Help understanding the use of polarities in KVL?

Discussion in 'General Electronics Chat' started by ArtemisFei, Mar 17, 2013.

  1. ArtemisFei

    Thread Starter New Member

    Mar 17, 2013
    It's been bothering me for a long time, enough that I decided to ask you helpful folks about it.

    I was doing a problem about Thevenin equivalents, and their use of KVL in my notes confused me.

    In this example of my notes, they used KVL to find the voltage across the 6 ohm resistor and therefore the open voltage.

    But, I have a problem with their use of KVL. With this one, they made a clockwise current and added up all the voltage drops.


    But, when I use KVL and make it counterclockwise, my KVL expression becomes totally wrong.

    So, when I do my KVL counterclockwise adding voltage drops, starting from the bottom right. I hit the 6 ohm resistor first, and current goes in the direction of a drop, so I get +6i. Then I hit the voltage source as a drop, so that's also +2i. Then, another resistor as a drop, so that's another +6i. And finally, a drop at the voltage source, so that's +20.

    My final expression is (6i+2i+6i+20=0), which gives me a completely different current value.

    What did I do wrong with my KVL?

    Thanks for any help guys!
  2. antonv

    Active Member

    Nov 27, 2012
    You did not take the polarities into account somehow. Going counter-clockwise I would have said -6i+2i-6i+20=0