Question

In an office a tube light of 40W,a fan of 75W, and a cooler of 150W have been installed. If all these appliances are used for 8 hours a day, calculate the energy consumed per day in commercial unit of energy.

Answer 1

Given:
Tube light of 40W, A fan of 75W, a cooler Of 150W.
Formula:
Electrical energy =E=Power XTime
E=total power consumed x Total time
Let us calculate total Power consumed:
⇒40+75+150
⇒265W
E=265x8
⇒2120WH
1KWh=1000wH
∴1WH=1/1000 Kw
Electrical energy =E=2120/1000=2.120KWH.
∴ Electrical energy consumed by all appliances in one day=2.120KWH

Answer 2

given,
power of tube light = 40 watt
power of fan = 75 watt
power of coolar = 150 watf

total energy consume =total power x time
= (40 + 75 + 150)watt x 8hour
=265 watt x {8 x 60 x 60 } sec
=265 x 8 x 3600 joule
=7632000 joule
we know ,
Kwh is commercial unit .
1 kwh = 3.6 x 10^6 joule
hence
7632000 joule = 7632000/3.6 x10^6 kwh
= 2.12 kwh (ans )