Question B.25: In Figure B.15, if R1=2.2 kΩ,R3=18 kΩ,v1=120 mV and v2=−40 m...

Using the summing amplifier equation in Chapter 12, we calculate:

\begin{aligned}v_{3} &=-\frac{R_{3}}{R_{1}} v_{1}-\frac{R_{3}}{R_{2}} v_{2} \text { or } 0=-\frac{18}{2.2}(120)-\frac{18}{R_{2}}(-40) \\\frac{120}{2.2} &=\frac{40}{R_{2}} \\R_{2} &=\frac{2.2 \times 40}{120}=0.733  \mathrm{k} \Omega\end{aligned}

Thus, the nearest answer is \mathrm{d}.