Consider the RC circuit shown below: What is the characteristic time constant of the circuit? What is the maximum charge on the capacitor? How long does it take after the switch is closed to charge the capacitor to within 63% of its maximum charge Solution a) Time constant is given by, T = R*C = 100*3.2*10^-6 = 3.2*10^-4 s b\') maximum charge, Qo = C*Vmax = 3.2*10^-6*15 = 4.8*10^-5 C c) The chrarge on the capacitor is given by : Q = Qo*(1 - e^(-t/RC)) Now, Q = 63 percent of Qo = 0.63*Qo So, 0.63*Qo = Qo*(1 - e^(-t/RC)) = e^(-t/RC) = 0.37 So, -t/RC = ln(0.37) = -0.994 So, t = RC*0.994 = 3.2*10^-4*0.994 = 3.18*10^-4 s <--------answer.