Problem 6.5-3

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.

We have V₂ = V₁, but we know that V₂ = 0 V so V₁ = 0 V

We have i₀ = i1 + i2, where i1 is the current from above the output node, and i2 is the one from under.

We have V₀ = – 6. 10³ *i₁ = – 6. 10³ *i₂ so we have i1 = i2.

As we have a virtual ground at the inverting node, we can find the equivalent circuit for the left part:

The two resistor in parallel give a Re1 = 3 kΩ resistor, in series with the upper 6 kΩ one.

This give Re2 = 9 kΩ

We get 12 V for 9 kΩ thus a current of I = 12/9 mA = 4/3 mA

So we have i1 = 2/3 mA

Finally we have i₀ = i1 + i2 = 2*i1 = 4/3 mA = 1.33 mA

And V₀ = – 6. 10³ *i₁ = – 4 V