Putting the values of this example into Eqs. (8.4-13) and (8.4-14) gives



$$

1=\left(\frac{R_1}{R_1+R_2}\right) 2+\left(\frac{R_2}{R_1+R_2}\right) V_{\mathrm{REF}}

$$



and



$$

0=\left(\frac{R_1}{R_1+R_2}\right)(-2)+\left(\frac{R_2}{R_1+R_2}\right) V_{\mathrm{REF}}

$$





Solving these two equations gives $3 R_1=R_2$ and $\mathrm{V}_{\mathrm{REF}}=0.667 \mathrm{~V}$.