An ideal operational amplifier is to be connected as a differential amplifier as shown in the diagram below. Nominally, $$R_1=R_3=5kOmega$$ and $$R_2=R_4=200kOmega$$

The gain for an input $$V_2$$, with the input $$V_1$$, grounded is most nearly:

A.  20
B.  35
C.  40
D.  41

When an operational amplifier is set up as a differential amplifier, it amplifies the voltage difference between the inverting and non-inverting terminals of the op-amp. This is shown in the following equation (which assumes $R_1=R_3 and R_2=R_4$):

$V_0=\frac{R_2}{R_1}(V_2-V_1)$

Since our problem states that $V_1$ is grounded, its voltage is equal to 0 and we can rewrite the equation as:

$V_0=\frac{R_2}{R_1}\left(V_2\right)$

Therefore, $\raisebox{1ex}{$R_2$}\!\left/ \!\raisebox{-1ex}{$R_1$}\right.$ is our gain factor:

$\frac{R_2}{R_1}=\frac{200k\Omega }{5k\Omega }=40$

Hence, our gain is equal to 40, which is answer C.