High School

Electrical Energy Practice Problems

1. How much energy is converted when a 110 kW appliance is used for 3.0 hours?

2. How much does it cost to light six 100 W lightbulbs for 6 hours if the price of electrical energy is $0.009 per kilowatt-hour?

3. A refrigerator operates an average of 10 hours a day. If the power of the refrigerator is 700 W, how much electrical energy does the refrigerator use in one day?

4. A TV with a power rating of 200 W uses 0.8 kWh of electrical energy in one day. For how many hours was the TV on during this day?

5. A hair dryer has a power of 1,200 W. How much electrical energy does the hair dryer use in three minutes?

Answer :

  1. The amount of energy converted is 330 kWh
  2. The cost of lighting the bulbs is $1.944
  3. The amount of electrical energy the refrigerator consumes one day is 7 kWh
  4. The TV was on for 4 hours
  5. A hairdryer will use 0.06 kWh in 3 minutes.

Recall that: Energy (E) = Power (P) x time (t)

1. Energy conversion of a 110 kW appliance used for 3.0 hours:

[tex]$$ E = P \times t = 110 \, \text{kW} \times 3.0 \, \text{hours} = 330 \, \text{kWh} $$[/tex]

2. Cost of lighting six 100-40W lightbulbs for 6 hours:

[tex]$$ \text{Cost} = \text{Number of bulbs} \times \text{Power per bulb} \times \text{Time} \times \text{Cost per unit energy} $$[/tex]

[tex]$$ = 6 \times 60 \, \text{W} \times 6 \, \text{hours} \times 0.009 \, \$/\text{kWh} = \$1.944 $$[/tex]

3. Electrical energy used by a refrigerator in one day:

[tex]$$ E = P \times t = 700 \, \text{W} \times 10 \, \text{hours} = 7 \, \text{kWh} $$[/tex]

4. Operating time of a TV that uses 0.8 kWh in one day:

[tex]$$ t = \frac{E}{P} = \frac{0.8 \, \text{kWh}}{200 \, \text{W}} = 4 \, \text{hours} $[/tex]

5. Electrical energy used by a hair dryer in three minutes:

[tex]$$ E = P \times t = 1200 \, \text{W} \times 3/60 \, \text{hours} = 0.06 \, \text{kWh} $$[/tex]

In these calculations, 1 kW = 1000 W and 1 hour = 60 minutes. The cost of electricity is given in dollars per kilowatt-hour (\$/kWh), so the power of the appliances and the time they operate are converted to kW and hours, respectively.

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