Question

An electric kettle consumes 1 KW of electric power when operated at 220V. A fuse wire of what must be used for it?

Answer 1

P = 1KW = 1000 W
V = 220 V

P = V×I
1000 = 220×I
I = 4.54 A

P = V×(V/R)
P = V^2/R
1000 = (220)^2/R
R = 48.4 ohm

hence the fuse wire used should withstand the 48.4 A of current.

Answer 2

The fuse wire should be of 48.4 ohm of resistance.

Given:

Power of electric kettle = 1 KW  

Potential difference = 220 V

Solution:  

So, we know that,

P = 1 KW = 1000 W  

And V = 220 V  

From the formula P = V $\times$ I  

From which, we can find the current,

P = V $\times$ I  

1000 = 220 $\times$ I  

$\Rightarrow I=\frac{1000}{220}$

$I=4.54 \ A$

Also,  

$I=\frac{V}{R}$

$I=\frac{220}{R}$

$R=\frac{220}{4.54}$

$R=48.4 \ ohm$

So, the fuse wire should be of 48.4 ohm of resistance.