# Write $V_{A}$ in terms of $V_{in}$ and the resistor values algebraically in a Op-Amp

Question

Consider the Op-Amp circuit in Fig. Q1(a). Write $V_{A}$ in terms of $V_{in}$ and the resistor values algebraically.

My Try

• Case 1 :

I see wire(+) as opened since it draw almost 0 current ideally. So, we have

$$V_A=V_{in}\left(\frac{R_3}{R_3+R_4}\right)$$

• Case 2 :

Transform the circuit as if the following one. Since wire(+) and wire(-) ideally have no potential difference, I shorted them together.

simulate this circuit – Schematic created using CircuitLab

$$R_{13}=\left(\frac{1}{R1}+\frac{1}{R3}\right)^{-1}\space \space \space \space and\space \space \space \space R_{24}=\left(\frac{1}{R2}+\frac{1}{R4}\right)^{-1}$$

$$R_{eq}=R_{13}+R_{24}$$

Therefore, $$V_A=V_{in}\left(\frac{R_{13}}{R_{eq}}\right)$$

My Questions

1. For Case 1 and Case 2, which one is correct? Or, both are wrong?
2. How to find the current I from $V_{A}$ to $V_{out}$? Can I calculate it from the following equation?

$$I=\frac{0-V_A}{R_1}+\frac{V_{in}-V_A}{R_2}$$

• You missed R5 and Vout in case 2 – nidhin Jul 12 '14 at 11:35
• Oh~...If I include R5 and Vout in the circuit, it will become so complicated... :c – Casper Jul 12 '14 at 11:37

Case 2 is wrong conceptually. In the actual circuit, resistors $R_3$ and $R_4$ are series connected - all of the current through $R_4$ is through $R_3$.