Solve a the 200 pf 300 pf and 400 pf capacitors are

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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 0 e 2 t / RC 5 I 0 e 2 1 86.1 m s 2 / 1 188 m s 2 5 0.633 I 0 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 . t 5 RC 5 1 125 V 21 1.50 m F 2 5 188 m s t 5 t . 1 / e i 5 I 0 e 2 t / RC . q 5 Q final 1 1 2 e 2 t / RC 2 t 5 RC . t 5 RC 5 1 20.0 V 21 19.2 pF 2 5 384 ps C eq 5 10.0 pF 1 9.23 pF 5 19.2 pF. C s I 5 E R 5 50.0 V 20.0 V 5 2.50 A V 20 1 V 30 1 V 40 5 50.0 V, V 40 5 q 40 C 40 5 461 pC 40.0 pF 5 11.5 V. V 30 5 q 30 C 30 5 461 pC 30.0 pF 5 15.4 V. V 20 5 q 20 C 20 5 461 pC 20.0 pF 5 23.1 V. V 10 5 50.0 V. q 20 5 q 30 5 q 40 5 461 pC. 1 50.0 V 2 5 461 pC. q 5 5 C 5 V 5 5 1 9.23 pF 2 C s q 10 5 C 10 V 10 5 1 10.0 pF 21 50.0 V 2 5 500 pC. C s 5 9.23 m F. 1 C s 5 1 20.0 pF 1 1 30.0 pF 1 1 40.0 pF . C s i 5 I 0 e 2 t / RC . t 5 RC , q 5 C / V . i 5 0.05 I 0 . e 2 t / t 5 0.05 t t 5 3.0. 2 t t 5 ln 1 0.05 2 . e 2 t / t 5 0.05. 0.95 5 1 2 e 2 t / RC . q 5 0.95 Q final i 5 I 0 e 2 t / RC . q 5 Q final 1 1 2 e 2 t / RC 2 t 5 RC , I 0 5 E R 5 12.0 V 75.0 V 5 0.160 A E 5 I 0 R . q 5 0, I 0 Current, Resistance, and Direct-Current Circuits 19-17
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Solve: (a) The graphs are the same as in Figure 19.33a and b in the textbook and are given in Figure 19.71a. (b) The graphs of i and q versus t for a discharging capacitor are sketched in Figure 19.71b. Figure 19.71 19.72. Set Up: There is a single current path so the current is the same at all points in the circuit. Assume the current is counterclockwise. Solve: (a) Apply the loop rule, traveling around the circuit in the direction of the current: Our calculated I is positive so I is counterclockwise, as we assumed. (b) 19.73. Set Up: The power for a resistor R is The power for an emf is Solve: (a) The loop rule for a counterclockwise direction gives (b) The net power output of the battery is (c) In the emf of this battery electrical energy is being consumed at the rate and in the internal resistance of the battery electrical energy is being consumed at the rate (d) Chemical energy is being converted to electrical energy at a rate of in the 12.0 V battery and electrical energy is being converted to chemical energy at a rate of in the 4.0 V battery. The net rate of conversion of chemical energy to electrical energy is The overall rate of dissipation of electrical energy (conversion to heat) is The two rates are equal, as required by conservation of energy.
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