Question

Suppose that the cost of electrical energy is $0.15 per kilowatt hour and that your electrical bill for 30 days is $80. Assume that the power delivered is constant over the entire 30 days. What is the power in watts? If a voltage of 120 V supplies this power, what current flows? 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

Answer:

a) 740 W b) 6.2 A c) 8.1%

Explanation:

We need first to get the total energy spent during the 30 days, that can be calculated as follows:

1 month = 30 days = 720 hr

If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:

80 $/mo  =  0.15 $/Kwh*x Kwh/mo

Solving for the total energy spent in the month:

$x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh$

Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:

$P = \frac{E}{t} = \frac{533.3 kWh}{720 h} = 0.74 kW = 740 W$

b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:

$I =\frac{P}{V} =\frac{740W}{120V} = 6.2A$

c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:

P = 740 W - 60 W = 680 W

The new value of the energy spent during the entire month will be as follows:

E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo

The reduction in percentage regarding the total energy spent can be calculated as follows:

ΔE = $\frac{533.3-490}{533.3} = 0.081*100= 8.1%$

⇒ ΔE(%) = 8.1%

$%(E) =\frac{533.3-490}{533.3} = 0.081 * 100 = 8.1%$