Discussion
The output voltage V0 for the circuit shown in the figure is:
Answer: A
The output is \(V_0 = V_1 - V_2\).
This is a difference amplifier. The non-inverting input sits on a divider of two equal resistors R, so \(V_+ = V_1/2\). By superposition, the inverting stage has gain \(-5R/5R = -1\) for \(V_2\) and the non-inverting stage has gain \(1 + 5R/5R = 2\) for \(V_+\):
\[ \begin{aligned} V_0 &= \left(1 + \frac{5R}{5R}\right)\frac{V_1}{2} - \frac{5R}{5R}V_2 \\ &= V_1 - V_2 \end{aligned} \]
\(2(V_1 - V_2)\) forgets that the divider halves \(V_1\). \(V_1 - 2V_2\) doubles the inverting gain, and \(V_2 - V_1\) has the sign of the inputs reversed.
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