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If an electric iron of 1600 W is used for 45 minutes everyday, find the electric energy consumed in a leap year?

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If an electric iron of 1600 W is used for 45 minutes everyday, find the electric energy consumed in a leap year?
posted Mar 6, 2017 by anonymous

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2 Answers

+1 vote

1600*45*366/(1000*60) kwh =439.2 kWh

answer Mar 7, 2017 by Kewal Panesar
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Energy consumed = 1600*60*45*366= 1581.12 MJ.
1581.12/(3600*1000) = 439.2 kWh

answer Mar 6, 2017 by Tejas Naik



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