Solution:
When a cooler of 1500 W, 200 V and a fan of 500 W and 200 V are to be used from a household supply then the rating of the fuse to be used is 10 A.It is given that the power and voltage of the cooler and fan is 1500 W, 200 V and 500 W, 200 V respectively.
The total power of the household is equal to the sum of the individual powers. Therefore,
P=P1+P2
⇒P=1500+500
⇒P=2000 W
The current drawn from the supply is given as
I=PV
⇒I=2000200
⇒I=10 A
Hence, the rating of the fuse is 10 A.