solutions_chapter19

# E f since and v is the same for all three resistors

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(e) (f) Since and V is the same for all three resistors, the resistor with the smallest R dissipates the greatest power. 19.51. Set Up: Let For a series network, the current is the same in each resistor and the sum of voltages for each resistor equals the battery voltage. The equivalent resistance is Solve: (a) (b) (c) the same as for each resistor. (d) Note that (e) (f) Since and the current is the same for each resistor, the resistor with the greatest R dissipates the greatest power. Reflect: When resistors are connected in parallel, the resistor with the smallest R dissipates the greatest power. 19.52. Set Up: For resistors in parallel, the voltages are the same and the currents add. so For resistors in series, the currents are the same and the voltages add. R eq 5 R 1 1 R 2 . R eq 5 R 1 R 2 R 1 1 R 2 , 1 R eq 5 1 R 1 1 1 R 2 P 5 I 2 R 1 3.18 A 2 2 1 4.80 V 2 5 48.5 W. P 3 5 I 2 R 3 5 P 2 5 I 2 R 2 5 1 3.18 A 2 2 1 2.40 V 2 5 24.3 W. P 1 5 I 2 R 1 5 1 3.18 A 2 2 1 1.60 V 2 5 16.2 W. V 1 1 V 2 1 V 3 5 28.0 V. 1 3.18 A 21 4.80 V 2 5 15.3 V. V 3 5 IR 3 5 V 2 5 IR 2 5 1 3.18 A 21 2.40 V 2 5 7.63 V. V 1 5 IR 1 5 1 3.18 A 21 1.60 V 2 5 5.09 V. I 5 3.18 A, I 5 V R eq 5 28.0 V 8.80 V 5 3.18 A R eq 5 1.60 V 1 2.40 V 1 4.80 V 5 8.80 V P 5 I 2 R . R 1 1 R 2 1 R 3 . R eq 5 R 3 5 4.80 V . R 2 5 2.40 V , R 1 5 1.60 V , P 5 V 2 R P 3 5 V 2 R 3 5 1 28.0 V 2 2 4.80 V 5 163 W. P 2 5 V 2 R 2 5 1 28.0 V 2 2 2.40 V 5 327 W. P 1 5 V 2 R 1 5 1 28.0 V 2 2 1.60 V 5 490 W. I 1 1 I 2 1 I 3 . I 5 V R eq 5 28.0 V 0.80 V 5 35.0 A. I 3 5 V R 3 5 28.0 V 4.80 V 5 5.8 A. I 2 5 V R 2 5 28.0 V 2.40 V 5 11.7 A. I 1 5 V R 1 5 28.0 V 1.60 V 5 17.5 A. R eq 5 0.80 V . 1 R eq 5 1 1.60 V 1 1 2.40 V 1 1 4.80 V P 5 V 2 R . 1 R eq 5 1 R 1 1 1 R 2 1 1 R 3 . R eq R 3 5 4.80 V . R 2 5 2.40 V , R 1 5 1.60 V , I 2 5 V R 2 5 120 V 90.0 V 5 1.33 A. 90.0 V I 1 5 V R 1 5 120 V 40.0 V 5 3.00 A. 40.0 V I 5 V R eq 5 120 V 27.7 V 5 4.33 A R eq 5 1 40.0 V 21 90.0 V 2 40.0 V 1 90.0 V 5 27.7 V 19-10 Chapter 19

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Solve: The rules for combining resistors in series and parallel lead to the sequences of equivalent circuits shown in Figure 19.52. In Figure 19.52c, This is the current through each of the resistors in Figure 19.52b. Figure 19.52 Note that is the voltage across and across so and is the voltage across and across so and 19.53. Set Up: For resistors in parallel, the voltages are the same and the currents add. so For resistors in series, the currents are the same and the voltages add. Solve: The rules for combining resistors in series and parallel lead to the sequence of equivalent circuits shown in Figure 19.53. In Figure 19.53c, In Figure 19.53b, the voltage across each resistor is 48.0 V, so and Note that Then in Figure 19.53a, and Figure 19.53 19.54. Set Up: We can use the power rating in to find the resistance of a bulb. Then we can find the voltage across each bulb when they are connected in series and in parallel. Solve: (a) The voltage across each bulb is (b) In parallel, the voltage across each bulb is 120 V and the power dissipated by each is 100 W. 19.55. Set Up: The power rating of each lightbulb is the power consumed if a voltage of 120 V is applied across the bulb. In series, the individual voltages are less than the line voltage, and each bulb consumes less than its rated P 5 V 2 R 5 1 60 V 2 2 144 V 5 25 W V 1 5 IR 5 60 V. I 5 V R eq 5 120 V 288 V 5 0.417 A. R eq 5 2 R 5 288 V .
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