In the general case of an unloaded two-resistor voltage divider in which we don't have a drawing, I would describe the situation as having two resistors, Ra and Rb, in series. The voltage at the end of Ra and Va and the voltage at the end of Rb is Vb. The output, Vo, is the voltage at the junction of Ra and Rb. With this we have all that is needed to analyze the circuit in terms of these quantities (and without resorting to assertions about the resistor on the bottom always being at some place in some regurgitated formula).

The current flowing from Vb to Va is (Vb - Va)/(Ra + Rb). The voltage at Vo is then simply the voltage at one end plus the voltage drop/gain due to that current flowing through the resistance between that end and Vo. Start at whichever end you like, the results will be compatible. Either

Vo = Va + [(Vb - Va)/(Ra + Rb)]·Ra

or

Vo = Vb - [(Vb - Va)/(Ra + Rb)]·Rb

Both of these yield

Vo = (Va·Rb + Vb·Ra)/(Ra + Rb)

Notice the nice symmetry of this result -- not "top" or "bottom". Also note how it reduces trivially to the classic voltage divider equation you are referring to when Va or Vb happens to be zero.

EDIT: Fixed errors pointed out by anhnha -- thanks!

Hello again,

Well if you look around on the web you'll almost always see the voltage divider depicted with just two resistors and one is connected to the source voltage and one is connected to ground for simplicity. That's how we end up with:
Vout=Vin*(R2/(R1+R2))

where the factor:
R2/(R1+R2)

is called the "divider ratio".

So it just looks like you went out of your way to complicate this simple idea by insisting that we must include another second voltage source in the problem when in most cases (and in particular the one shown originally in this thread and the one under question) there is no such secondary voltage.

But dont take my word for it, look around on the web see what you can find.

Also, TOP and BOTTOM resistors are the "friendly" names and in my experience with other people in real life we've always called them that, and they always understood what it meant. This is because the TOP resistor almost always appears on top schematically, and the BOTTOM resistor almost always appears on the bottom schematically. They are drawn like that because that's the way most people like to show voltage dividers. That doesnt mean that they MUST be drawn that way, but most people draw them that way. Yes, the TOP resistor could be drawn on the bottom and the BOTTOM resistor could be drawn on the top, but that's a physical view not an electrical view. Even if they were drawn like that, we could still call the resistor on the bottom the TOP resistor and the resistor on the top the BOTTOM resistor. This comes from the electrical view NOT the graphical view and that might be the point you are missing here.
In fact, even if the voltage source was negative we could still refer to the resistor connected to the source as the TOP resistor in the voltage divider.
If you look at ANY voltage divider, you'll always be able to identify one resistor as having a particular function ALL THE TIME and the other resistor has a different function FOR ANY CIRCUIT with a voltage divider. Thus by giving them 'friendly' names we always know which one is which without even having a schematic. For a funny but true example, we could call one TOM and the other JERRY and the one called TOM could always be the one on 'top' (connected to the voltage source). Think about this and see if you can recognize this fact, and of course if you have some ideas on a better set of friendly names you might mention that too.

Now let me give the formula in light of this view:
Vout=Vin*BOTTOM/(TOP+BOTTOM)

and you can see that the BOTTOM resistor is in the top of the equation.
Now if we called these two resistors TOM and JERRY then we would have:
Vout=Vin*TOM/(TOM+JERRY)

and this would AALLWWAAYYSS hold in EEVVEERRYY circuit

Also, there is no reason to drag in a secondary voltage as that just makes the explanation more complicated for someone who doesnt even now how a basic voltage divider with one source works yet, and when the original circuit was a question about a single source voltage divider.
However, one of the secondary issues that comes up more frequently is the voltage divider with a LOAD connected to the bottom resistor. The loading effect is often explored and depicted in the equation.

The choice is always yours of course, and you've shown your formula now, so you should be happy

Also, to you and/or anyone else that celebrates this day, have a good holiday

 

Hello again,

Well if you look around on the web you'll almost always see the voltage divider depicted with just two resistors and one is connected to the source voltage and one is connected to ground for simplicity. That's how we end up with:
Vout=Vin*(R2/(R1+R2))

where the factor:
R2/(R1+R2)

is called the "divider ratio".

So it just looks like you went out of your way to complicate this simple idea by insisting that we must include another second voltage source in the problem when in most cases (and in particular the one shown originally in this thread and the one under question) there is no such secondary voltage.

I answered YOUR question where you SPECIFICALLY asked for the GENERAL CASE in which we don't have a diagram or know anything about the voltages at each end of the two resistors. NOW you are bitching because I didn't assume one end was connected to ground for simplicity, thereby making it a special case?

Really?

 

I answered YOUR question where you SPECIFICALLY asked for the GENERAL CASE in which we don't have a diagram or know anything about the voltages at each end of the two resistors. NOW you are bitching because I didn't assume one end was connected to ground for simplicity, thereby making it a special case?

Really?

You suggested this long before i did.

You just like to argue which is fine with me, if it leads to better information.

Buy hey, post whatever formula you would like too. We all have some favorite formulas, i realize that too.