solutions_chapter19

# R a r a 5 263 3 10 2 8 v m r c 5 172 3 10 2 8 v m r 5

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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

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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 0 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: At Solve: (a) At and (b) so 19.92. Set Up: Zero current through the galvanometer means the current through N is also the current through M and the current through P is the same as the current through X . And it means that points b and c are at the same potential, so Solve: (a) The voltage between points a and d is so and Using these expressions in gives and (b) X 5 MP N 5 1 850.0 V 21 33.48 V 2 15.00 V 5 1897 V X 5 MP / N . NX 5 PM N 1 P 1 X 2 5 P 1 N 1 M 2 . E N 1 M N 5 E P 1 X P . I 1 N 5 I 2 P I 2 5 E P 1 X . I 1 5 E N 1 M E , I 1 N 5 I 2 P . I 2 I 1 C 5 t R 5 6.00 s 5.00 3 10 5 V 5 1.20 3 10 2 5 F. t 5 RC R 5 E i 5 400.0 V 0.800 3 10 2 3 A 5 5.00 3 10 5 V . E 5 iR t 5 0, t 5 RC . q 5 0. t 5 0, E 5 iR 1 q C . t 5 t 5 1 4600 V 21 0.800 3 10 2 9 F 2 5 3.68 3 10 2 6 s 5 3.68 m s. q 5 iRC 5 1 0.250 A 21 4600 V 21 0.800 3 10 2 9 F 2 5 9.20 3 10 2 7 C 5 0.920 m C. t 5 0, t 5 t 5 RC . 1 / e iR 5 q C .
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