Q. 5 B4.1( 22 Votes )

Solve the followi

Answer :

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


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