I have the following circuit for which I need to find the time constant:
Would it be correct in assuming that the only relevent resistances are R2 and R4? So then the time constant would simply be (R2||R4)C? Or am I missing something obvious?
Thanks
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\$\begingroup\$ Assume that the capacitor is charged to a certain voltage. Now imagine it discharging, which components can influence that discharge time? Indeed R1 and R3, R5 cannot influence that. Indeed R2 and R4 can. So yes Tau = (R2||R4) C \$\endgroup\$– BimpelrekkieOct 16, 2019 at 7:08
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2 Answers
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Yes, that's pretty much it.
R1 is irrelevant because it is shorted out by a voltage source, which has zero impedance.
R3||R5 is irrelevant because it is in series with a current source, which has infinite impedance.
answered Oct 16, 2019 at 11:51
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The time constant of your circuit can be determined by turning the excitation signal off. If the input is the voltage source, then turn it to 0 V while the current-source is set to 0 A. A 0-V source is replaced by a short circuit while a 0-A current source is simply open-circuited. Then, remove the capacitor and "look" through its connections: what resistance do you "see"?
By inspection, without writing a single line of algebra, you immediately see that the resistance is simply \$R_2||R_4\$ leading to a time constant equal to: \$\tau=(R_2||R_4)C_1\$. As explained with the fast analytical circuits techniques or FACTs, the circuit features a pole located at \$\omega_p=\frac{1}{\tau}=\frac{1}{(R_2||R_4)C_1}\$
answered Oct 16, 2019 at 14:46