solutions_chapter19

Solve end-to-end for each cable parallel now for each

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Unformatted text preview: Solve: end-to-end: for each cable. parallel: Now for each cable. L is doubled and A is halved compared to the other configuration, so and and The least resistance is for the cables in parallel. R eq 5 0.416 V . 1 R eq 5 1 R c 1 1 R a 5 1 0.688 V 1 1 1.052 V R a 5 4 1 0.263 V 2 5 1.052 V . R c 5 4 1 0.172 V 2 5 0.688 V L 5 1.00 3 10 3 m R eq 5 0.172 V 1 0.263 V 5 0.435 V R a 5 r a L A 5 1 2.63 3 10 2 8 V # m 21 0.50 3 10 3 m 2 0.500 3 10 2 4 m 2 5 0.263 V R c 5 r c L A 5 1 1.72 3 10 2 8 V # m 21 0.50 3 10 3 m 2 0.500 3 10 2 4 m 2 5 0.172 V L 5 0.50 3 10 3 m 1 R eq 5 1 R c 1 1 R a . R eq R eq 5 R c 1 R a . r a 5 2.63 3 10 2 8 V # m. r c 5 1.72 3 10 2 8 V # m R 5 r L A . 1 115.2 V 2 / 2 5 57.6 V. 6.0 V 19.2 V 1 96.0 V 5 115.2 V. 20.0 V 4.0 V 4.0 V 1 2.4 A 21 8.0 V 2 5 19.2 V. 8.0 V 16.0 V 16.0 V 8.0 V 18.0 V 9.0 V 12.0 V 6.0 V 6.0 V 24.0 V 8.0 V 4.0 V 20.0 V 6.0 V 20.0 V 2.4 A x x y x y x y y a ( a ) ( b ) ( c ) ( d ) a 8.0 V . 19-22 Chapter 19 Reflect: The parallel combination has less equivalent resistance even though both cables contain the same volume of each metal. 19.88. Set Up: Because of the polarity of each emf, the current in the resistor must be in the direction shown in Figure 19.88. Let I be the current in the 24.0 V battery. Figure 19.88 Solve: The loop rule applied to loop (1) gives: The junction rule then says that the current in the middle branch is 2.00 A, as shown in Figure 19.88b. The loop rule applied to loop (2) gives: and 19.89. Set Up: There is no current in the middle branch because there is not a complete conducting path for that branch. There is only a single current in the circuit, as shown in Figure 19.89. To find start at point b and travel to point a. Figure 19.89 Solve: Going from a to b through the 12.0 V battery gives Or, going from a to b through the 8.0 V battery gives and V a 2 V b 5 2 2.0 V 1 1 0.444 A 21 5.00 V 2 5 1 0.22 V. V a 2 I 1 2.00 V 2 2 I 1 1.00 V 2 2 8.0 V 2 I 1 2.00 V 2 1 10.0 V 5 V b V a 2 V b 5 12.0 V 2 10.0 V 2 1 0.444 A 21 4.00 V 2 5 1 0.22 V. V a 1 I 1 2.00 V 2 1 I 1 1.00 V 2 2 12.0 V 1 I 1 1.00 V 2 1 10.0 V 5 V b . I 5 0.444 A. 12.0 V 2 8.0 V 2 I 1 2.00 V 1 2.00 V 1 1.00 V 1 2.00 V 1 1.00 V 1 1.00 V 2 5 0. I a b I I I 12.0 V 10.0 V 1.00 V 8.0 V 1.00 V 1.00 V 3.00 V 2.00 V 2.00 V 2.00 V 1.00 V + + + V ab 5 V a 2 V b , E 5 8.6 V. 1 E 2 1 1.80 A 21 7.00 V 2 1 1 2.00 A 21 2.00 V 2 5 I 5 3.80 A. 1 24.0 V 2 1 1.80 A 21 7.00 V 2 2 I 1 3.00 V 2 5 0. I 24.0 V 24.0 V 3.80 A 2.00 A 1.8 A 1.80 A 7.00 V 2.00 V 3.00 V 7.00 V 2.00 V 3.00 V 1 1 2 1 2 2 ( a ) ( b ) + + + + E E 7.00 V Current, Resistance, and Direct-Current Circuits 19-23 The voltmeter reads 0.22 V. is positive, so point a is at higher potential. Reflect: Since there is no current in the middle branch, the two resistances in that branch could be removed without affecting the potential difference The 10.0 V emf doesn’t affect the current but does affect 19.90. Set Up: The charge is reduced to of its maximum value when Solve: (a) At (b) 19.91. Set Up:19....
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Solve end-to-end for each cable parallel Now for each cable...

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