Question

answer you willl be brain listed

a house hold use 3 bulb each for 6 hour 2 electric fans of 80w each for 8 hour and refrigerator 18 hour a day.calculate the electricity​

Answer 1

Answer:

$power \: of \: bulb(p1) = 80 \: w \\ power \: of \: fan (p2)= 80 \: w \\ \\ for \: bulb \\ work(w)= \frac{pt}{1000} = \frac{3 \times 80 \times 6}{1000} =1.44 \: kwh \\ \\ for \: fan \\ work(w2) = \frac{pt}{1000} = \frac{2 \times 80 \times 8}{1000} = 1.28 \: kwh \\ \\ for \: refrigerator(w3) = \frac{pt}{1000} = \frac{1 \times 80 \times 18}{1000} = 1.44 \: kwh$

now,

total work done

$= (1.44 + 1.28 + 1.44) \: kwh = 4.16 \: kwh = 9792000 \: j$