Am I doing this ideal op-amp problem correctly?

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The discussion revolves around solving an ideal op-amp problem using voltage division and the node-voltage method. The user calculates the output voltage (V0) based on the inverting input terminal and expresses concern about the value derived for V3. A response clarifies that using superposition simplifies the problem, allowing for the calculation of each input's contribution to the output. The correct output for V3 is determined to be 2.5V3, emphasizing the importance of grounding other voltage sources during the analysis. The conversation highlights the effective use of superposition in multi-input op-amp circuits.
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Homework Statement


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Homework Equations


Vp=Vn
Ip=In=0

The Attempt at a Solution


By voltage division:
Vp = 8/(8+8) V3
Vp = 1/2V3
Vn = 1/2V3

By node-voltage method at the inverting input terminal:
(Vn-V1)/4 + (Vn-V2)/4 + (Vn- V0)/8 = 0
Multiply equation by 8
2(Vn-V1) + 2(Vn-V2) + Vn-V0 = 0
V0 = Vn+2Vn-2V1+2Vn-2V2
V0 = 5Vn-2V1-2V2
Substitute in 1/2V3 for Vn

V0 = 5/2V3-2V1-2V2

Is this the correct answer? I'm pretty sure about the V1 and V2 part, because it matches the summing amplifier circuit equation of -(Rf/R1*V1+Rf/R2*V2). However, for V3, I don't see how its resistor composition would yield an answer of 5/2. Could someone clarify?

Thanks
 
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One easy way to work problems with multiple inputs is to take one input at a time, get the corresponding output, then add all three outputs. So for your worrisome V3 input the output is

Vo3 = V+(1 + Rf/Ri) = 0.5V3(1 + Rf/Ri) = 0.5V3(1 + 8/2) = 2.5V3

Also,
Vo1 = -8/4 v1
Vo2 = -8/4 v2

Then by superposition,
Vo = Vo1 + Vo2 + Vo3.

Note that when you do that the other voltage sources are set to zero volts (grounded), not left 'floating'.
 
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