Question

) A homeowner is considering putting solar panels on the roof of his house. The installed cost of putting 3 kW of solar panels is $6000 and the panels come with a 25 year guarantee. The panels would be able to meet the average monthly electrical consumption of 850 kW-hrs for the house. a) If the homeowner has the $6000 available for the project, what would the cost of electricity from the power company need to be greater than ($/kW-hr) to make the project viable if other investments are providing 8% interest. ($0.0545/kW-hr) b) If the homeowner had to borrow the $6000 from the bank at 5% interest for 10 years (monthly payments) what would the cost of electricity need to be greater than in $/kWhr from the power company to make the project viable if other investments are providing 8% interest. ($0.0476/kW-hr)

Answer 1

Answer:

a) If the homeowner has the $6000 available for the project, what would the cost of electricity from the power company need to be greater than ($/kW-hr) to make the project viable if other investments are providing 8% interest. ($0.0545/kW-hr)

we can use the present value of an annuity formula:

PV = monthly savings x annuity factor

• PV = $6,000
• Annuity factor, 300 periods, 0.6667% = 129.52005

monthly savings = $6,000 / 129.52005 = $46.3249

price of kW-hr = $46.3249 / 850 = $0.054499851 ≈ $0.0545

b) If the homeowner had to borrow the $6000 from the bank at 5% interest for 10 years (monthly payments) what would the cost of electricity need to be greater than in $/kWhr from the power company to make the project viable if other investments are providing 8% interest. ($0.0476/kW-hr)

the monthly payment to cover the loan = PV / annuity factor

• PV = $6,000
• Annuity factor, 120 periods, 0.4167% = 94.28033

monthly payment = $6,000 / 94.28033 = $63.64

price of kW-hr = $63.64 / 850 = $0.074870588 ≈ $0.0749