Check out this java simulator for resistors:

http://www.falstad.com/circuit/e-resistors.html

I'm using this to practice Kirchoff's laws. I have a quick question:

When you load this page, by default, the 100, 800, and 200 ohm resistor switches are closed.

Now, close the 400 ohm then 600 ohm switch. Ok, that's fine.

Finally, close the lone switch at the bottom right. I can't wrap my head around why the 400 and 600 ohm resistor pathways are impeded by throwing the final switch?

I have a guess.. by throwing the final switch, Kirchoff's laws effectively splits the outgoing current from the node below the 400 ohm resistor into 3 equal chunks instead of 2. This split creates too little mA's to pass through the 600 and 200 ohm resistors, and as a result the mA of the entirecircuit increases because these resistors are no longer in play?

Hope that makes sense. I haven't finished reading the nerd kit yet and don't want to start with building stuff until I understand the concepts..

Cheers, Amed

Remember, Current will always take the path of least resistance. In this case, the switch, effectively bypassing the resistors by making an electrical short across them. Also, since those resistors are now effectively out of the circuit, the current flow through the circuit increases as if that portion of the circuit wasn't even there.

Rick