Volts x Resistance = A Thing?

Hello,
I have this voltage divider circuit:

Vin = 5V
---
 |
 <  R1 = 10K
 >
 |
 +----o Vout
 |
 <  R2 = 510R
 >
 |
---
 -  GND

and this formula for finding Vout:

Vout = (Vin x R2) / (R1 + R2)

My question is about Vin x R2. What does the number resulting from Volts x Resistance represent? Using my circuit, I get the following:

5 x 510 = 2250 / (10000 + 510) = 0.24V

What is 2250? is it just a thing, or is it something? I’m guessing some kind of resistance?

Thanks
Karl.

The idea here is that the output voltage of the divider is proportional to the resistor across which the output is taken and inversely proportional to the total resistance of the divider. In other words the output is proportional to R2/(R1+R2) and this is how it is supposed to be read. Trying to read just a portion of the formula eg R2xV1 makes no sense.

Hi,

No, it hasn’t any phisical meaning, is just an operand that appears in the middle of the calculation.

Best Regards

You are not going to like the answer but its VΩ (or kg2 m4 A-3 s-6 in fully SI). Not a very useful quantity on it’s own. But if you divide it by Ω again you end with V :slight_smile:

The basic equation is V=IR, which also gives you I = V/R.
(I being the single current that flows through the combined series resistors.)

So Vout = I * R2
and I = Vin / (R1+R2)
giving Vout = (Vin / (R1+R2)) * R2 = (Vin * R2) / (R1+R2)
Which is the same as your equation. The Vin*R2 is just an artifact of the algebra, and has nothing to do with the physics (taken on its own.)

Vout = (Vin x R2) / (R1 + R2)

Despite what the others say an analysis of units in an equation can give you an insight into what is going on.

For a start the left hand side is in volts so that means the right hand side has to be in volts as well. Vin is in volts and R is in ohms But R2 / (R1 + R2) is in ohms top and bottom and thus the units of ohms cancel out just leaving the units of volts. So the equation is correct with regard to units volts = volts.

Now for a bonus prove that the units of capacitance multiplied by the units of resistance gives the units of time.

Now for a bonus prove that the units of capacitance multiplied by the units of resistance gives the units of time

Of course it does, since t=R*C :grin:

Watcher: Of course it does, since t=R*C :grin:

Please show your working.

That is just a statement not a proof.

For extra credit, show that the units on both sides of E = mc2 are equal.

Pete

el_supremo:
For extra credit, show that the units on both sides of E = mc2 are equal.

Pete

Not sure what you are getting at but yes this can be done. This is because that equation was derived, not just guessed.

wikipedia: A farad has the base SI representation of: s4 × A2 × m−2 × kg−1
and:

s4 × A2 × m−2 × kg−1 * kg * m2 * s-3 * A -2 =
s4 * s-3 × A2 * A -2 × m−2 * m2 × kg−1 * kg = s QED

-----------------------------------------------------

show that the units on both sides of E = mc2 are equal.

What’s to show? Ek= 0.5 mv2 is essentially the definition of kinetic energy; mc2 has exactly the same units.

Well a bit mathematical with little feel for what is going on, I'll give you a better explanation in the morning, it's late here.

Now for a bonus prove that the units of capacitance multiplied by the units of resistance gives the units of time.

Should you are pointing to the time factor that characterizes a RC circuit, yes, it has a phisical meaning. I (humbly) doubt that you can find something similar (even being indulgent; even with a properly rested mind) in the issue of the VxR case.

Best regards.

Well the trouble with reply #10 is that it starts out with a statement that already contains the answer:-

A farad has the base SI representation of: s4 × A2 × m−2 × kg−1

But how to prove that?
Well as we all know Time Constant = R * C
but in terms of units we have t = R * C
now take the right hand side and break it down. From ohms law we know that R = V / I and we also know that Capacitance C = Q / V where Q is the charge in coulombs and V is voltage. So in units that right hand side is

(V/I) * (Q / V) when this is rearranged this is equal to (Q V ) / ( I V)
With a V on the top and bottom it cancels and just leaves Q / I
Now Q is charge and I is current which is the flow of charge. One amp is the flow of one coulombs per second so we can replace I to give us:-
Q / Q t -1 because anything “per” is the reciprocal or one over.
Again there is a Q on the top and the bottom so they cancel leaving
1/ t -1
Which when you take t to the top simply becomes
t

Therefore the units of the right hand side of the equation are the units of time.
QED

. I (humbly) doubt that you can find something similar (even being indulgent; even with a properly rested mind) in the issue of the VxR case.

No it was not V * R, it was V * (R/R) the two Rs cancel leaving just the units of voltage.

But he’s not talking about UR (of VR for some weird parts of the world :p)… He’s talking about R*C.

But for the points :smiley:

[ohm] = [V] / [A] = [V] / ([C] / [s]) = [V] * [s] * 1 / [C]
[F] = [C] / [V] = [C] * 1 / [V]

[ohm] * [F] = [V] * [s] * 1 / [C] * [C] * 1 / [V] = [s]

1/x conversion to make it more clear in ascii…

UR doing that for UR gives you V J s C-2 which is kind of abstract :stuck_out_tongue:

Damn, Mike was fist :stuck_out_tongue:

No it was not V * R, it was V * (R/R) the two Rs cancel leaving just the units of voltage.

No, It was VxR (¿U?; what does "U" stand for?. (V)oltage much better)

My question is about Vin x R2. What does the number resulting from Volts x Resistance represent? Using my circuit, I get the following:

Regards

vffgaston:
No, It was VxR

Sorry no read that first post again:-

and this formula for finding Vout:

Vout = (Vin x R2) / (R1 + R2)

My question is about Vin x R2. What does the number resulting from Volts x Resistance represent? Using my circuit, I get the following:

In this function the units the R cancel. So he never has the units of VR he has them of VR/R which is V.

If you want to know what units VR are in then substitute V/I for the R to give
V*(V/I)
Which is V2/I
Or volts squared per amp

If you want to know what units VR are in then substitute V/I for the R to give V*(V/I) Which is V2/I Or volts squared per amp

For my understanding the OP asks whether the VxR product has a phisical meaning or not, not just an issue of units coherence.

Best regards.

Yep, and like I said in my fist post it does not.

And yeay, U is what we Europeans use. We like to separate symbol and unit from each other. Like, why do you use I for current and not A? Would make more sens if I follow your reasoning. ;) But he, just kidding ;) VS vs the world :p