Question

5. A 250 Q resistor is in series with 200 pF .capacitor. These are connected across a 20 V battery. The switch is turned on, and the capacitor begins to fill. After the capacitor is completely filled, the battery is taken out of the circuit and the leads to the battery are shorted together. (a) (6 pts) What is the time constant, the maximum current, and the maximum charge on the capacitor? (b) (6 pts) When t = 0.04 seconds, find the current through the resistor, the voltage on the capacitor, the voltage on the resistor, and the charge on the capacitor. (C) (8 pts) When the current through the resistor is 0.02A, find t.

D. When the current is .02 A find the voltage and fhe charge of the capacitor

Answer 1

1- (A) The time time constant is, T= RC = (250)(200x106) = 0.05 s =5x10-2 s The maximum current is, V R 20 250 = 0.083 A The maximum charge on capacitor is, Q=CV =(200x10^)(20) = 0.004 C =4x10+ c (b) The charging RC circuit is, Q=21-e RC = (4x104) 20)(1-2 * = 2.20x10+ c The voltage across the capacitor is, V-8 с 2.20 x 10+ 200x107 =1.1 V The voltage across the capacitor is, V=V-V. = 20-1.1 = 18.9 V

1. 
   I= dQ dt = le RC 0.04 =(0.083) e 0.05 = 0.037 A (c) The current trought voltage is, ਕੋਟ I=1, 0.02 = 0.083e 0.05 0.24 = e 0.05 0.05 t In(0.24)= 0.05 t -1.423 = 0.05 t = 0.071 s (d) The charge of the capacitor is, Q=2 1-e RC 0.071 0.05 = (4x10+) 1-2 = 3.03x10+ 0 The voltage across the capacitor is, C 3.03x10-3 200 x 100 = 15.15 V