Answer:
Following are the solution to the given question:
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.
[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]
Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]
The electricity used is continuously 694W over 30 days.
If just resistor loads (no reagents) were assumed,
[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]
Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]
This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.
