solutions_chapter19

V v 5 ir eq 5 1 0.316 a 21 35.0 v 2 5 11.1 v r eq 5

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Unformatted text preview: V V 5 IR eq 5 1 0.316 A 21 35.0 V 2 5 11.1 V. R eq 5 5.00 V 1 10.0 V 1 20.0 V 5 35.0 V . I 5 Å P R 20 5 Å 2.00 W 20.0 V 5 0.316 A. P 5 I 2 R R eq 5 R 1 1 R 2 1 R 3 . P 100 5 400 W. P 75 5 300 W P 60 5 240 W, V 5 240 V V 2 P 100 5 100 W. P 75 5 75 W P 60 5 60 W, V 5 120 V P 100 5 1 0.208 A 2 2 1 144 V 2 5 6.23 W. P 75 5 1 0.208 A 2 2 1 192 V 2 5 8.31 W P 60 5 I 2 R 60 5 1 0.208 A 2 2 1 240 V 2 5 10.4 W, I 5 V R eq 5 120 V 576 V 5 0.208 A. R eq 5 R 60 1 R 75 1 R 100 5 576 V . R 100 5 V 2 P 100 5 1 120 V 2 2 100 W 5 144 V . R 75 5 V 2 P 75 5 1 120 V 2 2 75 W 5 192 V R 60 5 V 2 P 60 5 1 120 V 2 2 60 W 5 240 V , P 5 V 2 R 5 I 2 R . 19-12 Chapter 19 19.58. Set Up: The circuit diagram is given in Figure 19.58. The junction rule has been used to find the magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown currents. Figure 19.58 Solve: The loop rule applied to loop (1) gives: The loop rule applied to loop (2) gives: 19.59. Set Up: The circuit is sketched in Figure 19.59. When we go around loop (1) in the direction shown there is a potential rise across the 200.0 V battery, so there must be a drop across R and the current in R must be in the direction shown in the figure. Similar analysis of loops (2) and (3) tell us that currents and must be in the directions shown. The junction rule has been used to label the currents in all the other branches of the circuit. Figure 19.59 Solve: (a) Apply the loop rule to loop (1): (b) Loop (2): I 2 5 160.0 V 40.0 V 5 4.00 A 1 160.0 V 2 I 2 1 40.0 V 2 5 0. R 5 1 200.0 V I 1 5 1 200.0 V 10.0 A 5 20.0 V 1 200.0 V 2 I 1 R 5 0. 40.0 V 20.0 V 200.0 V 160 V 1 1 2 1 4 2 1 2 2 1 3 2 R I 1 I 5 I 2 I 1 1 I 2 I 2 1 I 5 + + I 5 I 2 E 2 5 20.0 V 2 1.00 V 2 2.00 V 2 4.00 V 2 6.00 V 5 7.0 V. 1 20.0 V 2 1 1.00 A 21 1.00 V 2 2 1 2.00 A 21 1.00 V 2 2 E 2 2 1 2.00 A 21 2.00 V 2 2 1 1.00 A 21 6.00 V 2 5 E 1 5 20.0 V 2 1.00 V 1 4.00 V 1 1.00 V 2 6.00 V 5 18.0 V. 1 20.0 V 2 1 1.00 A 21 1.00 V 2 1 1 1.00 A 21 4.00 V 2 1 1 1.00 A 21 1.00 V 2 2 E 1 2 1 1.00 A 21 6.00 V 2 5 1.00 V 1.00 V 1.00 V 2.00 V 4.00 V 1.00 A 1.00 A 6.00 V 20.0 V E 1 E 2 2.00 A 1 1 2 + + + Current, Resistance, and Direct-Current Circuits 19-13 Loop (3): reads reads reads reads Reflect: The sum of potential changes around loop (4) is The loop rule is satisfied for loop (4) and this is a good check of our calculations. 19.60. Set Up: The circuit diagram is given in Figure 19.60. The junction rule has been used to find the magnitude and direction of the current in the upper branch of the circuit. There are no remaining unknown currents. Figure 19.60 Solve: (a) The junction rule gives that the current in R is 2.00 A to the left. (b) The loop rule applied to loop (1) gives: (c) The loop rule applied to loop (2) gives: E 5 24.0 V 1 18.0 V 5 42.0 V. 2 1 4.00 A 21 6.00 V 2 1 E 2 1 6.00 A 21 3.00 V 2 5 0....
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V V 5 IR eq 5 1 0.316 A 21 35.0 V 2 5 11.1 V R eq 5 5.00 V...

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