Chapter 27: CircuitsProblem 1A wire of resistance 5.0 Ωis connected to a battery whose emf Eis 2.0 V and whoseinternal resistance is 1.0 Ω. In 2.0 min, how much energy is (a) transferred from chemical toelectrical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipatedas thermal energy in the battery?Answer(a) 80 J; (b) 66.7 J; (c) 13.3 JSolution(a)The energy transferred from chemical to electrical in the battery isU=Pt=εit=2tRtotWhere the total resistance for the wire and internal resistance in series isRtot= 5.0 Ω+ 1.0 Ω= 6.0 ΩThereforeU=2.0V( )2∗2min6Ω=80J(b)The thermal energy in the 5.0-Ωwire isU=Pt=i2Rwiret=Rtot⎛⎝⎜⎞⎠⎟2Rwiret=2.0V6.0Ω⎛⎝⎜⎞⎠⎟2∗5.0Ω ∗2.0min=66.7J(c)The “missing” 13.3 J = 80 J – 66.7 J is dissipated as thermal energy because ofthe internal resistance.Problem 3A 5.0 A current is set up in a circuit for 6.0 min by a rechargeable battery with a 6.0 V emf.By how much is the chemical energy of the battery reduced?Answer10.8 kJThe energy transferred is the product of the power and the time and the power is the productof the current and the potential difference.U = iEt = (5.0 A)*(6.0 V)*(6.0 min) = 10.8 kJ
27-2Chapter 27: CircuitsProblem 4A standard flashlight battery can deliver about 2.0 W-h of energy before it runs down. (a) Ifa battery costs 80¢, what is the cost of operating a 100-W lamp for 8.0 h using batteries?(b) What is the cost if power provided by an electric utility company, at 12 cents per kW-h,is used?Answer(a) $320. (b) 4.8¢.Solution(a)The total amount of energy required to operate a 200-W lamp for 8.0 h isU = P*t = (100 W)*(8 h) = 800 W-h.To obtain 800 W-h from flashlight batteries would require 400 batteries (n = 800 W-h/2.0W-h). If each battery cost 80 cents then the total cost would becost = 400($0.80) = $320. This is not a cheap way to light your home!(b)The cost from the power company at the rate of 6 cents per kw-h iscost = ($ 0.06/kw-h)*(800 W-h) = 4.8¢.But it is difficult to keep the power company in the glove compartment of your car.Problem 12(a) In Fig. 27-29 what value must R have ifthe current in the circuit is to be 1.0 mA?Take E1= 2.0 V, E2= 3.0V, and r1= r2= 3.0Ω. (b) What is the rate at which thermalenergy appears in R?Answer(a) 994 Ω. (b) 9.94 ×10-4W.E1E2r1r2RiFigure 27-29(a)We assume that the current is going counter-clockwise in the circuit and apply theloop rule. Traversing the loop in a clockwise sense beginning at the upper left corner- ir2+E2- E1- ir1- iR = 0orR=ε2−1i−r2−r1=3.0V−2.0V1.0mA−3.0Ω −3.0Ω=994Ω.(b)The power dissipated in R isP = i2R = (1.0 mA)2(994 Ω) = 9.94 ×10-4W
Chapter 27: Circuits27-3Problem 22A solar cell generates a potential difference of 0.10 V when a 500 Ωresistor is connectedacross it and a potential difference of 0.15 V when a 1000-Ωresistor is substituted. Whatare (a) the internal resistance and (b) the emf of the solar cell? (c) The area of the cell is 5.0cm2and the rate per unit area at which it receives energy from light is 2.0 mW/cm2. What is
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