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A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volt are to be used from a household supply. The rating of fuse to be used is
- a)2.5 A
- b)5.0 A
- c)7.5 A
- d)10 A
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volt are to be us...
Total power used, P - P1 + P2 = 1500 + 500 = 2000 W.
Current drawn from the supply,
Current drawn from the supply,
Most Upvoted Answer
A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volt are to be us...
Given:
Cooler rating = 1500 W, 200 V
Fan rating = 500 W, 200 V
To calculate the rating of the fuse required, we need to add the power ratings of both the cooler and fan.
Total power = 1500 W + 500 W = 2000 W
Now, we need to calculate the current drawn by the devices using the formula:
Current = Power/Voltage
Current drawn by cooler = 1500 W/200 V = 7.5 A
Current drawn by fan = 500 W/200 V = 2.5 A
Total current drawn = 7.5 A + 2.5 A = 10 A
Therefore, the rating of the fuse required to run both the devices from the household supply is 10 A, which is option D.
Cooler rating = 1500 W, 200 V
Fan rating = 500 W, 200 V
To calculate the rating of the fuse required, we need to add the power ratings of both the cooler and fan.
Total power = 1500 W + 500 W = 2000 W
Now, we need to calculate the current drawn by the devices using the formula:
Current = Power/Voltage
Current drawn by cooler = 1500 W/200 V = 7.5 A
Current drawn by fan = 500 W/200 V = 2.5 A
Total current drawn = 7.5 A + 2.5 A = 10 A
Therefore, the rating of the fuse required to run both the devices from the household supply is 10 A, which is option D.
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A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volt are to be us...
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A cooler of 1500 W, 200 volt and a fan of 500 W, 200 volt are to be used from a household supply.The rating of fuse to be used isa)2.5 Ab)5.0 Ac)7.5 Ad)10 ACorrect answer is option 'D'. Can you explain this answer?
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