Question

How does this answer came? Plz tell in DETAIL.​

Answer 1

Answer:

Here P(power) = 3000W

and PD(potential difference) = 230 V

Since, P = IV

So, I = P/V = 3000/230

Then I = 13.04 A

But for safe working of this appliance the fuse wire must allow more than 13.04 A of current.

Hence, the cable connected to it should be capable to carry at least 15 A of current.

✌️Hope it helps!!!