Q.Determine the output voltage when $V_1 = -V_2 = 1\text{ V}$.

The difference amplifier, or subtractor, produces an output proportional to the difference of two independent input signals. One signal $V_1$ is applied through a resistor to the inverting $(-)$ input with feedback resistor $R_f$; the other signal $V_2$ is applied through a resistor to the non-inverting $(+)$ input, which is also returned to ground through a resistor.

The output is found by superposition — treating each input's effect separately (the inverting input acts as an inverting amplifier, the non-inverting input acts through the divider and non-inverting gain) and adding the results:

When the resistor pairs are matched (the inverting side $R_1=R_f$ and the non-inverting side $R_2=R_3$), this collapses to the clean, symmetric result:

so the circuit is a unity-gain subtractor. For example, with $V_1=1\ \text{V}$ and $V_2=-1\ \text{V}$ the output is $-1-1=-2\ \text{V}$; with $V_1=-1\ \text{V}$ and $V_2=1\ \text{V}$ it is $1-(-1)=+2\ \text{V}$.

Matching the resistors is essential: any mismatch lets part of a common-mode signal appear at the output, degrading the rejection of noise. The subtractor is therefore the core of instrumentation amplifiers, which measure a small difference sitting on top of a large common-mode voltage.