# How do I define the voltage drops polarity when applying KVL?

Let the circuit be:

Question: Find Ix by applying KVL.

Correct Answer: Starting from node ‘f’ and going clockwise, KVL gives: 12 – 3Ix – 6Ix = 0, so that Ix = 12/9 = 4/3 A.

Problem: This seems to be a bit confusing. According to my knowledge, we can choose the signs of the resistor and we'll have the same answer.

It seems that the answerer picked those signs + --3Ω-- - and - --6Ω-- + . When Ix passed through the negative pole of the battery, we gave a (+) sign, when Ix passed through the positive pole of 3Ω and 6Ω, we gave a (-) sign.

I picked those signs. - --3Ω-- + and - --6Ω-- +. Starting from node ‘f’ and going clockwise, KVL gives: 12 + 3Ix – 6Ix = 0, so that Ix = 12/3 = 4A (the correct answer is 4/3 A).

Question: why the answerer chose the signs of the resistors this way? Why is the answer 4A wrong and 4/3 A correct? If it has to do with the passive sign convention, I would appreciate if someone would explain it to me?

• I*R is a voltage - depending on how you define the direction of the CURRENT this product can be positive or negative. Resistors are not defined by direction, you cannot just assign them positive and negative values at random. Sep 4 '14 at 15:29

Strictly speaking, it's not the sign of the resistors that can be changed, it's the sign (direction) of the current $I_x$. Resistors always have positive resistance but you can choose the current in either direction as long as you are consistent.
Suppose you chose $I_x$ in the opposite direction. Starting from node $f$ and moving counterclockwise, you have a positive voltage across the $6\Omega$ resistor, another positive voltage across the $3\Omega$ resistor, and a positive voltage across the 12V supply. This gives:
$I_{x}(6\Omega + 3\Omega) + 12 = 0$
which results in $I_{x} = -4/3$A. The sign is different, indicating that the current actually flows in the opposite direction (i.e. in the original direction shown in the image).
The reason your 4A answer is wrong is that you weren't consistent with the $I_x$ direction for the two resistors -- you had $I_x$ flowing in the opposite direction through the $3\Omega$ resistor.