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Unformatted text preview: Chapter 27 1. (a) The cost is (100 W 8.0 h/2.0 W h) ($0.80) = $3.2 2 . (b) The cost is (100 W 8.0 h/10 3 W h) ($0.06) = $0.048 = 4.8 cents. 2. The chemical energy of the battery is reduced by E = q , where q is the charge that passes through in time t = 6.0 min, and is the emf of the battery. If i is the current, then q = i t and E = i 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. 3. 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 is the emf of the battery, then E = q . Equating the two expressions for E and solving for t , we obtain (120A h)(12.0V) 14.4h. 100W q t P = = = 4. (a) The energy transferred is U Pt t r R = = = = 2 2 2 0 2 0 60 50 80 ( . ( . min)( . . V) s/ min) 1.0 J (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 . . ( . )( . min)( V 1.0 s/ min) J. (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. 5. (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 Kirchhoffs loop rule: 1 iR 2 iR 1 2 = 0. We solve for i : 1091 CHAPTER 27 i R R = = = 1 2 1 2 12 6 0 8 0 050 V V 4.0 A . . . . 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 = (0.50 A) 2 (4.0 ) = 1.0 W, (c) and for R 2 , P 2 = (0.50 A) 2 (8.0 ) = 2.0 W. If i is the current in a battery with emf , then the battery supplies energy at the rate P =i provided the current and emf are in the same direction. The battery absorbs energy at the rate P = i if the current and emf are in opposite directions. (d) For 1 , P 1 = (0.50 A)(12 V) = 6.0 W (e) and for 2 , P 2 = (0.50 A)(6.0 V) = 3.0 W. (f) In battery 1 the current is in the same direction as the emf. Therefore, this battery supplies energy to the circuit; the battery is discharging. (g) The current in battery 2 is opposite the direction of the emf, so this battery absorbs energy from the circuit. It is charging. 6. The current in the circuit is i = (150 V 50 V)/(3.0 + 2.0 ) = 20 A. So from V Q + 150 V (2.0 ) i = V P , we get V Q = 100 V + (2.0 )(20 A) 150 V = 10 V. 7. (a) The potential difference is V = + ir = 12 V + (0.040 )(50 A) = 14 V....
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 Fall '08
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 Physics, Charge, Energy

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