solutions_chapter19

Set up solve a b p increases when r decreases so the

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Set Up: Solve: (a) (b) P increases when R decreases, so the 100 W bulb has less resistance. Reflect: The bulb with smaller R draws more current and this more than compensates for smaller R in 19.38. Set Up: and Solve: (a) Yes, this is very dangerous. (b) We can also calculate P as or VI. These expressions also give 14.4 W. 19.39. Set Up: and energy is the product of power and time. Solve: Reflect: The energy delivered depends not only on the voltage and current but also on the length of the pulse. 19.40. Set Up: The heater consumes 540 W when Solve: (a) so (b) so (c) The cost is (d) Assuming that R remains P is smaller by a factor of 19.41. Set Up: Solve: (a) The energy supplied is (b) Reflect: The battery stores a large amount of energy and it takes time to return this energy to the battery in order to recharge it. 19.42. Set Up: For an emf, For a resistor, Solve: (a) and The rate at which chemical energy is being converted to electrical energy within the battery is (b) (c) Note that P E 5 P r 1 P R . P R 5 I 2 R 5 1 2.0 A 2 2 1 5.0 V 2 5 20.0 W P r 5 I 2 r 5 1 2.0 A 2 2 1 1.0 V 2 5 4.0 W P E 5 VI 5 1 12.0 V 21 2.0 A 2 5 24.0 W. I 5 2.0 A. 12.0 V 2 I 1 1.0 V 1 5.0 V 2 5 0 P 5 I 2 R . P 5 VI . t 5 energy P 5 2.7 3 10 6 J 0.45 3 10 3 J / s 5 6.0 3 10 3 s 5 100 minutes VIt 5 1 12.6 V 21 60 A 21 3600 s 2 5 2.7 3 10 6 J. P 5 VI . Energy 5 Pt . 1 110 / 120 2 2 . P 5 V 2 R 5 1 110 V 2 2 26.7 V 5 453 W. 26.7 V , 1 0.540 kWh 21 7.2 cents / kWh 2 5 3.9 cents. Energy 5 1 0.540 kW 21 1 h 2 5 0.540 kWh. I 5 P V 5 540 W 120 V 5 4.50 A P 5 VI R 5 V 2 P 5 1 120 V 2 2 540 W 5 26.7 V P 5 V 2 R Energy 5 Pt . V 5 120 V. V 5 IR . P 5 I 2 R 5 V 2 R 5 VI . Energy 5 Pt 5 1 40 W 21 10 3 10 2 3 s 2 5 0.40 J. P 5 1 500 V 21 80 3 10 2 3 A 2 5 40 W. P 5 VI I 2 R P 5 V 2 R 5 1 120 V 2 2 1.0 3 10 3 V 5 14.4 W. I 5 V R 5 120 V 1.0 3 10 3 V 5 0.12 A. P 5 V 2 R . V 5 IR I 2 R . P 5 I 2 R . R 5 V 2 P 5 1 120 V 2 2 100 W 5 144 V R 5 V 2 P 5 1 120 V 2 2 60 W 5 240 V P 5 V 2 R Current, Resistance, and Direct-Current Circuits 19-7
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19.43. Set Up: A kWh is power of 1 kW for a time of 1 h. Solve: (a) In 3.0 yr the bulbs are on for compact bulb The energy used is The cost of this energy is One bulb will last longer than this. The bulb cost is $11.00, so the total cost is $19.08. incandescent The energy used is The cost of this energy is Six bulbs will be used during this time and the bulb cost will be $4.50. The total cost will be $39.54. (b) The compact bulb will save (c) Reflect: The initial cost of the bulb is much greater for the compact fluorescent bulb but the savings soon repay the cost of the bulb. The compact bulb should last for over six years, so over a 6 year period the savings per year will be even greater. 19.44. Set Up: The total resistance is the resistance of the person plus the internal resistance of the power supply. Solve: (a) (b) (c) The resistance of the power supply would need to be 19.45. Set Up: Solve: (a) (b) (c) 19.46. Set Up: For resistors in parallel, For resistors in series, These rules may have to be applied in several ways. Solve: (a) All four resistors are in parallel. so (b) The 75 and 25 resistors are in parallel. so their equivalent is The and resistors are in series, so (c) The 13 and 15 resistors are in series, so their equivalent is The 32 28 and 14 resistors are in parallel, so and their equivalent is The 72 and 45
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