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Unformatted text preview: 1067 Chapter 27 1. (a) The energy transferred is U Pt t r R = = + = + = ε 2 2 2 0 2 0 60 50 80 ( . ( . min)( . . V) s / min) 1.0 J Ω Ω (b) The amount of thermal energy generated is ′ = = + F H G I K J = + F H G I K J = U i Rt r R Rt 2 2 2 2 0 50 50 2 0 60 67 ε . . ( . )( . min) ( V 1.0 s / min) J. Ω Ω Ω (c) The difference between U and U' , which is equal to 13 J, is the thermal energy that is generated in the battery due to its internal resistance. 2. If P is the rate at which the battery delivers energy and ∆ t is the time, then ∆ E = P ∆ t is the energy delivered in time ∆ t . If q is the charge that passes through the battery in time ∆ t and ε is the emf of the battery, then ∆ E = q ε . Equating the two expressions for ∆ E and solving for ∆ t , we obtain (120A h)(12.0V) 14.4h. 100W q t P ε ⋅ ∆ = = = 3. The chemical energy of the battery is reduced by ∆ E = q ε , where q is the charge that passes through in time ∆ t = 6.0 min, and ε is the emf of the battery. If i is the current, then q = i ∆ t and ∆ E = i ε ∆ t = (5.0 A)(6.0 V) (6.0 min) (60 s/min) = 1.1 × 10 4 J. We note the conversion of time from minutes to seconds. 4. (a) The cost is (100 W · 8.0 h/2.0 W · h) ($0.80) = $3.2 ×1 2 . (b) The cost is (100 W · 8.0 h/10 3 W · h) ($0.06) = $0.048 = 4.8 cents. 5. (a) The potential difference is V = ε + ir = 12 V + (50 A)(0.040 Ω ) = 14 V. (b) P = i 2 r = (50 A) 2 (0.040 Ω ) = 1.0×10 2 W. (c) P' = iV = (50 A)(12 V) = 6.0×10 2 W. CHAPTER 27 1068 (d) In this case V = ε – ir = 12 V – (50 A)(0.040 Ω ) = 10 V. (e) P r = i 2 r =(50 A) 2 (0.040 Ω ) = 1.0 ×1 2 W. 6. The current in the circuit is i = (150 V – 50 V)/(3.0 Ω + 2.0 Ω ) = 20 A. So from V Q + 150 V – (2.0 Ω ) i = V P , we get V Q = 100 V + (2.0 Ω )(20 A) –150 V = –10 V. 7. (a) Let i be the current in the circuit and take it to be positive if it is to the left in R 1 . We use Kirchhoff’s loop rule: ε 1 – iR 2 – iR 1 – ε 2 = 0. We solve for i : i R R = + = + = ε ε 1 2 1 2 12 6 0 8 0 050 V V 4.0 A . . . . Ω Ω A positive value is obtained, so the current is counterclockwise around the circuit. If i is the current in a resistor R , then the power dissipated by that resistor is given by 2 P i R = . (b) For R 1 , P 1 = 2 1 i R = (0.50 A) 2 (4.0 Ω ) = 1.0 W, (c) and for R 2 , P 2 = 2 2 i R = (0.50 A) 2 (8.0 Ω ) = 2.0 W. If i is the current in a battery with emf ε , then the battery supplies energy at the rate P =i ε provided the current and emf are in the same direction. The battery absorbs energy at the rate P = i ε if the current and emf are in opposite directions. (d) For ε 1 , P 1 = 1 i ε = (0.50 A)(12 V) = 6.0 W (e) and for ε 2 , P 2 = 2 i ε = (0.50 A)(6.0 V) = 3.0 W....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics, Energy, Thermal Energy

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