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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 |
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Answer» 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: HOPE THIS HELPS YOU BETTERPLZ MARK ME AS BRAINLIESTGIVE THANKS =TAKE THANKS |
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