Question

In a house, four 60 W electric bulbs are lighted for 2 hours and two 100 W bulbs are lighted for 4 hours everyday. Calculate the energy consumed in the house for 30 days.​

Answer 1

Answer:

Bulb 1 is four in numbers:

Energy = Power × time

Power P_{1}P

1

= 60 W

Time t_{1}t

1

= 2 hours

Energy consumed in one day (W_{1})(W

1

) = P_{1} \times t_{1} \times 4P

1

×t

1

×4

=60 W \times 2 h \times 4=60W×2h×4

= 480 W

Bulb 2 is 2 in numbers:

Power P_{2}P

2

= 100 W

Timet_{2}t

2

= 4 hours

Energy consumed in one day (W_{2})(W

2

) = P_{2} \times t_{2} \times 2P

2

×t

2

×2

= 100 W \times 4 h \times 2=100W×4h×2

= 800 W

Therefore, electric energy consumed in one day = W1 + W2

= 480 + 800 = 1280 W

Electric energy consumed in 30 days

=1280 \times 30=1280×30

= 38400 W = 38.4 kW

The energy consumed in the house for 30 days is 38.4 kW.