In a circuit that has a 12volt power supply and two 2- ohm resistors, R1 and R2, in series, the pd across each resistor is obviously 6v. If a 3rd resistor (also 2-ohm)R3, is placed in parallel across R2 then the pd across them changes to the following: R1=8v R2=4v= R3 (parallel) However, when I am explaining pd to my students I ask them to imagine they are a single electron, that has to make its way thru the resistors, and has to do alot of work, and loses elec potential energy as a result. This works very well for series circuits, but the above calculations blow my analogy out of the water, because I inevitably get the “well, if it is a single electron and only takes ONE pathway, then why does it seem easier to get thru the parallel portion of the circuit? Ie. In the original series circuit the electron had to expend half “energy” to get thru the 2-ohm resistor, but just cos a new one is introduced, it eases the pathway thru the second, and now it spends 2/3 of its energy going thru the first resistor and only 1/3 going thru the second.” In other words, just because another route has opened, this should have no bearing on how hard it is to get thru the very FIRST resistor R1. Also, if an electron takes ONE route (R2), it is completely unaware of the parallel route (R3), and should still have to work just as hard to get thru the R2 as it originally did. I can explain it properly when we are talking about coulombs because I say that the overall work done by the coulomb is “shared” along the two paths (but I am not even totally happy with that explanation myself!) but surely it should work for individual electrons just as much, and yet each electron takes ONLY ONE path! Is it because the sheer number of electrons going thru has been reduced so that somehow reduces the work an individual electron has to do, since it is less “crowded” en route? I understand all the Ohm’s Law calculations, but I really would like a conceptual explanation on a single-electron level that would also explain it in terms of a charge having to battle its way across a crowded resistor surface. Sorry to have been so longwinded. I have asked several teachers and I have not got any satisfactory response, and if u can help me out at all, I would really apprec.