Question

An electric kettle of 2kW works for 2h daily. Calculate the (a) energy consumed in SI unit and commercial unit (b)cost of running it in the month of june at the rate of 3.00RS per unit

Answer 1

we know that Energy = Power x time so, E = 2kW x 2hr so, E = 4kwh per day per month E = 4kwh x 30days so, E = 120kWh and cost of running per month = 120kWh x Rs.3/kWh so, = Rs. 360

Answer 2

Given:

Power consumed = 2 kW.

SOLUTION:

Electric kettle is a device which uses electricity consumption for cooking purpose . The characteristic rating of this device is given by the power consumed by it.

According to the question given,

We know that the formula for Energy = Power × time  

So, $E=2 \mathrm{kW} \times 2 \mathrm{hr}$

So, Energy consumption per day will be be  E = 4 kWh per day

For energy consumption per month $\mathrm{E}=4 \mathrm{kWh} \times 30 \mathrm{days}$

So, E = 120 kWh and  

Cost of running per month = Power x commercial rate of consumption.

$\cos t=120 \mathrm{kWh} \times \mathrm{Rs} \cdot 3 / \mathrm{kWh}$=Rs. 360.