Question

Example E 42 v (a) Right after S is closed, find the current through each resistor. (b) 7 seconds after the switch has been closed, find the charge on the capacitor. (c) Switch is reopened. After how much time does the charge reduce by 50%?

Answer 1

The working principle for this question is to know the charging and discharging transients of RC circuit with d.c. input.

The solution is illustrated below.

@) Right after the switch s ů closed, the capacitor acts as a short circuit, so the circuit can be redrawn as, 8302 625 E=424 - 8 + 2 = 1032 8 + 6 X 3 6+ 3 I : 42-4.2 A 30, Ig = 4:2A Using current division in 62832 Kesistous, 16 = 3 * 4.2 = 1.4 A +3 Iz = 6 * 4.2 =28A nd the charge on capacitor after 7 seconds, we will find the Rin (Thevenin resistance) seen at terminals of c. To find Rin, redraw ckt as, . Rth = 3+8x6 = 6.43.2 8+6 So, time constant, I = Rth c = 6.43 X 4 X10-6 25.7 * 1026 sec. Now Q = EC(1-e-t/Rth C) = 42 X 4X10-6(1--7/25-7x10-6). = 42 x 4x10-6 (1-0) Q = 168re

(6) Fore discharging, the equation is given as, $ = 8c 9 -t/RT C where ec is the initial charge. For 50% discharging, Q-lee-t/RTC EC 2 → In ( ) = at Irinei > -0.693 = -t 25.781006 st 0.693825.7 x 10-6 sec t : 17.81 Usec.