An electric iron of 750 W is used for S hours per day. Then the energy consumed in one day by the iron is:
A). 6 units
B). 600 units
C). 0.6 units
D). 60 units
6 units
Solution:
Given: Power = 750 W and Time = 8 hours
Total energy consumed = Power × Time
Total energy consumed = 750W × 8H = 6000WH (Watt Hour)
We know that 1KW = 1000W
So, Total energy consumed = 6 KWH (Kilo Watt Hour)
1 unit = 1 KWH
So, Total energy consumed = 6 units
Answer : A). 6 units
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