I'm having difficulty understanding voltage drop and hope someone can explain it to me in a way I can grasp.

I made a basic circuit on a breadboard to proof my math and its correct but I don't think I understand whats going on.

12v circuit w/2 resistors in series (1k ohm [R1] and 10k ohm [R2]).

So we have about 1ma current and 11k resistance.

Vout would be between the two resistors.

Thus Vout = (R2/R1+R2)*V == 10.9v (around 11v).

Reversing the resistors on the circuit of course gave me around Vout of around 1 volt.

I seem to be missing what "voltage drop" really is and whats going on. In other words, why does a 10k resistor with a "voltage drop" of 10.9v yield 10.9v when tested on the meter? If we have 12v and we "drop" 10.9v, I should then get a reading of 1.1v !!

To confound things further, when the circuit is reversed how is it possible that a 1k ohm resistor @ R2 giving me just 1 volt? My point here is how is a resistor with 1/10th of the resistance of the 10k resistor, giving me only 1v and the one that is 10x stronger (10k ohm) allowing 10x the voltage to pass thru it (10.9v) ?!?!?!

It seems like the 10k ohm one is allowing 10.9 v to pass thru it and the 1k ohm resistor is allowing only 1.1v to pass thru it ?!?! I would think it'd be the exact opposite!!!

Keep in mind I'm aware of electron flow (vs conventional) and viewing it that way... as is the formula focused on R2 to give Vout.

Am I asking this correctly?

I made a basic circuit on a breadboard to proof my math and its correct but I don't think I understand whats going on.

12v circuit w/2 resistors in series (1k ohm [R1] and 10k ohm [R2]).

So we have about 1ma current and 11k resistance.

Vout would be between the two resistors.

Thus Vout = (R2/R1+R2)*V == 10.9v (around 11v).

Reversing the resistors on the circuit of course gave me around Vout of around 1 volt.

I seem to be missing what "voltage drop" really is and whats going on. In other words, why does a 10k resistor with a "voltage drop" of 10.9v yield 10.9v when tested on the meter? If we have 12v and we "drop" 10.9v, I should then get a reading of 1.1v !!

To confound things further, when the circuit is reversed how is it possible that a 1k ohm resistor @ R2 giving me just 1 volt? My point here is how is a resistor with 1/10th of the resistance of the 10k resistor, giving me only 1v and the one that is 10x stronger (10k ohm) allowing 10x the voltage to pass thru it (10.9v) ?!?!?!

It seems like the 10k ohm one is allowing 10.9 v to pass thru it and the 1k ohm resistor is allowing only 1.1v to pass thru it ?!?! I would think it'd be the exact opposite!!!

Keep in mind I'm aware of electron flow (vs conventional) and viewing it that way... as is the formula focused on R2 to give Vout.

Am I asking this correctly?

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