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I know I have got a trivial question but I can't find a proper explanation on the internet. In the image below I know that the formula for the cutoff frequency is f = 1/(2*PI*(R1//R2)*C). One way to get the above formula is to find parallel impedance of R2 and C1. Then apply a voltage divider formula for R1 and R2//C1 and then we can derive the cutoff frequency. But I was wondering is there a better way to get this formula by finding the equivalent resistance of R1 and R2 first and then substituting this equivalent resistance into simple RC filter formula? Is there any method that would allow me to calculate the equivalent resistnce of R1 and R2 first and then use standard RC filter formula? Thanks in advance enter image description here

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  • \$\begingroup\$ What about this way? electronics.stackexchange.com/questions/377467/… \$\endgroup\$ – G36 Oct 4 at 19:30
  • \$\begingroup\$ Both resistors in parallel, then T=RC. \$\endgroup\$ – Marko Buršič Oct 4 at 19:40
  • \$\begingroup\$ Yes thevenins theorem seems to work but I am a bit confused about vth as it is equivalent thevenins voltage. Even though we don't really need that voltage for the formula so I am a bit baffled whether we can only use thevenins equivalent resistance? \$\endgroup\$ – SGS333 Oct 4 at 19:47
  • \$\begingroup\$ Because you are only interested in finding the pole frequency and not the "voltage level", you only need to care about the resistance seen by the capacitor \$\endgroup\$ – G36 Oct 4 at 20:05
  • \$\begingroup\$ Thanks very much \$\endgroup\$ – SGS333 Oct 4 at 20:06
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You can use mesh analysis as alternative, in time domain as well as in s-domain (Laplace).

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You do calculate the divider first. $$\dfrac{V_{out}}{V_{in}}=\dfrac{R2}{R1+R2}$$

Then according to Thevenin you do calcualte equivalent resistance \$R=R1 || R2\$. Finally \$\tau=RC\$, you do calculate C.

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  • \$\begingroup\$ Thank you I think get it now \$\endgroup\$ – SGS333 Oct 4 at 20:04

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