Solution 1.30 Monthly charge = $6 First 250 kWh @ $0.02/kWh = $5 Remaining 2,436–250 kWh = 2,186 kWh @ $0.07/kWh= $153.02 Total = $164.02 Solution 1.31 In a household, a business is run for an average of 6 hours per day. The total power consumed by the computer and its printer is 230 watts. In addition, a 75-W light runs during the same 6 hours. If their utility charges 11.75 cents per kWhr, how much do the owners pay every 30 days? Total energy consumed over every 30 day period = 30[(230+75)6] = 54.9 kWhr Cost per 30 day period = $0.1175x54.9 = $6.451 Solution 1.32 i = 20 µA q = 15 C t = q/i = 15/(20x10-6) = 750x103 hrs Solution 1.33 i= dq → = − q ∫ idt = 2000 × 3× 10 3 = 6 C dt Solution 1.34 (a) Energy = ∑ pt = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2 = 10 kWh (b) Average power = 10,000/24 = 416.7 W Solution 1.35 energy = (5x5 + 4x5 + 3x5 + 8x5 + 4x10)/60 = 2.333 MWhr Solution 1.36 A battery can be rated in ampere-hours or watt hours. The ampere hours can be obtained from the watt hours by dividing watt hours be a nominal voltage of 12 volts. If an automobile battery is rated at 20 ampere-hours, (a) what is the maximum current that can be supplied for 15 minutes? (b) how many days will it last if it is discharged at a rate of 2 mA? (a) I = 20/0.25 = 80 amps. (b) days = (20/0.002)/24 = 416.7 days. Solution 1.37 A total of 2 MJ are delivered to an automobile battery (assume 12 volts) giving it an additional charge. How much is that additional charge? Express your answer in amperehours. Solution 2,000,000 = w = pt = vit = 12it = 12(charge) or charge = 2x106/12 = 1.6667105 coulomb = 1.6667105 Coulomb x 1 hour/3,600 seconds = 46.3 ampere-hour. charge = 46.3 ampere-hours. Solution 1.38 P = 10 hp = 7460 W W = pt = 7460 × 30 × 60 J = 13.43 × 106 J Solution 1.39 W = pt = 600x4 = 2.4 kWh C = 10cents x2.4 = 24 cents