Discussion
The output voltage (\(V_{o}\)) of the following Op-amp circuit is :
In the circuit, \(V_{2}\) = 4 V is applied through \(R_{3}\) = 1 kΩ to the inverting (-) input, with a feedback resistor \(R_{4}\) = 1 kΩ from the output \(V_{o}\) to that input; \(V_{1}\) = 2 V is applied through \(R_{1}\) = 1 kΩ to the non-inverting (+) input, which is connected to ground through \(R_{2}\) = 1 kΩ.
Answer: B
The output voltage is - 2 V.
This is a difference amplifier. The non-inverting input sees \(V_{1}\) through the divider \(R_{1}, R_{2}\), and the gain to the output is set by \(R_{3}\) and \(R_{4}\):
\[ \begin{aligned} V_{+} &= V_{1}\frac{R_{2}}{R_{1} + R_{2}} = 1\ \text{V} \\ V_{o} &= \left(1 + \frac{R_{4}}{R_{3}}\right)V_{+} - \frac{R_{4}}{R_{3}}V_{2} \\ &= 2(1) - 4 = -2\ \text{V} \end{aligned} \]
With all resistors equal, the circuit simply gives \(V_{1} - V_{2}\). The option +2 reverses the subtraction, and \(\pm 4\) uses only one input.
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