Chapter27 - 1. (a) The energy transferred is (2.0 V) 2 (2.0...

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1. (a) The energy transferred is UP t t rR == + = + = ε 22 20 60 50 80 (. (.m i n ) ( . . V) s/ min) 1.0 J ΩΩ (b) The amount of thermal energy generated is + F H G I K J = + F H G I K J = Ui R t Rt 2 2 2 2 0 60 67 . . )(. m in )( 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.
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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 . Equating the two expressions for E and solving for t , we obtain (120A h)(12.0V) 14.4h. 100W q t P ∆= = =
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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 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.
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4. (a) The cost is (100 W · 8.0 h/2.0 W · h) ($0.80) = $3.2 × 1 0 2 . (b) The cost is (100 W · 8.0 h/10 3 W · h) ($0.06) = $0.048 = 4.8 cents.
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(d) In this case V = ε ir = 12 V – (50 A)(0.040 ) = 10 V. (e) P r = i 2 r =(50 A) 2 (0.040 ) = 1.0 ×1 0 2 W. 5. (a) The potential difference is V = + ir = 12 V + (50 A)(0.040 ) = 14 V. (b) P = i 2 r = (50 A) 2 (0.040 ) = 1.0×10 2 W. (c) P' = iV = (50 A)(12 V) = 6.0×10 2 W.
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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.
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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 2 = 0. We solve for i : i RR = + = + = 12 12 6 0 80 050 VV 4.0 A . . .. ΩΩ 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 PiR = . (b) For R 1 , P 1 = 2 1 iR = (0.50 A) 2 (4.0 ) = 1.0 W, (c) and for R 2 , P 2 = 2 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 = 1 i = (0.50 A)(12 V) = 6.0 W (e) and for 2 , P 2 = 2 i = (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.
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8. (a) The loop rule leads to a voltage-drop across resistor 3 equal to 5.0 V (since the total drop along the upper branch must be 12 V). The current there is consequently i = (5.0 V)/(200 ) = 25 mA. Then the resistance of resistor 1 must be (2.0 V)/ i = 80 . (b) Resistor 2 has the same voltage-drop as resistor 3; its resistance is 200 .
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9. (a) Since R eq < R , the two resistors ( R = 12.0 and R x ) must be connected in parallel: R RR R R x x x x eq == + = + 300 12 0 12 0 .
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter27 - 1. (a) The energy transferred is (2.0 V) 2 (2.0...

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