solutions_chapter19

19.33 set up for a resistor and solve(a(b 19.34 set

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Unformatted text preview: 19.33. Set Up: For a resistor, and Solve: (a) (b) 19.34. Set Up: For a resistor, and Solve: (a) (b) 19.35. Set Up: Solve: (a) (b) 19.36. Set Up: Power is energy per unit time. The power delivered by a voltage source is Solve: (a) (b) Energy 5 Pt 5 1 300 W 21 3.0 3 10 2 3 s 2 5 0.90 J P 5 1 25 V 21 12 A 2 5 300 W. P 5 V ab I . Energy 5 1 1.17 W 21 1.5 h 21 3600 s / h 2 5 6320 J P 5 1 9.0 V 21 0.13 A 2 5 1.17 W Energy 5 Pt . P 5 VI . P 5 V 2 R 5 1 120 V 2 2 9.0 3 10 3 V 5 1.6 W V 5 IR 5 1 0.0183 A 21 15 3 10 3 V 2 5 275 V. I 5 Å P R 5 Å 5.0 W 15 3 10 3 V 5 0.0183 A. V 5 IR . P 5 I 2 R 5 V 2 R 5 VI R 5 V I 5 15.0 V 21.8 A 5 0.688 V I 5 P V 5 327 W 15.0 V 5 21.8 A V 5 IR . P 5 VI V ab V ab V ab 5 E ; R S ` . I S R S I S ` I 5 E R . r 5 V ab R E (a) R I (b) V ab R E (c) V ab V ab S E . R S ` , V ab 5 0. R 5 0, V ab 5 IR 5 E R R 1 r 5 E 1 1 r / R . r 5 E I 5 5.00 V 2.5 A 5 2.00 V R 5 0, R 5 0. I 5 2.5 A V ab 5 E 2 Ir . I 5 E R 1 r . I 5 24.0 V 6.00 V 5 4.00 A, R 1 r 5 6.00 V . R 5 V ab I 5 21.2 V 4.00 A 5 5.30 V . V ab 2 IR 5 r 5 E 2 V ab I 5 24.0 V 2 21.2 V 4.00 A 5 0.700 V . V ab 5 E 2 Ir V ab 5 E 2 Ir . V ab 1 2 19-6 Chapter 19 19.37. 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 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 ....
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19.33 Set Up For a resistor and Solve(a(b 19.34 Set Up For...

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