Question

in a house four 60W bulbs are lighted for 2 hours and two 100W bulbs are lighted for 4 hours everyday.calculate the energy consumed in the house for 30 days.

Answer 1

Energy = power * time
2 hours energy consumed = 60*2 = 120 J
4 hours energy consumed = 100*4 = 400 J

Hours in 30 days = 24*30 = 720 hours

Energy consumed in 720 hours(30 days) = (120+400)*720 = 520*720 = 374400 J

Answer 2

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

To find:

The energy consumed in the house for 30 days as per the given.

Solution:

Bulb 1 is four in numbers:

Energy = Power × time

Power $P_{1}$ = 60 W

Time $t_{1}$  = 2 hours

Energy consumed in one day $\left(W_{1}\right)$ = $P_{1} \times t_{1} \times 4$

$=60 W \times 2 h \times 4$

= 480 W

Bulb 2 is 2 in numbers:

Power $P_{2}$ = 100 W

Time$t_{2}$ = 4 hours

Energy consumed in one day $\left(W_{2}\right)$ = $P_{2} \times t_{2} \times 2$

$= 100 W \times 4 h \times 2$

= 800 W

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

= 480 + 800 = 1280 W

Electric energy consumed in 30 days                

$=1280 \times 30$

= 38400 W = 38.4 kW

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