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Question

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

enter image description here


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.

schematic

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?

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


Thank you for your help.

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  • \$\begingroup\$ You missed R5 and Vout in case 2 \$\endgroup\$ – nidhin Jul 12 '14 at 11:35
  • \$\begingroup\$ Oh~...If I include R5 and Vout in the circuit, it will become so complicated... :c \$\endgroup\$ – Casper Jul 12 '14 at 11:37
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Since the op-amp inputs are (ideally) open circuits, case 1 is correct.

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\$.

However, in your case 2 schematic, this is not the case; there can be a non-zero current through the wire connected the two resistor branches.

Consider the following instead:

schematic

simulate this circuit – Schematic created using CircuitLab

Now, this gives the same result as case 1.

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