Solve the following examples.

If a 1200 W electric iron is used daily for 30 minutes, how much total electricity is consumed in the month of April?

Unit or kilowatt hour kWh is commercial unit of Energy, measured mainly in case of electrical appliances, if an appliance of 1 kW or 1000 W(Spending 1000 J of energy per second) is operated for 1 hour then 1kWh of energy or 1 unit of energy is used

So energy consumed in unit is given as

E = P × t

Where E is the energy consumed by appliance, P is the power of appliance in Kilo Watt (kW), t is the time for which appliance is operated in hours

Here power of the appliance is

P = 1200 W

Converting it to kW

1 kW = 1000 W

i.e. 1 W = 1/1000 kW, so power is

P = 1200 × 1/1000 kW = 1.2 kW

Time for which appliance is used daily is 30 min

1 hour = 60 min

i.e. 1 min = 1/60 hour

i.e. appliance is used daily for

30 × 1/60 hour = � hour

Appliance is used daily in month of April i.e. for 30 Days

So time for which appliance is used is

t = � hour × 30 = 15 hours

putting value of P in t in equation to find energy consumed

E = 1.2 kW × 15 hours = 18 kWh

i.e. 18 kWh or 18 Unit of energy in form of electricity has been used in month of April