Solve Ideal Op Amp Question: KCL & KVL

[jackstock]

Homework Statement

Homework Equations

KCL & KVL.

The Attempt at a Solution

Using KVL:

-9 + 10kI₁ + 6 = 0
-10 + 20kI₂ + 6 = 0
12 + 30kI₃ + 6 = 0
40kI₄ + 6 = 0

:. I₁ = 0.3mA
I₂ = 0.2mA
I₃ = -0.6mA
I₄ = -0.15mA

Now here's where the problem is. When I take KCL at the bottom node:

I₁ + I₂ + I₃ + I₄ = I₆ [the op amp is ideal so the current through the 6V source is zero]
I₆ = -0.25mA

:. V_o = -0.25m*20k = -5V [which is apparently incorrect]

However, when I take KCL at the inverting input to the op amp:

I₁ + I₂ + I₃ + I₄ = I₅
I₅ = -0.25mA

Using KVL again:
V_o - 6 - 0.25m*100k = 0
V_o = 31V [which is apparently correct]

So my question is why do I get different answers depending on which node I take KCL at? Is there something obvious which I'm missing? Thank you in advance for your help.

 

[The Electrician]

It's because I1, I2, I3, I4 and I6 are not the only currents into the bottom node. The opamp's power supply also provides a current into the bottom node. You will notice that the opamp's output absorbs (or supplies) the sum of I5 and I6, which ultimately finds its way to the power supply. So there is a current into the bottom node not shown on the schematic.

But, summing currents into the inverting input works because there are no hidden currents into that node.

 

[jackstock]

Thank you, that makes sense. For future reference, is there a way to identify where these currents from the power supply will turn up so this doesn't happen again?

 

[The Electrician]

The power supply will be connected to the opamp V+ and V- pins (or just V+ if a single supply voltage is used) and the "hidden" currents will be injected into the circuit wherever the supply return is connected. Don't use that node (probably the reference or "ground" node) as a node for summing currents.