# Calculating an First Order RC Circuit Time Constant

I apologize for the basic nature of this question, I am trying to figure out how to calculate the RC Time Constant of this circuit I know you have to approach the circuit from the capacitor to find its time constant. I think the Req would be R1 + R2 + (R3||R4) because when finding Req for the Capacitor you just open it, but am very confused and not entirely sure.

Any help would be huge!

• It seems to me that, by inspection, R2 must be in parallel with the equivalent resistance of the other three resistors. Think about it, if all the other resistors were removed, C1 could still discharge through R2 and thus, R2 cannot be in series with any of the other resistors. Nov 1, 2017 at 3:12
• In other words, to calculate the time constant, you have to treat the voltage source as a short circuit. You end up with R2 || (R1 + (R3||R4)) from the point of view of the capacitor. Nov 1, 2017 at 3:52 Reducing a voltage source to 0 V is the same as replacing the source in the circuit by a short circuit. If you would have a current source instead, reducing to 0 A would be similar to open-circuit it. Now, by a simple inspection - meaning that in your head you connect an ohm-meter across the capacitor connecting terminals - what resistance do you "see"? As correctly pointed out in the comments section, you can see that the resistance is equal to: $R=R_2||(R_1+R_3||R_4)$. From this value, the circuit time constant is simply $\tau=C_1[R_2||(R_1+R_3||R_4)]$ and the pole is $\omega_p=\frac{1}{\tau}$. Should you look at the transfer function linking the voltage across $R_3||R_4$ to the input source, I can tell that there is a zero and it is located at $\omega_z=\frac{1}{R_2C_1}$. These are the basis for the fast analytical techniques that are described here.