# 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
1
0
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.

(6i-2i+6i-20=0)

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 Member

Nov 27, 2012
149
27
You did not take the polarities into account somehow. Going counter-clockwise I would have said -6i+2i-6i+20=0