Question

A 1.7-liter electric kettle rated power of 2200 W will take roughly 780 seconds to boil the water. The kettle is operated at 240 V, used in Malaysia. Determine: i) the current when the kettle boils the water (for a single operation). ii) the average energy consumption for boiling the water in Joule and kWh for twenty (20) times of operation.

Answer 1

Part (i)

The current flowing through the kettle can be given as,

`I=(P)/(V)`

Substitute the known values,

`\begin{gathered} I=\frac{2200\text{ W}}{240\text{ V}}(\frac{1\text{ A}}{1\text{ W/V}}) \\ \approx9.17\text{ A} \end{gathered}`

Thus, the current flowing in the kettle is 9.17 A.

Part (ii)

The average energy consumption for boiling the water is,

`E=Pt`

Substitute the known values,

`\begin{gathered} E=(2200\text{ W)(780 s)(}\frac{1\text{ J}}{1\text{ Ws}}) \\ \approx1.72*10^7\text{ J} \end{gathered}`

Thus, the average energy consumption for boiling the water is

`1.72*10^7\text{ J}`