What will be the energy consumption of two 300 W bulbs, three 100 W fans, and one 1200 W Refrigerator for continuous operation of 30 hours?

The correct answer is 63 kWh.

  Important Points

• Electric Power - The rate of doing work is called power. This is also the rate of consumption of energy. The rate at which electric energy is dissipated or consumed in an electric circuit is also termed electric power.

• The power P is given by 
  • P = VI Or P = I2R = V2 / R.
• The S.I. unit of electric power is the watt (W).
• It is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V. 
• Thus, 1 W = 1 Volt × 1 ampere = 1 V A.
• The unit 'watt' is very small. Therefore, in actual practice, we use a much larger unit called 'kilowatt'. it is equal to 1000 watts.
• Electric energy is the product of power and time, the unit of electric energy is 'watt-hour' ( W h).
•   1 watt-hour is the energy consumed when 1 watt of power is used for 1 hour. The commercial unit of electric energy is a kilowatt-hour (KW h), commonly known as Units.
• 1 KW h = 1000 watts × 3600 seconds = 3.6 × 106 watt second = 3.6 × 106 joule (J).​​

Key Points

• CONCEPT:  Electric energy = Power × Time 
• CALCULATION: ​
• p1 =  Two 300 watt bulb = 600 watt
• p2 = Three 100 watt fans = 300 watt
• p3 = one 1200watt refrigerator = 1200 watt 
• power( p1 + p2 + p3 ) = 2100 Watt = 2.1 kW
• time = 30 hours ​
• electric energy = ?

​BY FORMULA: 

• Electric energy = power × time 
• Electric energy = 2.1 kW × 30 h 
• Electric energy = 63 kWh
• So, the power consumption is 63 kWh.