Last year someone asked an interesting but off-topic
question on math
How voltage drop occurs in a resistor?
So I was getting confused over a thought, say I have a resistor and we
have 10V source across it, now when electrons flow from negative end
to the positive end they lose 10 joules of energy as they flow through
the resistor, now with all the weird and crazy quantum stuff happening
inside the wires how can we exactly predict how much energy an
electron is going to lose that precise amount of energy when it passes
through the resistor, can’t it do this - lose 5J of energy inside the
resistor and continue with 5J energy left ?
Well, that is a good question, and I didn't really know the answer
because I know very little about physics. But I thought about it and
thought maybe I had the explanation, so I replied in the comments.
The question was later deleted, but I saw it again recently and
thought that my explanation was worth preserving, so here it is:
I think the answer is going to look like this: if you throw a handful
of fine sand, all the grains identical, off of a 100m cliff, some
grains might fall straight down, some might be caught by the wind and
whirled all about, but every single one will lose exactly !!100g/m!!
joules per kilogram, where !!g!! is the acceleration of gravity in meters
per second squared and !!m!! is the mass of a grain.
Three hours later:
Considering this more, I think the analogy is pretty good. The 10J
loss (actually 10eV, but the principle is the same) is not a property
of the resistor but of the power source. You can replace the resistor
with any other resistor — bigger, smaller, different shape, different
material, whatever — and the energy loss per electron will not
change. The electrons lose 10eV each because the power source is a 10V
source. This does not mean that it imparts 10eV to each electron; it
means that the energy difference between its two terminals is 10eV per
electron. The energy measurement is relative.
“Per electron” in the next-to-last sentence is wrong, but I think the
rest of it is pretty good. In particular, my point about how the
energy loss is not a property of the resistor. The energy loss is the
same, regardless of the resistance; this is Kirchoff's voltage
If you replace the resistor with one of lower resistance, the energy
drop through it is still exactly the same (per electron), but the
voltage source is able to push electrons through it at a greater rate;
this is Ohm's law.