# Pot diff. part deux

xyx
Hi,

Sorry to bring this up again but I've been looking so long for an answer to a new question that arrised.
This is how I see it and tried to explain it, so plz correct me if I'm wrong.

example 01: 1 bulb in a simple circuit : 12v ,
bulb 01 : 2ohm
expln: potential energy drops by 12 v over bulb; Ok np so far.

example 02: 2 bulbs in series: 12v ,
bulb 01 : 2 ohm , 8v
bulb 02 : 1 ohm , 4v

But here's my problem: THE DIFFERENCE IN V. (POTENTIAL DROP) OVER BULB 01 IN EXAMPLE 01(12v.) AND BULB 02 IN EXAMPLE 02(8V).
Which one is the right explaination?

EXPLAINATION 01:

The energy needed per electron to pass through bulb's 01 ions is still the same so each electron loses the same energy per coulomb but there are less passing so the drop in energy is less over bulb 01.
Since there are less q/s(I)(more energy to spend per electron) every electron starts with more energy (more then with just one light bulb)so it also can get past bulb 02.
But this is a bit in contradiction (to my textbook) with the fact that the potential drop is said to be 8v per coulomb on bulb 01, so the [del]E/q, the drop per electron(loss in energy to go through bulb 01), in bulb 01 is in my above explaination still 12v but since there are less electron the v-meter measures less namely 8v.

EXPLAINATION 02:

The energy needed to pass through light bulb 01 IS less then in example 01 because I is smaller and therefore there is less energy loss in the same bulb(bulb 01) for some sort of atomic reason I don't get(maybe because the electrons are moving less fast)?

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xyx
correction on line 12: I meant "and bulb 01 in example 02(8v)".

EXPLAINATION 01:
The energy needed per electron to pass through bulb's 01 ions is still the same
I think there is no 'energy needed to pass through'.
Imagine you operate a bulb at very low voltage, so that it does not glow.
You will find that there is a current. In fact, the bulb is a good conductor when it's cold.
Yes, the current is small then. But not because electrons don't have the 'energy needed to pass thru'. The current is small because the electrons move very slowly. Remember, current=charge/time.
so each electron loses the same energy per coulomb but there are less passing so the drop in energy is less over bulb 01.
I think this is false. Voltage is energy/charge. So, in example 02 each electron does in fact deposit less energy in bulb 01, than it does in example 01. If you so will, it 'saves' some energy for bulb 02. Yes, there are less passing but this has nothing to do with how the 12V distribute over both bulbs.
Since there are less q/s(I)(more energy to spend per electron) every electron starts with more energy (more then with just one light bulb)so it also can get past bulb 02.
No. In both examles, each electron starts with exactly the same energy.
But this is a bit in contradiction (to my textbook) with the fact that the potential drop is said to be 8v per coulomb
The potential drop is just 8V = 8joule/coulomb. If your textbook uses the unit 'volt per coulomb', then it is wrong.
on bulb 01, so the [del]E/q, the drop per electron(loss in energy to go through bulb 01), in bulb 01 is in my above explaination still 12v but since there are less electron the v-meter measures less namely 8v.
No. The voltmeter measures volts. Which is basically energy per electron. The initial energy per electron is a sole property of the source, no matter what circuitry you connect. (If it's an ideal source, which is the case here).

EXPLAINATION 02:
The energy needed to pass through light bulb 01 IS less then in example 01
There is no 'energy needed'. There only is 'energy deposited' per electron. And that's in fact less in example 02 than in example 01.
because I is smaller
You're getting closer. The electron deposits less energy in bulb 01 because it moves slower. (And I is smaller also because of this).
and therefore there is less energy loss in the same bulb(bulb 01) for some sort of atomic reason I don't get
The 'atomic reason' is, if you so will, collisions between electrons and atoms. If the electron is slower, it loses less energy in a collision.
(maybe because the electrons are moving less fast)?
Yes.