Question

a house is fitted with 5 electric bulbs of 60 watts each, 2 electric bulb of 100 watt each, one electric iron of 500 watt, 4 fans of 75 watt and a heater of 1.1 kilowatt. if bulbs and fan works for 6 hour a Day and electric iron and heater work for 2 hours a day , find the electricity bill for the month of August. Electrical energy is supplied at 2.00 rupees per unit Rs 1600 Rs. 649 Rs 496 Rs 2400

Answer 1

Answer:

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

Explanation:

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