A little theory (Kirkehoff's voltage law)

[Herr Masel]

So I am beginning to get serious with learning electronics myself (at least now, I may take up a more formal approach), so I went right back to the beginning, with an electricity/electronics introduction book of the open university that a friend loaned me. It is very simple and easily approachable, but what I can't understand is this: take a (really) simple circuit - a power supply and two resistors in series. The combined "backwards" voltage - if this is the name in english, I'm not sure - of the two resistors is equal to the forward voltage of the power supply, that much is very easy, but how do you calculate the way this load is divided? I mean, say the power supply is some random number like ohh, 9v , the backwards voltage on resistors could be 1v and 8v, 2x4.5v, 3.4v and 5.6v etc. you get the idea. I must of missed something but I can't figure this one out and it is bothering me, even though it might be of scant importance from a practical point of view. Thanks alot.

[Herr Masel]

Oops, V=IR. The backwards voltage on each resistor is determined by the current and it's resistance. Sorry, I will delete this thread later.