Question

chapter 8 RC electricity

8.1.2 Prelab Exercise 1 1. If the experimental set up in Fig. 8.2 has a 12.0V battery and a 10pF capacitor, determine the time it would take for the fully charged capacitor to decrease the voltage from 12.0V to 8.0V (switch will be in the discharging position A-C). 2. The two resistors, 1k2 and 10M2, in this circuit (Fig.8.2) are different by many orders of magnitude. Explain why? 81

Answer 1

V_0 = 12.0V C = 10 mu F = 10 times 10^-6 F V = 8.0V and R = 10 M ohm = 10 times 10^+ 6 ohm In discharging situation, voltage across the capacitor is defined as V = V_0 e^-t/RC \r\n 8V = 12V e^-t/10 times 10^+ 6 ohm times 10 times 10^-6 F 8/12 = e^-t/100 sec -t/100 sec = ln(2/3) t = -100 sec ln(0.67) t = 40.5 sec (b) Larger resistance is taken be curve, It take large time to discharge the charge from the capacitor due to this larger resistance. So, current decrease slowly. and we get proper time to observe it