# chapt27 - Chapter 27 Circuits Problem 1 A wire of...

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Chapter 27: Circuits Problem 1 A wire of resistance 5.0 Ω is connected to a battery whose emf E is 2.0 V and whose internal resistance is 1.0 Ω . In 2.0 min, how much energy is (a) transferred from chemical to electrical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery? Answer (a) 80 J; (b) 66.7 J; (c) 13.3 J Solution (a) The energy transferred from chemical to electrical in the battery is U = Pt = ε it = 2 t R tot Where the total resistance for the wire and internal resistance in series is R tot = 5.0 Ω + 1.0 Ω = 6.0 Ω Therefore U = 2.0 V ( ) 2 2min 6 Ω = 80 J (b) The thermal energy in the 5.0- Ω wire is U = Pt = i 2 R wire t = R tot 2 R wire t = 2.0 V 6.0 Ω 2 5.0 Ω ∗ 2.0min = 66.7 J (c) The “missing” 13.3 J = 80 J – 66.7 J is dissipated as thermal energy because of the internal resistance. Problem 3 A 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? Answer 10.8 kJ The energy transferred is the product of the power and the time and the power is the product of the current and the potential difference. U = i E t = (5.0 A)*(6.0 V)*(6.0 min) = 10.8 kJ

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27-2 Chapter 27: Circuits Problem 4 A standard flashlight battery can deliver about 2.0 W-h of energy before it runs down. (a) If a 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 is U = 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/2.0 W-h). If each battery cost 80 cents then the total cost would be cost = 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 is cost = (\$ 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 if the current in the circuit is to be 1.0 mA? Take E 1 = 2.0 V, E 2 = 3.0V, and r 1 = r 2 = 3.0 Ω . (b) What is the rate at which thermal energy appears in R? Answer (a) 994 Ω . (b) 9.94 × 10 -4 W. E 1 E 2 r 1 r 2 R i Figure 27-29 (a) We assume that the current is going counter-clockwise in the circuit and apply the loop rule. Traversing the loop in a clockwise sense beginning at the upper left corner - ir 2 + E 2 - E 1 - ir 1 - iR = 0 or R = ε 2 1 i r 2 r 1 = 3.0 V 2.0 V 1.0 mA 3.0 Ω − 3.0 Ω = 994 Ω . (b) The power dissipated in R is P = i 2 R = (1.0 mA) 2 (994 Ω ) = 9.94 × 10 -4 W
Chapter 27: Circuits 27-3 Problem 22 A solar cell generates a potential difference of 0.10 V when a 500 Ω resistor is connected across it and a potential difference of 0.15 V when a 1000- Ω resistor is substituted. What are (a) the internal resistance and (b) the emf of the solar cell? (c) The area of the cell is 5.0 cm 2 and the rate per unit area at which it receives energy from light is 2.0 mW/cm 2 . What is

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## This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.

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chapt27 - Chapter 27 Circuits Problem 1 A wire of...

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