Problem 6.6-10

As usual, we call V₁ the inverting input node voltage and V₂ the non-inverting input node voltage.

The two currents entering the op amp are equal to zero.

At the non-inverting node we have:

– (V₂ + 4)/4. 10³  + 2. 10^-3 = 0

thus 8 = V₂ + 4 so V₂ = V₁ = 4V

From the KVL we get V₀ = – 24. 10³ *i₀ – 8. 10³*i₀ + 3

And V₁ = – 8. 10³*i₀ + 3 = V₂ = 4V

Thus i₀ = – 1/8 mA

So we get V₀ = 24/8 + 8/8+ 3 = 7V