Question

A cooler of 1500 W, 200 V and a fan of 500 W, 200 V are to be used from a household supply. What is the rating of the fuse to be used in the circuit?

Answer 1

Answer:

the fuse should be 10A

Explanation:

We know that power p is given by

p=iv⇒i= p /v

So current in cooler is  1500 /200

 =7.5 A

and power drawn by fan is  500 /200

 =2.5 A.

So the total current drawn is 7.5+2.5=10A. Hence the fuse should be 10A

Answer 2

Given :

Power of cooler = 1500 W

Power of fan = 500 W

Voltage of cooler= 200V

Voltage of fan = 200 V

To find :

Rating of the fuse

Solution :

Power (p)=iv⇒i= p /v where i=current

Therefore,.current in cooler is  1500 /200

=7.5 A

power drawn by fan is  500 /200

=2.5 A.

Therefore,the total current drawn is 7.5+2.5=10A.

Hence, the rating of the fuse to be used in the circuit should be 10A.

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