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:
[tex]x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh[/tex]
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:
[tex]P = \frac{E}{t} = \frac{533.3 kWh}{720 h} = 0.74 kW = 740 W[/tex]
b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:
[tex]I =\frac{P}{V} =\frac{740W}{120V} = 6.2A[/tex]
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 = [tex]\frac{533.3-490}{533.3} = 0.081*100= 8.1%[/tex]
⇒ ΔE(%) = 8.1%
[tex]%(E) =\frac{533.3-490}{533.3} = 0.081 * 100 = 8.1%[/tex]
