The Power Point is HERE

Let's take a nasty looking circuit:

We Start By Labeling the Current in AND out of EACH Junction Point

Normally I'd draw arrows at each junction, but that would make this beastie just too cluttered, so we're going to assume right arrows and down arrows at each junction

Let's pick off the easy series resistors first. The resistors on both sides of I₈ are fair game... lets combine those

Which gives us:

 

Which Gives Us

So Let's Reduce Those

Which Gives Us

Notice I₃ & I₅ are now In Parallel... Notice How the Position of I₉ makes us WAIT before including it in our reduction!

Let's Redraw That to More Easily See Them That Way

Now We'll Reduce 'em.... Which Leaves I₃ & I₉ in Series...

So Add those UP to make 12

Almost Done! Two More ( I₃ & I₄) in Parallel

 

We add Those Two Leaving Just 2 More In Series... easy peazy!

2Ω + 4Ω = Et Voila!!