Question

an electric refrigerator rated 400w operates 8 hours a day 4 bulbs each rated 60w are lighted 6 hours a day each an iron box rated 600w is used 2 hours a day what is the cost of the energy to operate in m/o September if rate of a unit is Rs 3?​

Answer 1

Solution:

Refrigerator

• Power = 400 W

we need to convert W into KW in order to do this simply divide by 1000

$\implies\mathtt{ P = \dfrac{400}{1000} = 0.4 \:KW}$

• Power = 0.4 KW

• Time operated = 8 hr

we know that ,

$\implies\mathtt{Energy = Power \times time }$

$\implies\mathtt{E_r = P_r \times t_r }$

$\implies\mathtt{E_r =0.4\times 8 }$

$\implies\mathtt{E_r = 3.2\: KWh }$

________________

Bulbs :

• Power = 60 W

we need to convert W into KW in order to do this simply divide by 1000

$\implies\mathtt{ P_b = \dfrac{60}{1000} = 0.06 \:KW}$

• Power = 0.06 KW

• Time operated = 6 hr

Total no of bulbs = 4

we know that ,

$\implies\mathtt{E_b= P_b \times t_b\times n_b }$

$\implies\mathtt{E_b= 0.06\times 6\times 4 }$

$\implies\mathtt{E_b = 1.44 \: KWh }$

___________________

Iron box:

• Power = 600 W

we need to convert W into KW in order to do this simply divide by 1000

$\implies\mathtt{ P_i = \dfrac{600}{1000} = 0.6 \:KW}$

• Power = 0.6 KW

• Time operated = 2hr

we know that ,

$\implies\mathtt{E_i = P_i \times t_i }$

$\implies\mathtt{E_r = 0.6\times 2 }$

$\implies\mathtt{E_r = 1.2 \:KWh}$

_______________

Total energy consumed in one day

$\implies\mathtt{E_r +E_b+E_i}$

$\implies\mathtt{3.2 +1.44+1.2}$

$\implies\mathtt{5.84 \:KWh}$

____________

Total no of days in September= 30

Energy consumed by all the appliances in 30 days

= 5.84 x 30

=  175.2 KWh

Energy consumed by all the appliances in 30 days =   175.2 KWh

______________

Total cost of the appliances to operate in the month of September  

= 175.2 x 3

= Rs 525 .6

Total cost of the appliances to operate in the month of September is Rs 525.6