Here, for the electrical applicance, it is given, Power $=5 \mathrm{KW}=5000 \mathrm{W}$ and potential difference (voltage) $=200 \mathrm{V}$ We know, $\; P = VI$or $\; I = \frac{P}{V} = \frac{5000}{200} = 25 \mathrm{A}$i.e. the appliance will draw a current of $25 \mathrm{A}$ . So, a fuse of current rating $8 \mathrm{A}$ cannot be used as it cannot allow a current of 25 A to flow through it.