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

Question

I am taking a circuits class and this example problem is really confusing me. I am asked to find \$ v_{o} \$ and \$ i \$ for the circuit below.

In their solution, the book starts from the 12 V source and traverses the loop clockwise. The last element in the loop is the 6 Ohm resistor. Since the current encounters the negative terminal first and moves from low potential to high potential, I figured the sign would be negative and the voltage for that element would be \$ -6i \$ making the final equation

$$ -12 + 4i + 2v_{o} - 4 - 6i $$

But they give it a positive sign. Why?

When they apply Ohms law in the second step they correctly assign the voltage a negative sign. Why do they not give it a negative sign when using KVL?

Answer 1

It is not an error. With the polarity of the current chosen, the voltage will drop across the resistor as you go clockwise. Since your book is summing voltage drops around the loop, it gets a positive sign. When looking at that resistor for KVL, ignore \$v_0\$.

Perhaps a reformulation will clarify.

Note that I have tossed out \$v_0\$ and introduced \$v_R\$ with opposing polarity (that is, \$v_R = -v_0\$). As a consequence, the voltage on the dependent source is now \$-2 v_R\$, rather than \$2 v_0\$.

KVL around the loop, summing voltage drops:

$$ -12 + 4i + (-2 v_R) - 4 + 6i = 0$$

By Ohm's law (note the lack of a negative here): $$ v_R = 6 i$$

Subsituting, we obtain $$ -12 + 4i + (-2 \cdot 6i) -4 + 6i = 0$$ $$ -12 -2i = 0$$ $$ i = -8 A $$

Notice that reversing the polarity of the dependent source's test voltage did not change the resistor terms in the KVL, only the dependent source term.

Answer 2

The schematic is:

^{simulate this circuit – 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:

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

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.