It has been said that Vout = Va/2 I'm not sure why this would be. Could someone please tell me why?

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  • \$\begingroup\$ Because the impedances form a voltage divider. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 9 '16 at 4:48
  • \$\begingroup\$ And just what exactly would have you believe that half of the impedances are present at node Va? \$\endgroup\$ – gorge Jun 9 '16 at 5:35
  • \$\begingroup\$ What exactly does that mean "half of the impedances are present at node Va?" Assuming no load at Vout, it is very obvious that Vout = Va / 2 because the two capacitors of the same value (C) form a voltage divider that cuts the input (Va) in half. \$\endgroup\$ – mkeith Jun 9 '16 at 6:54

At any frequency, the impedance offered by those two capacitors will be same (because both have same capacitance). Since same current flows through them, the voltage drop across them will also be the same = Va/2.

  • \$\begingroup\$ The only two components with the same current flowing through them is the capacitor and resistor in series near Vin \$\endgroup\$ – gorge Jun 9 '16 at 9:24
  • \$\begingroup\$ these two rail systems confuse me a bit but after converting it into a single line i understand now. i.e. vin-------C1------VOUT--------C2--ground \$\endgroup\$ – gorge Jun 9 '16 at 9:30
  • \$\begingroup\$ Do you have any tips on how to not be confused by things like this? \$\endgroup\$ – gorge Jun 9 '16 at 9:34
  • \$\begingroup\$ by vin above i actually mean Va by the way \$\endgroup\$ – gorge Jun 9 '16 at 9:36
  • \$\begingroup\$ @gorge same current flows through \$R_1\$ and \$C_1\$. At node Va, this splits into two. \$V_a/R_2\$ flows to ground through the resistor and \$V_a/(\frac{1}{jwC} + \frac{1}{jwC}) \$ flows through the capacitors \$C_2\$ and \$C_3\$ (left most). \$\endgroup\$ – nidhin Jun 9 '16 at 9:46

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