Consider an RC-circuit with R = 2 kOhm\'s, C = 500 uF, and E = 12 V and a switch
ID: 1270651 • Letter: C
Question
Consider an RC-circuit with R = 2 kOhm's, C = 500 uF, and E = 12 V and a switch. At t = 0 the capacitor is empty and the switch is closed so the charging begins.
(a) What is the current through the resistor just after the switch is closed?
(b) What will be the current in the circuit after a long time has passed?
(c) Calculate the time constant of the circuit.
(d) What will be the charge on the capacitor after a long time has passed?
(e) How long does it take the capacitor get charged to 75% of its maximum capacity?
Explanation / Answer
a) Imax = V/R = 12/2000 = 0.006 A = 6 mA
b) I = 0
c) T = R*C = 500*10^-6*2000 = 1 s
d) Qmax = c*E = 500*12 = 6000 micro C or 6 mC
e) Q = Qmax*(1-e^(-t/T))
0.75*Qmax = Qmax(1 - e^(-t/T))
0.75 = 1 - e^(-t/T)
e^(-t/T) = 1 - 0.75
t = -T*ln(0.25)
= -1*ln(0.25)
= 1.386 s