When the switch is closed at t 0s, the capacitors in the figure are charged. a) What is the time constant of this circuit? 400 250 30UF 1OuF Reg = 4052 + 2582) + 150 + 1092 + 150 = 55.380 40UF Ceq = + = 20uF 30 F + 10 F 40 F TRC = RegCeg = 1.11ms 100 150 b) At what time has the current in the 150 resistor decayed to half the value it had immediately after the switch was closed? i (t) = ige-t/ReqCeq In ()= In e-"/Reg Cea] 1 = ige t/ReqCeq In (7 =- Reg Ceq = e-t/Req Ceq 1 = - In ( Req Cea = 0.77ms A 9V battery is added to the circuit as shown in the picture. If the capacitors are not charged, right after the battery is connected ... 409 250 c) What is the initial current through the 150 resistor? MMMM 30UF 10UF Capacitors are not charged: Q=0 - Vc=0 Loop rule (cw): 150 40UF AV = Vc+VR+ VB=0 0 + Regli - 9V = 0 109 150 - 1; = 9V/Reg = 9V/55.389 = 0.16A d) How long does it take for the current to drop to 1/e times its initial value? i (t) = ige t/ReqCeq In 4) = In e-/ReqCeq = ige-t/ReqCeq In (1) - In (e) = - t RegCea = e-t/ReqCeq t = RegCea = 1.11ms