Solve the followi

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