Consider the attached circuit. I know that no current passes throw R(2) given the 9V drop on R(1) which consumes all the voltage V(2). Please consider my following question as more of a physics type; why are electrons which are being driven toward the V(2) source have to spend the same amount of work as those being driven towards the V(1) source that is 9v. I know that KVL has to be valid in this circuit. but I want to know the reason if there is any. This came to my mind as to the fact that if there was a simple voltage divider circuit with one source and two or more resistors, voltage drops would be divided among the resistors and would sum up to the voltage supplied by KVL; in other words.... the voltage source did whatver ever it takes inorder to drive electrons form its negative to its positive terminal. Why wouldn't electrons do the same in this circuit?
http://postimage.org/image/2ugq56zr8/
http://postimage.org/image/2ugq56zr8/