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I can understand the first case above, that seems simple as I can simply measure across the ground and V+, but I am getting very confused for the second case.

Is the capacitor and the resistor in parallel?

  • \$\begingroup\$ Huh? What is a "look in voltage"? \$\endgroup\$ – Olin Lathrop Nov 20 '14 at 14:49
  • \$\begingroup\$ Well, sorry for my wording. I meant how do i calculate what the ratio of V+/V0 is \$\endgroup\$ – Raaj Nov 20 '14 at 14:52

Okay, if you assume no load at all (infinite impedance on V+), the series capacitor gives you an arbitrary offset voltage corresponding to the initial condition of the capacitor. Since there is no load, the capacitor voltage will remain the same forever. Other than that arbitrary voltage you can ignore the capacitor.

If you assume some resistive load, then I think you can calculate it similarly to the first example.

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  • \$\begingroup\$ yes imagine thats an input to an op amp. Ic, so that means that the equation will be the same as above? \$\endgroup\$ – Raaj Nov 20 '14 at 15:06
  • \$\begingroup\$ Same as the first one without the right hand capacitor. Remember, the ideal op-amp input is like an open circuit so the right hand capacitor will not be in parallel with the right-hand resistor. \$\endgroup\$ – Spehro Pefhany Nov 20 '14 at 15:09
  • \$\begingroup\$ yeap. this makes sense \$\endgroup\$ – Raaj Nov 20 '14 at 15:17

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