An electric refrigerator rated 400 W operates 8 hours/day. What is the cost of energy required to operate it for 30 days at Rs 2.50 per kW h?

CONCEPT:

Power:

• The rate of work done by an electric current is called power. It is denoted by P. The SI unit of power is the watt (W).
• Power dissipation is given by:

⇒ Power (P) = V I = V2/R = I2 R

Where V is the potential difference across resistance, I is current flowing and R is resistance.

• Heat dissipation/work done/Electrical energy is given by:

⇒ Heat dissipated (H) = Power (P) × Time (t)

CALCULATION:

Given that: Power (P) = 400 W, Per day working time = 8 hrs, Total days = 30 days, and Cost per unit = 2.5 per kWh

⇒ Total time (t) = 8 × 30 = 240 hrs

• The total energy consumed by the refrigerator in 30 days would be:

⇒ Energy (H) = P × t = 400 × 240 hrs = 96000 Wh = 96 kWh

⇒ The cost of energy to operate the refrigerator for 30 days = total energy × cost per unit 

⇒ ​The cost of energy to operate the refrigerator for 30 days = 96 kW h × Rs 2.50 per kW h = Rs 240.00.

• Hence option 4 is correct.