solutions_chapter19

I 5 q 5 e c 5 1 500 v 21 6.00 3 10 2 6 f 2 5 3.00 3

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Unformatted text preview: i 5 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 q C 5 E i 5 i 5 E R 5 500 V 4500 V 5 0.111 A q 5 iR 5 E ; q 5 q 5 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. Solve: (a) The 20.0 pF, 30.0 pF and 40.0 pF capacitors are in series and their equivalent capacitance is given by After a long time the voltage across the 10.0 pF capacitor is 50.0 V. The charge on this capacitor is The voltage across is 50.0 V and This is the charge on each of the capacitors in series, so (b) Note that as it should. (c) (d) The 10.0 pF capacitor and are in parallel so their equivalent is Reflect: The charges on the capacitors start at zero and increase to their final values. The current starts at its maxi- mum value and then decays to zero. 19.70. Set Up: The time constant for the circuit is While charging, and In the discharging circuit both the charge and current decrease to of their initial values in time Solve: (a) gives and (b) (c) 19.71. Set Up: Equations (19.17) give i and q as functions of time during charging. During discharging, both i and q decay exponentially, as in Eq. (19.19) for i. t 5 t 5 188 m s i 5 I e 2 t / RC 5 I e 2 1 86.1 m s 2 / 1 188 m s 2 5 0.633 I t 5 2 RC 1 ln 3 1 2 1 / e 42 5 2 1 188 m s 21 2 0.458 2 5 1 86.1 m s e 2 t / RC 5 1 2 1 e 1 e 5 1 2 e 2 t / RC . q 5 Q final e q 5 Q final 1 1 2 e 2 t / RC 2 ....
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i 5 q 5 E C 5 1 500 V 21 6.00 3 10 2 6 F 2 5 3.00 3 10 2 3...

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