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So, I calculated Vo first by using I = VR, second by using voltage distribution but the results are different. I wonder which one is wrong and why.

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    \$\begingroup\$ In the 1st approach, you assumed that i flows through 40R resistor and generates the output voltage. This is wrong. I didn't check the rest, so there might be other mistakes. \$\endgroup\$ Nov 3, 2021 at 13:17
  • \$\begingroup\$ Assume that V1 and V2 are positive, what does that mean for the currents they inject into the net Va? The opamp tries to make Va = Vb, for that to happen it will apply a voltage Vc at its output. Will Vc need to be positive or negative given that V1 and V2 are both positive? \$\endgroup\$ Nov 3, 2021 at 13:17
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    \$\begingroup\$ You can ignore the input currents for an ideal op-amp. You can not ignore the output current, which is what it looks like you've done. The output current will be the sum of i and the current into 10 + 40 ohms and will, in fact, be much larger in magnitude than i. \$\endgroup\$ Nov 3, 2021 at 13:23
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    \$\begingroup\$ case 2 seems correct unless this note ... that 10 and 40 Ohm are a bit too low for op amp output (unless it is a power amplifier ? What about output impedance ? ) \$\endgroup\$
    – Antonio51
    Nov 3, 2021 at 13:32

1 Answer 1

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Well, notice that we have a system of equations using KCL:

$$ \begin{cases} \text{i}=\text{i}_1+\text{i}_2\\ \\ \text{i}_\text{out}=\text{i}+\text{i}_\text{opamp} \end{cases}\tag1 $$

Using Ohm's law and \$(1)\$, we can see that:

$$ \begin{cases} \frac{\text{V}_\text{a}-\text{V}_\text{c}}{2000}=\frac{\text{V}_1-\text{V}_\text{a}}{100000}+\frac{\text{V}_2-\text{V}_\text{a}}{100000}\\ \\ \frac{\text{V}_\text{c}}{10+40}=\frac{\text{V}_\text{a}-\text{V}_\text{c}}{2000}+\text{i}_\text{opamp} \end{cases}\tag2 $$

Now, in an ideal opamp circuit we know that:

$$\text{V}_+=\text{V}_-=\text{V}_\text{a}=\text{V}_\text{b}=0\space\text{V}\tag3$$

So, we get:

$$ \begin{cases} \frac{0-\text{V}_\text{c}}{200000}=\frac{\text{V}_1-0}{100000}+\frac{\text{V}_2-0}{100000}\\ \\ \frac{\text{V}_\text{c}}{10+40}=\frac{0-\text{V}_\text{c}}{200000}+\text{i}_\text{opamp} \end{cases}\tag4 $$

This leads to:

$$\text{V}_\text{o}=\frac{40}{40+10}\cdot\left(-2\left(\text{V}_1+\text{V}_2\right)\right)=-\frac{8}{5}\cdot\left(\text{V}_1+\text{V}_2\right)\tag5$$

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  • \$\begingroup\$ Thanks for the well-motivated downvote. \$\endgroup\$ Nov 4, 2021 at 9:52

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