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A transformer is working at its full load, and its efficiency is also maximum. The iron loss is 1000 watts. Then, its copper loss at half of the full load will be-
- a)250 Watt
- b)300 Watt
- c)400 Watt
- d)500 Watt
Correct answer is option 'A'. Can you explain this answer?
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Calculation of Copper Loss at Half Load
Definition: Copper loss is the power loss that occurs in the transformer due to the resistance of the copper wire used in the transformer winding.
Given:
- Transformer is working at full load
- Efficiency is maximum
- Iron loss is 1000 watts
To find: Copper loss at half of the full load
Solution:
- Let's assume that the full load copper loss is 'x' watts
- Since the transformer is working at maximum efficiency, copper loss and iron loss will be equal to each other at full load
- Therefore, total loss at full load = copper loss + iron loss = x + 1000 W
- Efficiency = Output power / Input power
- At maximum efficiency, output power = input power
- Therefore, copper loss = input power - iron loss
- Input power at full load = output power at full load + total loss at full load
- Input power at full load = output power at full load + x + 1000 W
- Since the transformer is working at maximum efficiency, output power at full load = input power at full load
- Therefore, input power at full load = 2(output power at full load) + 1000 W
- Let's assume that the full load output power is 'P' watts
- Therefore, input power at full load = 2P + 1000 W
- Copper loss at full load = input power at full load - iron loss
- Copper loss at full load = (2P + 1000 W) - 1000 W
- Copper loss at full load = 2P watts
- Now, we need to find the copper loss at half of the full load
- Output power at half load = P/2 watts
- Input power at half load = Output power at half load + Copper loss at half load
- Input power at half load = P/2 + Copper loss at half load
- Efficiency at half load = Output power at half load / Input power at half load
- Efficiency at half load = (P/2) / (P/2 + Copper loss at half load)
- Since the transformer is working at maximum efficiency, the efficiency at half load will also be maximum
- Therefore, (P/2) / (P/2 + Copper loss at half load) = 1
- Solving for Copper loss at half load, we get:
- Copper loss at half load = P/2
- Substituting the value of P = input power at full load/2, we get:
- Copper loss at half load = (2P)/2 = P/2 = (2(2P + 1000 W))/4 = (4P + 2000 W)/4 = P/2 + 500 W
- Substituting the value of P = input power at full load/2, we get:
- Copper loss at half load = (2P)/2 = P/2 = (2(2P + 1000 W))/4 = (4P + 2000 W)/4 = P/2 + 500 W
- Copper loss at half load = (2(2P/2) + 1000 W)/2 = P + 1000 W)/2 = (2P
Definition: Copper loss is the power loss that occurs in the transformer due to the resistance of the copper wire used in the transformer winding.
Given:
- Transformer is working at full load
- Efficiency is maximum
- Iron loss is 1000 watts
To find: Copper loss at half of the full load
Solution:
- Let's assume that the full load copper loss is 'x' watts
- Since the transformer is working at maximum efficiency, copper loss and iron loss will be equal to each other at full load
- Therefore, total loss at full load = copper loss + iron loss = x + 1000 W
- Efficiency = Output power / Input power
- At maximum efficiency, output power = input power
- Therefore, copper loss = input power - iron loss
- Input power at full load = output power at full load + total loss at full load
- Input power at full load = output power at full load + x + 1000 W
- Since the transformer is working at maximum efficiency, output power at full load = input power at full load
- Therefore, input power at full load = 2(output power at full load) + 1000 W
- Let's assume that the full load output power is 'P' watts
- Therefore, input power at full load = 2P + 1000 W
- Copper loss at full load = input power at full load - iron loss
- Copper loss at full load = (2P + 1000 W) - 1000 W
- Copper loss at full load = 2P watts
- Now, we need to find the copper loss at half of the full load
- Output power at half load = P/2 watts
- Input power at half load = Output power at half load + Copper loss at half load
- Input power at half load = P/2 + Copper loss at half load
- Efficiency at half load = Output power at half load / Input power at half load
- Efficiency at half load = (P/2) / (P/2 + Copper loss at half load)
- Since the transformer is working at maximum efficiency, the efficiency at half load will also be maximum
- Therefore, (P/2) / (P/2 + Copper loss at half load) = 1
- Solving for Copper loss at half load, we get:
- Copper loss at half load = P/2
- Substituting the value of P = input power at full load/2, we get:
- Copper loss at half load = (2P)/2 = P/2 = (2(2P + 1000 W))/4 = (4P + 2000 W)/4 = P/2 + 500 W
- Substituting the value of P = input power at full load/2, we get:
- Copper loss at half load = (2P)/2 = P/2 = (2(2P + 1000 W))/4 = (4P + 2000 W)/4 = P/2 + 500 W
- Copper loss at half load = (2(2P/2) + 1000 W)/2 = P + 1000 W)/2 = (2P
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A transformer is working at its full load, and its efficiency is also maximum. The iron loss is 1000 watts. Then, its copper loss at half of the full load will be-a)250 Wattb)300 Wattc)400 Wattd)500 WattCorrect answer is option 'A'. Can you explain this answer?
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A transformer is working at its full load, and its efficiency is also maximum. The iron loss is 1000 watts. Then, its copper loss at half of the full load will be-a)250 Wattb)300 Wattc)400 Wattd)500 WattCorrect answer is option 'A'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A transformer is working at its full load, and its efficiency is also maximum. The iron loss is 1000 watts. Then, its copper loss at half of the full load will be-a)250 Wattb)300 Wattc)400 Wattd)500 WattCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A transformer is working at its full load, and its efficiency is also maximum. The iron loss is 1000 watts. Then, its copper loss at half of the full load will be-a)250 Wattb)300 Wattc)400 Wattd)500 WattCorrect answer is option 'A'. Can you explain this answer?.
A transformer is working at its full load, and its efficiency is also maximum. The iron loss is 1000 watts. Then, its copper loss at half of the full load will be-a)250 Wattb)300 Wattc)400 Wattd)500 WattCorrect answer is option 'A'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A transformer is working at its full load, and its efficiency is also maximum. The iron loss is 1000 watts. Then, its copper loss at half of the full load will be-a)250 Wattb)300 Wattc)400 Wattd)500 WattCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A transformer is working at its full load, and its efficiency is also maximum. The iron loss is 1000 watts. Then, its copper loss at half of the full load will be-a)250 Wattb)300 Wattc)400 Wattd)500 WattCorrect answer is option 'A'. Can you explain this answer?.
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