Question

find the energy in units consumed in the month of June by four devices of power 500 Watt each working for 10hrs daily​

Answer 1

Given:-

• Power of each device = 500w → 0.5kw

• Time = 10h

• Number of Device = 4

• Number of Days = 30

To Find:-

• The energy Consumed in one month of June

Formulae used:-

• $\rm{energy\:Consumed = Power\times{Time}}$

Now,

→ $\sf{Energy\:Consumed\:each\:day = 0.5 * 4 * 10 = 20kwh}$

→ $\sf{Energy\:Consumed\:in\:30\:days = 20 × 30 = 600kwh}$

→ $\sf{ 600unit}$

Hence, The total energy consumed in unit is 600.

Answer 2

Given :

• Power consumed by each device = 500W = 0.5 kW

• Total days = 30

• Time 10 hours

• number of devices = 4

To Find :

• find the energy in units consumed in the month of June

Solution :

Energy consumed by each one in 10 hours

$\sf : \: \implies \: \: \: \: \: \: \: \: \: Power = \frac{Energy}{Time} \\ \\ \\ \:$

Substitute all values :

$\sf : \: \implies \: \: \: \: \: \: \: \: \: 0.5 = \frac{Energy}{10} \\ \\ \\ \sf : \: \implies \: \: \: \: \: \: \: \: \:Energy = 5kwh \:$

Therefore, power consumed by 4 such devices =

$\sf : \: \implies \: \: \: \: \: \: \: \: \: \: 4 \times 5 kWh \\ \\ \\ </p><p> </p><p> \sf : \: \implies \: \: \: \: \: \: \: \: \: 20 units$

the energy in units consumed in the month of June

$\sf : \: \implies \: \: \: \: \: \: \: \: \: 20 \times 30 \\ \\ \\ \sf : \: \implies \: \: \: \: \: \: \: \: \: 600 unit$