The switch is closed and then reopened after 1.0 s.
What is the voltage across the capacitor 1.0 s after the switch is reopened?
Immediately on switch closure, the circuit behaves like the one shown at the left.
This may be determined by determining the Thévenin generator for the circuit above.
Killing the source, the 5k resistor is found to be in parallel with the 20k resistor for an equivalent parallel resistance of 4k. This is in series with the remaining 3k resistor for a Thévenin resistance of 7k. The open circuit voltage will be the voltage across the 20k resistor; there is no drop across the 3k resistor because no current flows through it in the open circuit condition. The current through the 20k resistor is 9 V / 25k = 0.36 mA under the open circuit condition. Thus the Thévenin voltage is (0.36 mA)(20k) = 7.2 V. From the Thévenin generator, we find that the time constant for charging is R_{TH}C = (7k)(100 mF) = 0.7 s. Finally, the voltage across the capacitor after 1.0 s is 7.20 V [1 exp(1.0/0.7)] = 5.47 V. 

When the switch is reopened, the circuit behaves as the circuit at the left.
The time constant of discharging circuit (23k)(100 mF) = 2.3 s.
A 1.0 s discharge from 5.47 yields a voltage of 5.47 V exp(1.0/2.3) =3.54 V. 