Look at the circuit at Fig 1 At t 0 the switch S is closed

Look at the circuit at Fig. 1. At t = 0 the switch S is closed and the capacitor starts charging. What is the time constant for this circuit and how long will it take for the capacitor to charge to the half of the maximum voltage.

Solution

a)time constant = RC = 10*10^3 *5*10^-6 = .05 s

b) let t is time taken

V = V0(1-e^(-t/.05))

.5 =1- e^-(t/.05)

e^(-t/.05) =.5

t = .693 *.05 =.03465 s