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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!
To determine the time constant of this circuit, simply reduce the excitation to 0 and "look" at the resistance offered by the capacitor terminals when the cap. is removed. See the schematic below:
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.
Here is the most simple method for finding the time constant for such a circuit (without any transformations):
Imagine that the capacitor is charged and find the variuous resistive ways which allow discharging (ground node removed, short ciruit for the voltage source).
Then you will see that we have R2 in parallel to the series combination of R1 and R3||R4.
