Problem 6.4-3

We will use V₁ as the inverting input node voltage and V₂ as the noninverting input node voltage.

We have V₂ = – 2 V = V₁ and we have V₀ = voltage of the 8 kΩ resistor

So V₀ = – 8. 10³ * i₀ – 2

Moreover, we have V₁ = 12 – 4. 10³ * i₀ = – 2 V

Thus we have i₀ = 14/4 mA = 3.5 mA

And then V₀ = – 8. 10³ * i₀ – 2 = – 30 V