Question

Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off

Answer 1

Following are the solution to the given question:

Step-by-step explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

`\to (\$60)/((30 \ days)/(24\ hours)) = \$0.08 / kwh.`

Thus,
`(\$0.08)/(\$0.12) = 0.694 \ kW * 0.694 \ kW * 1000 = 694 \ W.`

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

`\to I = (P)/(V)= (694\ W)/(120\ V) = 5.78\ A`

Energy usage reduction percentage =
`((60\ W)/(694\ W) * 100\%)`

This bulb accounts for
`8.64\%` of the energy used, hence it saves when you switch it off.