‘A fuse is rated 8 A’. Can it be used with an electrical appliance of rating 5 kW, 200 V?

Solution:

Expression for the safe limit of current that can flow through the electrical appliance is

I = P / V

Now, substituting the given values of P and v we get –

=> I = 5000 / 200

Therefore, I = 25 A

As we know, 25 A is greater than 8 A.

As a result, such fuse cannot be used.