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?