# Op-amp voltage divider

I have troubles understanding the way the circuit below is solved. I figured out that the solution is: $$V_+=\frac{R_2}{R_1+R_2}V_1 \\ V_a=\frac{R_3+R_4}{R_4}V_+$$ But what is actually the reason that R2 is not considered when calculating V_a? After all V_a=V_3+V_2 forms a mesh right? So the solution would be: $$V_a=\frac{R_3+R_4||R_2}{R_4||R_2}V_+$$ I know that is wrong, but I can't grasp why. As V+=V- isn't R4 in parallel to R2? Probably I am not understanding something basic. Can someone help me out?

simulate this circuit – Schematic created using CircuitLab

• If you substitute V+ in eq2 for V+ from eq1, you get Va as a function of V1. Try that. – scorpdaddy May 12 '16 at 19:27
• Yes, I know, I wrote it that way to make it clearer what I mean, because what is bugging me is the 2nd part of calculating Va. – Daiz May 12 '16 at 19:29
• It's not really in parallel- no current flows between the inputs. Va is driven by the op-amp to an unknown voltage such that V+ == V-. – Spehro Pefhany May 12 '16 at 19:43
• In short, Eq 3 is wrong because it's constructing parallel resistor arrangements that aren't there in the given circuit. Eq 1 and Eq 2 are correct. And when you make the substitution you get the correct results. In order to discuss some mesh with a V_2 and V_3, perhaps you could post a drawing of a mesh with a V_2 and V_3 on it. – scorpdaddy May 16 '16 at 17:01