know:

  • VPH(primary)=6.6kV=6600VV_{PH (primary)} = 6.6 kV = 6600 V
  • VPH(secondary)219.39VV_{PH (secondary)} \approx 219.39 V

Thus: TR=6600219.3930:1TR = \frac{6600}{219.39} \approx 30 : 1

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