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I drew this low pass filter circuit to have a output voltage of 5v across capacitor at a frequency of 100Hz sine wave of amplitude 10v. But I am only getting 4.5 volt as output in steady state. I am doing simulation in LTspice on Ubuntu. What mistake I am making?

R = sqrt(3)/(2*pi*fC) - This is how I calculated R for a given C to have a half voltage drop across C

enter image description here

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  • \$\begingroup\$ The \$\sqrt 3\$ term is not part of the calculation for this type of circuit. \$\endgroup\$ – Peter Smith May 22 '16 at 14:03
  • \$\begingroup\$ How did you read this 4.5V ?? I get 4.9956V in my LTspice. \$\endgroup\$ – G36 May 22 '16 at 14:05
  • \$\begingroup\$ @G36 - My LTSpice gives somewhere around 4.60 volt. Don't know why? I did a transient analysis for 8 second. \$\endgroup\$ – InQusitive May 22 '16 at 15:12
  • \$\begingroup\$ @G36:- When I put exactly 200ms as you did in transient, I am getting the correct value. I think I am making something related to transient timing? Can't I use a large value for stop time in transient analysis, like 10 seconds? \$\endgroup\$ – InQusitive May 22 '16 at 15:16
  • \$\begingroup\$ @InQusitive try set Maximum Timestep at around 10us in transient analysis window. \$\endgroup\$ – G36 May 22 '16 at 15:34
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To get 5V peak from a 10V peak supply across \$1 \mu F\$ capacitor at 100Hz we need current equal to I = 5V/Xc = 5V/1.592kOhm = 3.1407mA and the voltage drop across resistor is $$V_R = \sqrt{(10V^2 - 5V^2)} = 8.66V peak $$
therefore $$R = \frac{8.66V}{3.1407mA} =2.757k\Omega $$ So yes, your calculations are correct.
And the simulation result look like this enter image description here

As you can see the peak voltage is 5V but RMS value is around 3.5V

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