# Which resistors are taken into account when evaluating the discharge of a capacitor

I'm trying to evaluate the time it takes the following mono-stable circuit to change the voltage of the capacitor from 1.25V to 2.5V after the Vin single pulse enters, and evaluate the recovery time of the circuit (given the voltage on the capacitor should be 5mV from zero for full recovery).

1. In the official solution, the tau is calculated as $\tau = (R1+R2)C1$ while i thought it should be only $R1$.
Why are both resistors taken into account? Isn't the charging voltage Vout charging only the upper side through R1?
What happens in the discharging process? Again is it discharged through both of them?

2. For the recovery time, the official solution states that after Q changes back from 1 to 0, the voltage in Vc drops from 2.5V to 1.25V (VDD=5V). I understand the the voltage on a capacitor should be continuous, but how did they calculate it?
Is it because the change in the voltage for R2 is 1.25V, so that was the 'drop' in voltage there?
Somehow I'm missing the point of calculating the voltage drop on the capacitors.

Many thanks!

• R1, C1, and R2 form a series RC circuit (assume no current into R terminal so open circuit) with the time constant in the solution. You can find more specific info about how it's charging/discharging here. You are basically just combining those two resistors in series. – Avid Pro Tool Jun 1 '18 at 6:39
• Totally, simple RC circuit, don't know how i missed that. I've edited my question, Many thanks! – DsCpp Jun 1 '18 at 6:47

$Z_{all} = Z_{r1} + Z_{c1} + Z_{r2}$