Why is the sign of the voltage on this resistor positive in this example from my textbook
The schematic is:







simulate this circuit &ndash; Schematic created using CircuitLab

I've applied the + and - signs consistently across the two resistors, given the current direction. (This is not to be considered the same thing as how the VCVS is applied. Note that the (+) of the VCVS is tied to the correct end of \$R_2\$ for its purposes.)

Running clockwise around the loop starting at ground, I get:

$$ 0\:\textrm{V} + 12\:\textrm{V} - i\cdot R_1 - 2\cdot\left(0\:\textrm{V}-V_3\right) + 4\:\textrm{V} - i\cdot R_2=0\:\textrm{V}$$

You also know that:

$$V_3 = 0\:\textrm{V}+i\cdot R_2$$

Putting these together I get:

$$\begin{align*}
12\:\textrm{V} - i\cdot R_1 + 2\cdot i\cdot R_2 + 4\:\textrm{V} - i\cdot R_2&amp;=0\:\textrm{V} \\ \\
16\:\textrm{V}&amp;= i\cdot \left(R_1 - 2\cdot R_2 + R_2\right)\\ \\ i&amp;=\frac{16\:\textrm{V}}{R_1 - R_2} = -8\:\textrm{A}
\end{align*}$$

It must then be the case that:

$$V_o = \left(0\:\textrm{V} - V_3\right) = \left(0\:\textrm{V} - \left[0\:\textrm{V}+i\cdot R_2\right]\right)=-i\cdot R_2 = 48\:\textrm{V}$$

So I guess I get the same results.