Question

If an electric iron of 1200 w is used for 30 minutes everyday, find electric energy consumed in the month of april.

Answer 1

We have,

P = 1,200 W = 1.2 kW       Usage time per day = 1/2 hrs

No. of days in April = 30 days

∴  Total time, t = 1/2 * 30 = 15 hrs

Energy consumed = Pt = 1.2 kW * 15 hrs

                                        = 18 kWh    

Hope this helps

Answer 2

$\textbf P = \dfrac{1200}{1000}$

= 1.2 kW

$\textbf Time = \dfrac{30}{60}$

= 0.5 hour

Electric Energy = Power × Time × Days

= 1.2 × 0.5 × 30

= 18 kWh