solutions_chapter19

# Resistors are in parallel so and their equivalent is

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resistors are in parallel so and their equivalent is This gives a network with 19 7.2 27.7 and 24 in series. The equivalent resistance is (d) The 12 13 and 14 resistors are in series and their equivalent is and the resistor are in parallel and their equivalent is given by so Then and the resistor are in series and R eq 5 14.1 V 1 11 V 5 25.1 V . 11 V R p R p 5 14.1 V . 1 R p 5 1 39 V 1 1 22 V , R p 22 V R s R s 5 12 V 1 13 V 1 14 V 5 39 V . V V V , R eq 5 19 V 1 7.2 V 1 27.7 V 1 24 V 5 77.9 V . V V V , V , R p2 5 27.7 V . 1 R p2 5 1 72 V 1 1 45 V V V R p1 5 7.2 V . 1 R p1 5 1 32 V 1 1 28 V 1 1 14 V V V , V , R s 5 13 V 1 15 V 5 28 V . V V R eq 5 9.0 V 1 18.8 V 1 18 V 5 45.8 V . 18 V 18.8 V , 9.0 V , R p 5 18.8 V . 1 R p 5 1 75 V 1 1 25 V V V R eq 5 2.9 V . 1 R eq 5 1 25 V 1 1 12 V 1 1 5.0 V 1 1 45 V R eq 5 R 1 1 R 2 1 c 1 R eq 5 1 R 1 1 1 R 2 1 c P 5 I 2 R 5 1 100 3 10 2 3 A 2 2 1 1.0 3 10 3 V 2 5 10 W V 5 IR 5 1 100 3 10 2 3 A 21 1.0 3 10 3 V 2 5 100 V R 5 1 5.0 V # m 21 1.6 m 2 p 1 0.050 m 2 2 5 1.0 3 10 3 V P 5 I 2 R . V 5 IR . R 5 r L A . 14 3 10 6 V 2 10 3 10 3 V 5 14 3 10 6 V 5 14 M V . R tot 5 V I 5 14 3 10 3 V 1.00 3 10 2 3 A 5 14 3 10 6 V . P 5 I 2 R 5 1 1.17 A 2 2 1 10 3 10 3 V 2 5 1.37 3 10 4 J 5 13.7 kJ I 5 V R tot 5 14 3 10 3 V 10 3 10 3 V 1 2000 V 5 1.17 A P 5 I 2 R . V 5 IR . R 5 V 2 P 5 1 120 V 2 2 23 W 5 626 V \$39.54 2 \$19.08 5 \$20.46. 1 \$0.080 / kWh 21 438 kWh 2 5 \$35.04. 1 100 W 21 4.38 3 10 3 h 2 5 4.38 3 10 5 Wh 5 438 kWh. 1 101 kWh 2 5 \$8.08. 1 \$0.080 / kWh 2 1 23 W 21 4.38 3 10 3 h 2 5 1.01 3 10 5 Wh 5 101 kWh. 1 3.0 yr 21 365.24 days / yr 21 4.0 h / day 2 5 4.38 3 10 3 h P 5 V 2 R . 19-8 Chapter 19

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19.47. Set Up: A network of N of the resistors in series has resistance and a network of N of the resistors in parallel has resistance Solve: (a) A parallel combination of two resistors in series with three others (Figure 19.47a). (b) Ten in parallel. (c) Three in parallel. (d) Two in parallel in series with four in parallel (Figure 19.47b). Figure 19.47 Reflect: There are other networks that also have the required resistance. An important additional consideration is the power dissipated by each resistor, whether the power dissipated by any resistor in the network exceeds the maximum power rating of the resistor. 19.48. Set Up: so the resistance is proportional to the length of the bar. The bars can be represented by the circuit shown in Figure 19.48. Figure 19.48 Solve: The two resistors in parallel have an equivalent resistance Then the three resistors in series have equivalent resistance 19.49. Set Up: For a parallel connection the full line voltage appears across each resistor. The equivalent resistance is and R eq 5 R 1 R 2 R 1 1 R 2 . 1 R eq 5 1 R 1 1 1 R 2 R 5 2 R 3 1 R 6 1 2 R 3 5 9 R 6 5 3 R 2 . 1 R / 3 2 2 5 R 6 . R / 3 R / 3 2 R / 3 2 R / 3 R 5 r L A 10.0 V 10.0 V 10.0 V 10.0 V 10.0 V ( a ) 10.0 V 10.0 V 10.0 V 10.0 V 10.0 V 10.0 V ( b ) 1 10.0 V 2 / N . N 1 10.0 V 2 Current, Resistance, and Direct-Current Circuits 19-9
Solve: (a) (b) (c) For the resistor, For the resistor, Reflect: If either resistor is removed, the voltage and current for the other resistor is unchanged. 19.50. Set Up: Let For a parallel network the full battery voltage appears across each resistor. The equivalent resistance is given by Solve: (a) and (b) (c) This is also equal to (d) The voltage is 28.0 V across each resistor.

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