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A 10kVA, 1150/115 V, two winding transformer has a full-load efficiency of 96% at 0.6 pf. The transformer is used as an 1150/1265 V auto-transformer. 1).Estimate the power conducted (in kw) by the auto-transformer at full load and unity pf. 2).Estimate the efficiency (in %) of the auto-transformer at full load and 0.6 pf.?
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A 10kVA, 1150/115 V, two winding transformer has a full-load efficienc...
1) Power conducted by the auto-transformer at full load and unity power factor (pf):

To estimate the power conducted by the auto-transformer at full load and unity power factor, we can use the formula:

Power (in kW) = (Apparent Power * Power Factor) / 1000

Given that the transformer is rated at 10kVA and the power factor is unity (1.0), we can calculate the power as follows:

Power (in kW) = (10kVA * 1.0) / 1000
= 10 kW

Therefore, the power conducted by the auto-transformer at full load and unity power factor is 10 kW.

2) Efficiency of the auto-transformer at full load and 0.6 power factor:

To estimate the efficiency of the auto-transformer at full load and 0.6 power factor, we can use the formula:

Efficiency = (Output Power / Input Power) * 100

However, the input power is not given directly. We can calculate it using the formula:

Input Power = Output Power / Efficiency

Given that the full-load efficiency of the transformer is 96% and the power factor is 0.6, we can calculate the efficiency as follows:

Input Power = 10 kW / (96/100)
= 10.42 kW

Using the calculated input power, we can now calculate the output power:

Output Power = Input Power * Efficiency
= 10.42 kW * (96/100)
= 10 kW

Finally, we can calculate the efficiency of the auto-transformer at full load and 0.6 power factor:

Efficiency = (10 kW / 10.42 kW) * 100
= 96%

Therefore, the efficiency of the auto-transformer at full load and 0.6 power factor is 96%.

Explanation:

The power conducted by the auto-transformer at full load and unity power factor is simply the apparent power, which is calculated by multiplying the rated apparent power (10kVA) by the power factor (1.0). This gives us a power of 10 kW.

The efficiency of the auto-transformer at full load and 0.6 power factor is calculated by dividing the output power by the input power, and then multiplying by 100 to get a percentage. However, the input power is not given directly, so we need to calculate it using the given efficiency and output power. Once we have the input power, we can calculate the output power by multiplying it by the efficiency. Finally, we can calculate the efficiency by dividing the output power by the input power and multiplying by 100.

In this case, the efficiency of the auto-transformer at full load and 0.6 power factor is found to be 96%, which indicates that 96% of the input power is being transferred to the output power.
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A 10kVA, 1150/115 V, two winding transformer has a full-load efficiency of 96% at 0.6 pf. The transformer is used as an 1150/1265 V auto-transformer. 1).Estimate the power conducted (in kw) by the auto-transformer at full load and unity pf. 2).Estimate the efficiency (in %) of the auto-transformer at full load and 0.6 pf.?
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A 10kVA, 1150/115 V, two winding transformer has a full-load efficiency of 96% at 0.6 pf. The transformer is used as an 1150/1265 V auto-transformer. 1).Estimate the power conducted (in kw) by the auto-transformer at full load and unity pf. 2).Estimate the efficiency (in %) of the auto-transformer at full load and 0.6 pf.? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 10kVA, 1150/115 V, two winding transformer has a full-load efficiency of 96% at 0.6 pf. The transformer is used as an 1150/1265 V auto-transformer. 1).Estimate the power conducted (in kw) by the auto-transformer at full load and unity pf. 2).Estimate the efficiency (in %) of the auto-transformer at full load and 0.6 pf.? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10kVA, 1150/115 V, two winding transformer has a full-load efficiency of 96% at 0.6 pf. The transformer is used as an 1150/1265 V auto-transformer. 1).Estimate the power conducted (in kw) by the auto-transformer at full load and unity pf. 2).Estimate the efficiency (in %) of the auto-transformer at full load and 0.6 pf.?.
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