# Voltage Drop

Discussion in 'General Electronics Discussion' started by fscii, Nov 13, 2012.

1. ### fscii

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Nov 13, 2012
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.

Last edited: Nov 13, 2012

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3. ### fscii

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Nov 13, 2012
I'm not sure what you mean.

In the wiki article, which I have read prior, using conventional theory Vin would be + and how would R2 be primarily responsible for the voltage drop or resulting voltage?

In electron theory, the ground symbol would be - and our source of electrons and thus the formula for Vdrop centering on R2, makes sense.

In physical reality, R2 is the first resistor to get hit with electrons (or Z2 in wiki article).

I think my question really has more to do with how Ohms law applies here.

Checking the Voltage Drop of R2...
Voltage = Current / Resistance... (V=IR)
and we know we have about 1.xx ma in my circuit @ 10k ohms on r2 =
10.9v

How can there' be 10.9v "dropped" if the resulting voltage is 10.9v?

12v - 10.9v dropped = 1.1v So shouldn't vout be 1.1v?

This is the source of my confusion

[Question: Is that wiki image conventional flow or electron flow??? I'm assuming its conventional and Vin = + dc which isn't sending the electrons but the ground is]

Last edited: Nov 13, 2012
4. ### CocaCola

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Apr 7, 2012
The short answer is that voltage is a measure of potential difference not the flow of electrons

Beyond that I have an almost 3 year old hanging on me and an infant cutting two teeth so hopefully someone that can actually type more then 3 words without being inundated with others needs, can elaborate if it doesn't click...

5. ### BobK

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Jan 5, 2010
Voltage drop is a somewhat unfortunate name for the phenomenon. If we say that a 10K resistor "drops" 10V at 1ma, what we are really saying is the the voltage across the resistor must be 10V in order for 1ma of current to flow though the resistor. The higher the resistace, the more voltage that is required to make the same current flow. This makes sense, if the resistor resists more, it will require more voltage to push the electrons through.

Bob

6. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
CC, have you researched how that happened?

Chris

7. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Short answer is It doesn't matter which convention is used. The results will be the same.

Chris

8. ### fscii

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Nov 13, 2012
ok Cocacola u got me lol yes potential not actual current. Hmmmm. Then why always R2 used to figure out the drop between the two? Why not R1? I figured it had to do where the source of electrons was coming from. I know it matters somewhat because if I reverse the two (e.g. conventional vs electron) the #'s reverse.

Ok Bob I think you understand my conundrum when talking about what is "voltage drop" and now you say it doesn't mean that it dropped X volts from the circuit but that X volts is required across that component to drive the current thru.

Ok but that also confuses me a bit. Between the two resistors we now have 10v. So between the + terminal and the 10v we have a 1kohm resistor. How do the remaining 10v vanish just because a 1k ohm resistor is between them

Can you see what I'm driving at? Ohms law says the 10k resistor drops 11v (when in reality it seems like it DROPS 1v hence 10v available @ Vout). Ohms law says the 1k resistor drops 1v (when on the meter it appears to drop the remaining 10v since we have 10v @ vout and kirchoffs law says the rest must drop [and only the 1k resistor remains to do this!]).

I'm sorry to ask so many noob questions about the same thing, but I know if I move on w/o completely understanding something, I'll regret it later and I really want to understand and enjoy electronics to make and repair cool things!

9. ### CocaCola

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Apr 7, 2012
Because we measure voltage that way AKA the positive side is a positive number and ground or the negative side is usually zero, or at least less than the positive side...

Dumbed down... Maybe it will help or maybe it will confuse

In the complete series circuit 11K of resistance 'consumes' all 12V, aka we start with 12 and end with 0... Think of the series circuit as a 12 to 0 ladder... Now, consider that 11K as eleven 1K resistors, or steps on the ladder... With each step away from the 12 at the top that ladder the step removes 11/12ths of a count...

Now, start at the top number of 12, before we jump any resistors your value is 12, now jump down 1 step towards zero, the result? Now jump two steps, the result? Now, jump all 11 steps, the result?

10. ### CDRIVEHauling 10' pipe on a Trek Shift3

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I've heard of this. It's called a "ladder schematic".

Chris

11. ### fscii

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Nov 13, 2012
Well if I put the black probe on - terminal (some say ground but its the true source of current!) and red probe on Vout = 11v. If I put the red probe on + (the real ground lol ty ben franklin!) and black on Vout = 1v. So kirchoffs law etc...

I'm not sure I understand the ladder thinking...

Because if we took 11 steps down we would be at one volt when crossing the 11k resistor, but we are at 11v. That's the source of my confusion

[keeping in mind black probe on - terminal, red probe on Vout = 11v]

There's some fundamental thing I'm not understanding or grasping here. ty for bearing with me. I truly appreciate it as much as I truly need to understand this.

12. ### CocaCola

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Apr 7, 2012
No we are at about 1V...

13. ### fscii

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Nov 13, 2012
But vout comes out to 11v not 1v

I attached a diagram of the circuit
(basically the same as: http://en.wikipedia.org/wiki/Voltage_divider )

I colored the wire segments differently to help (mostly myself) explain things.
When I put the black probe onto the - and red probe onto Vout I get 11v.
V=I*R and it works out to 11v but that doesn't explain voltage DROP cuz it only dropped 1v not 11v.

Vout = (R2 / (R1+R2)) * V == 11v

So how did it DROP 11v, when the meter reads 11v?

Original voltage 12v now if we DROP 11v it should read 1v

Keeping in mind the electrons going thru the 11k resistor I'd expect the big drop to happen there but we didn't lose 11v we lost 1v ?!

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14. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Study this simple circuit. You will note that there is no common or ground in this circuit. Flip it upside down or rotate it 180°. Nothing changes. Currents are all the same in a series circuit. Voltages and polarities are relative to the two points (nodes) measured.

Chris

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15. ### CocaCola

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Apr 7, 2012
That is because you only made one hop (1K) down from the +12 units of potential towards the 0 units of potential, thus 11V just like we expect... Again we are not concerned with the direction or flow of electrons... What you are looking for is the fraction of the starting potential between the two potentials...

You only dropped 11/12ths of a volt because you only made one hop down... Remember that 11 resistors 'consume' the entire 12 volts and it's equally 'consumed' by each resistor... Split the drops into the steps like I said...

As CDRIVE said flip the resistor network upside down and the reverse happens, as you had pictured...

Again forget about the 'direction' that the electrons are flowing and look at the units of potential on each end of the circuit...

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16. ### fscii

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Nov 13, 2012
If it doesn't matter which way the electrons are flowing, then why are we assuming positive gives 12v potential and negative 0? I'm missing that point.

I realize the ben franklin/conventional stuff is backwards so it helps me to often visualize what is really going on physically. I also understand potential (to a degree) meaning how badly the electrons want to go from negative to positive. That's why I couldn't grasp why the electrons crossing a 10k resistor didn't "drop" 11 volts but only 1v.

I guess the term "voltage drop" confuses me and I haven't had it defined properly but I think Bob came close in saying its an unfortunately confusing term to use and means that R2 has 10v across its connections.

Also, interestingly, if you put the black probe on - terminal and + probe on Vout you get the 11v. If you put the red probe on + terminal and black one on Vout you get 1v. An interesting view that potential is merely the difference between the two points and not absolute.

So is it true then that I could connect either a 10v device or a 1v device to Vout? Assuming the 1v device would treat Vout as - and other terminal connected to +... and if I wanted to use it as 10v I'd connect the + terminal to Vout and - terminal to - (ground)???

Thanks for all of your help guys I really appreciate it. I'm learning from the following books: Electronics for dummies and Make Electronics. Also the articles I found on this site.

There are other things that confuse me too lol I'll post them as I go and hopefully someday I'll be able to post answers to someone new once I understand it all

17. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
When measuring between any two nodes one node will always be zero with respect to the other node. The other node can be a positive or negative voltage with respect to the node we're calling zero.
The voltage drop across a resistor will be equal the resistance times the current through it or E= I*R.
The term may have been coined very early in the study of electron flow. Possibly when experimenters first noted that the voltage at the end of a long conductor run was lower than the source end of the conductors. Volta may have said " Hmmm... the voltage appears to have dropped." This guess is as good as anyone's I think.
Yes it is absolute, as per my first quote.

Chris

Last edited by a moderator: Nov 18, 2012
18. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
I'm not sure if this will help, but I have created an analogy for voltage, current and resistance that may help you understand their interrelationship more intuitively.
https://www.electronicspoint.com/newbie-questions-t248766.html#post1470672
This analogy is NOT compatible with the traditional analogy of voltage = pressure, current = flow, resistance = thin pipe, that is traditionally used to explain these quantities, but it explains voltage, current, resistors, voltage dividers and capacitors in a way that (I think) is easy to grasp.

19. ### fscii

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Nov 13, 2012
Thank you that is helpful in the voltage divider problem for sure!

I think I'm starting to understand it more, I'm glad I found this board. I am reading "Electronics for dummies", "Make Electronics" and trying to intermix the tutorials on this site: https://www.electronicspoint.com/electronics-tutorials-f101.html to gain an understanding of things.

Thanks again everyone for helping me with this!