Question

A house is fitted with 5 bulbs of 60 W each, 2 bulbs 100 W each, one electric iron of 220 V drawing 2 A of current, 5 fans 110 W each and a heater of 1kW. If bulbs and fans work for 6 hrs a day, iron and heater work 2 hrs a day, find the electricity bill for one month when energy is supplied at Rs 2.50 per unit

Answer 1

5 bulbs 60 watt each = 60×5 = 300Watt
2Bulbs 100watt each = 2×100= 200Watt
Bulbs total power = 500watt
Electric Iron power = V×I = 220×2 = 440 Watt
5 Fans 110 W each = 5× 110 = 550 W
Heater 1KW = 1000 W
fans & Bulbs total power = 500+550 = 1050 watt = 1.05 KW
In month power consumption when operates for 6 hrs a day = 1.05×6×30 = 189 KWH (units)
Iron & heater total power = 1000+440 = 1440W = 1.44 KW
In month power consumption when operates for 2 hrs a day= 1.44×2×30 = 86.4 KWH (units)
Total units = 189+86.4= 275.4
Total Electricity Bill = 275.4×2.5 = 688.5 Rupees
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