Question

A house uses 2 tubes 40w each for 6 hours a day,3 fans 50w each for 12 hours a day and a heater rated 1000w for 2 hours a day.how many units are consumed in a day?​

Answer 1

Step-by-step explanation:

1) For two tubes

power- 40w×2 = 80w

t - 6 hrs

E = p×t

E = 80×6v = 480wh = 0.480 Kwh

2) For fans

power = 50w×3 = 150w

t = 12 hrs

E = p × t

E = 150×12 = 1800wh = 1.8Kwh

3) For heater

power = 1000w

t = 2 hrs

E = p × t

E = 1000 × 2

E = 2000kw

E = 2 Kwh

Total Energy used that day = 0.480+1.8+2

= 4.28units

HOPE IT IS CORRECT

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Answer 2

Answer:

A house uses 2 tubes 40w each for 6 hours a day,3 fans 50w each for 12 hours a day, and a heater rated 1000w for 2 hours a day, units consumed in a day are 4280W

Explanation:

The number of tube lights  $=2$

power used by one tube light $=40W$

working time of tube light in a day $=6hrs$

total power consumption of tube light $=6*40*2=480W$

The number of fans $=3$

power used by one fan $=50W$

working time of a fan in a day $=12hrs$

total power consumption of the fan  $=12*50*3=1800W$

The number of heaters  $=1$

power used by one tube light $=1000W$

working time of tube light in a day $=2hrs$

total power consumption of tube light $=2*1000*1=2000W$

so the total power consumption of the home $=2000+1800+480=4280W=4.28KW$