An electric kettle is rated $750\ \mathrm{W}$; $220\ \mathrm{V}$. Can this kettle be used in a circuit which has a fuse of current rating $3\ \mathrm{A}$? Give reason for your answer.

As explained in the chapter under electric power, $P = VI$ (Eq. 11.19), so the operating current is $I = P/V$. For the kettle, $P = 750~\mathrm{W}$ and $V = 220~\mathrm{V}$:

$$I = \frac{750~\mathrm{W}}{220~\mathrm{V}} \approx 3.4~\mathrm{A}$$

The chapter explains the role of a fuse: 'It protects circuits and appliances by stopping the flow of any unduly high electric current... If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.' Since the kettle's normal working current ($\approx 3.4~\mathrm{A}$) already exceeds the fuse's rated value of $3~\mathrm{A}$, the fuse would melt (blow) as soon as the kettle is switched on, even under normal, non-faulty operation. Therefore this kettle cannot be used with a 3 A fuse; a fuse of higher rating (e.g., 5 A, as used for the 1 kW iron example in the chapter) would be needed.