Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarterof the time and the unit cost of electricity is $0.09/kWh, the electricity cost of this refrigerator per month (30 days) is(a) $3.56 (b) 55.18 (a 53.54 {0159.26 [6) $20.74

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Answer:

$5.184

Explanation:

The cost can be calculated using the formula: [tex]Cost = Load \ factor \times Number \ of \ hours \ \\M_{month} = M_{units} \times W\\[/tex]

Before using this, we require the following conversions:

320 W → kW:

[tex]\frac {320}{1000} = 0.32[/tex]

30 Days → Hours:

[tex]30 \times 24 = 720[/tex]

Using the above stated formula:

[tex]M_{month} = 0.09 \times 0.32 \times \frac{1}{4} \times 720 = 5.184[/tex]

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