# Brightness and the incorrect direction of current

I was thinking of a circuit similar to this:

$$(+)$$ $$\rightarrow R_1 \rightarrow R_2 \rightarrow$$ $$(-)$$

where $$R_1$$ and $$R_2$$ are lightbulbs. I have read that as the voltage drops at $$R_1$$, we then expect a lower voltage through $$R_2$$, thus, less brightness at the light bulb.

However, the conventional current direction is 'incorrect' as it is the negative charges that flow from the negative end to the positive end. Considering that 'direction', wouldn't we expect $$R_2$$ to be brighter than $$R_1$$, since true current passes through $$R_1$$ first?

How can I make sense of the true current direction and the voltage drops across the resistors?

Thank you guys for your help.

• "How can I make sense of the true current direction", consider the following 2 scenarios: 1, I add negative value to your bank account, 2 I take away some money from your account, they are the same concept aren't they? same thing with electron (-ve) moving in one direction, or "hypothetical" positive charges moving in the opposite. Commented Oct 4, 2021 at 20:41
• Conventional current flow is positive to negative. If the two light bulbs have the same ratings, then they will be equally bright.
– Ed V
Commented Oct 4, 2021 at 20:41

If R1 and R2 are he same kind of bulb they both will shine equally bright! The current in resistors in a row is everywhere the same. If R1=R2 the voltage at both is the same, half of the voltage between + and -, so what you have read was wrong or a misunderstanding.It does not matter if the charges which make the current are negative or positive, there are conductors where the moving charges are positive or even positive and negativ charges make the current.

There is no "passing first through". In steady state, charges pass through all parts of the circuit simultaneously. Remember that the circuit wires are not empty of charge-carrier electrons before you turn on the voltage source - rather, all already present free electrons are simultaneously set in motion as soldiers on march.

You are thinking about the voltages all wrong.

First of all, only the polarity but not the the magnitude of the voltages across the two resistors depends on the direction of the current.

Secondly, the current does not pass first through R1 and then through R2. The current through both resistors occurs essentially instantaneously causing the total voltage to divide between them per the equations

$$V_{R1}= \frac{R_1}{R_{1}+R_{2}}V_{T}$$

$$V_{R2}=\frac{R_2}{R_{1}+R_{2}}V_{T}$$

Where $$V_T$$ is the total voltage from + to - in your drawing.

Hope this helps.

. . . . we then expect a lower voltage through ????? R2 . . . .

This statement may be the reason for your question.

The (conventional = equivalent to the flow of positive charges) current through both resistors is the same as both resistors are connected in series.

The potential difference (voltage) is measured across each of the resistors.

The current flows from a point of higher potential to a point of lower potential.

If both resistors have the same resistance then the potential difference across each of the resistors is the same.

You should be under the misapprehension that a charged particle leaves the terminal of a battery with lost of energy, goes around an electrical circuit losing energy and then arriving at the other terminal of the battery with less energy.

What happens is that a charged particle leaves the terminal of a battery and is accelerated by an electric field thus gaining kinetic energy.
The charged particle then collides with an ion (charged atom of the conductor) and loses kinetic energy in the process which then means that the ion vibrates more.
The process then repeats itself so that the moving charge repeated gains and loses energy so in effect has the same kinetic energy when it leaves one terminal of the battery and arrives at the other terminal of the battery.