Question

In the op-amp circuit (ideal op-amp), find $V_o$ (rounded to one decimal place).

Hint:

With ideal op-amps, set $V_+=V_-$ and use node-voltage relations of the surrounding resistor network to solve for $V_o$.

Answer 1

Correct Answer: 1.9

At the inverting input, there is a divider of $3R$ (to $V_o$) and $3R$ (to ground), so \[ V_-=\frac{3R}{3R+3R}V_o=\frac{V_o}{2}. \] Let $V_+$ be the non-inverting input node. Writing nodal equations at $V_+$ and at the intermediate node of the $R$–$R$–$R$ feedback network gives (with superposition for the $1$ V source through $3R$) \[ V_+=\frac{V_{\!in}}{3}+\frac{V_o}{3}\qquad (V_{\!in}=1~\text{V}). \] With ideal negative feedback, $V_+=V_-$, hence \[ \frac{1}{3}+\frac{V_o}{3}=\frac{V_o}{2} \Rightarrow V_o=2V_{\!in}=2.0~\text{V}. \]