solutions_chapter19

# 1 150 v 2 i 2 1 700 v 2 2 1 200 a 21 500 v 2 5 0 i 1

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1 15.0 V 2 I 2 1 7.00 V 2 2 1 2.00 A 21 5.00 V 2 5 0. I 1 5 Å 20.0 W 5.00 V 5 2.00 A. P 5 I 2 R 15.0 V E 7.00 V 2.00 V 5.00 V I 1 1 1 2 + + I 2 V Current, Resistance, and Direct-Current Circuits 19-15

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Solve: (a) gives The loop rule applied to loop (1) gives: (b) The loop rule applied to loop (2) gives: The emf is 46.1 V and the polarity of the battery is opposite to what is shown in the figure in the problem; the terminal is adjacent to the resistor. 19.63. Set Up: The time constant is Solve: 19.64. Set Up: Solve: RC has units of 19.65. Set Up: The loop rule gives Solve: (a) The initial value of q is (b) Reflect: The larger the value of R, the smaller the initial current and the longer it takes the charge to decrease to of its initial value. 19.66. Set Up: The loop rule gives Just after the circuit is completed, After a long time the capacitor is fully charged and Solve: (a) so the voltage across the capacitor is zero. (b) so the voltage drop across the resistor is 500 V. (c) (d) (e) (a) so and the voltage across the capacitor is 500 V. (b) so the voltage across the resistor is zero. (c) (d) 19.67. Set Up: The time constant is Kirchhoff’s loop rule applied to the charging circuit gives at all times. The current and the charge on the capacitor as functions of time are given in Equations (19.17). Solve: (a) (b) when so Q final 5 E C 5 1 12.0 V 21 6.00 m F 2 5 72.0 m C E 5 Q final C . i S 0, q 5 Q final t 5 RC 5 1 75.0 V 21 6.00 3 10 2 6 F 2 5 4.5 3 10 2 4 s 5 450 m s E 5 iR 1 q C t 5 RC . i 5 0 q 5 E C 5 1 500 V 21 6.00 3 10 2 6 F 2 5 3.00 3 10 2 3 C 5 3.00 mC. i 5 0 q C 5 E i 5 0 i 5 E R 5 500 V 4500 V 5 0.111 A q 5 0 iR 5 E ; q 5 0 q 5 0 i 5 0. q 5 0. E 5 iR 1 q C . 1 / e t 5 RC 5 1 8.00 3 10 5 V 21 0.250 3 10 2 9 F 2 5 2.00 3 10 2 4 s 5 0.200 ms i 5 q RC 5 60.0 3 10 2 9 C 1 8.00 3 10 5 V 21 0.250 3 10 2 9 F 2 5 3.00 3 10 2 4 A 5 0.300 mA. q 5 60.0 nC. q C 5 iR . t 5 RC . V # F 5 1 V A 21 C V 2 5 C A 5 C C / s 5 s. 1 F 5 1 C / V. 1 V 5 1 V / A. C 5 t R 5 2.00 s 500.0 V 5 4.00 3 10 2 3 F 5 4.00 mF t 5 RC . 13 V 1 E 5 25 V 2 18 V 2 53.1 V 5 2 46.1 V. 2 1 2.0 A 21 3.0 V 1 6.0 V 2 1 25 V 2 1 1.77 A 21 17 V 2 2 E 2 1 1.77 A 21 13 V 2 5 0. I 3 5 I 1 2 I 2 5 2.0 A 2 0.233 A 5 1.77 A. I 2 5 25 V 2 18 V 30.0 V 5 0.233 A. 2 1 2.0 A 21 3.0 V 2 2 1 2.0 A 21 6.0 V 2 1 25 V 2 I 2 1 10.0 V 1 19.0 V 1 1.0 V 2 5 0. I 1 5 Å P R 5 Å 24 J / s 6.0 V 5 2.0 A. P 5 I 2 R 19-16 Chapter 19
(c) The current has its largest value when so The graphs are of the same form as in Figures 19.33a and b in the textbook. Reflect: The maximum charge on the capacitor does not depend on the resistance. The initial current is the same as the current produced when the resistor alone is connected to the battery. 19.68. Set Up: The time constant is so and Solve: (a) gives Three time constants are required. (b) so 19.69. Set Up: After a long time the current has dropped to zero, there is no potential drop across the resistor and the full battery voltage is applied to the capacitor network. Just after the switch is closed the charge and voltage across each capacitor is zero and the battery voltage equals the voltage drop across the resistor. The time constant is where C is the equivalent capacitance of the network.

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