Question

in the office room of mohan father a tube 40 watt , a fan of 75 watt,a cooler of a 150 watt have been installed if all this applicant are used for 8 hour a day .calculate the energy consumed per day and commercial unit of energy?

Answer 1

40w= 40/1000kw=0.04kw
75w=75/1000kw=0.075kw

Energy= power x time
=(0.04+0.075) x 8h
=0.115kw x 8h
=0.92kwh

In kilo watt hour
1Kwh=3.6 x 10^6
0.92Kwh=0.92 X 3.6 x 10^6=3.332x 10^6

Answer 2

Power = $\frac{Energy}{Time}$

Energy = Power x Time
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1kW = 1000 W
Power of tube = 40 W = 0.04 kW

Time = 8 hr

Energy consumed by the tube = 0.04 x 8 =0.32 kWh
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Power = 75 W = 0.075 kW

Time= 8 hr
Energy consumed by the fan = 0.075 x 8
= 0.6 kWh
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Power = 150 W = 0.15 kW

Time = 8 hrs

Energy consumed by the cooler = 0.15 x 8
= 1.2 kWh
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Total energy consumed = 0.32 + 0.6 + 1.2
= 2.12 kWh

Commercial unit of energy is kWh

SI Unit of energy is Joule (J).

1 kWh = 3.6 × 10⁶ Joules

2.12 kWh = 3.6× 10⁶ × 2.12 = 7632000 Joules