# Basic Voltage

#### EdgarW

Joined Mar 6, 2017
2
DC 28 volts single bulb circuit; could someone kindly explain how the single bulb can stop the voltage at the bulb.
The voltage before the bulb is clearly 28 volts and the voltage after the bulb is clearly 0 volts; how does the bulb lite and then stop the voltage.
Didn't have any trouble finding text to say this will happen, but not how the bulb stops the voltage. Thanks

#### LowQCab

Joined Nov 6, 2012
3,580
The Bulb has what is known as "Resistance".
There has to be something in the Circuit to limit how much Current can flow
or there would be "infinite" Current, and maybe even a Fire or Explosion.

All this is governed by "Ohms-Law",
( that is until You start getting into more advanced concepts ).

Ohms Law is simply a set of Math-Equations.
.
.
.
.

.
.
.
.

#### MrChips

Joined Oct 2, 2009
29,841
DC 28 volts single bulb circuit; could someone kindly explain how the single bulb can stop the voltage at the bulb.
The voltage before the bulb is clearly 28 volts and the voltage after the bulb is clearly 0 volts; how does the bulb lite and then stop the voltage.
Didn't have any trouble finding text to say this will happen, but not how the bulb stops the voltage. Thanks
Your wording is not quite right. The bulb does not stop voltage.
The bulb is a resistor. It resists current flow.

If you apply any voltage across the bulb, the current flowing in the circuit is given by Ohm's Law:

I = V / R

For example if V = 28V and R = 100 ohms,
the current I = 28V / 100Ω = 0.28A or 280mA

#### DickCappels

Joined Aug 21, 2008
10,104
What you see depends on where your probes are placed.

#### BobTPH

Joined Jun 5, 2013
8,101
The battery (or other voltage source,) has two terminals. One of the terminals is 0V and the other is 28V.

Now, you connect the 0V terminal to one side of the bulb and the 28V terminal ti rhe other side of the bulb. Neither voltage changes. Nothing “stopped the voltage.”. They both remained the same.

So, what is it you think should happen? Should they both become 0V. Or maybe 28V? And if so why?

#### transconductance

Joined Jun 29, 2019
80
Voltage is an illusion we create. It is the difference in potential between two points and we choose the points arbitrarily. The difference in potential is the important part because it's that difference that creates a force on the electrons and makes them move.

What's really flowing isn't voltage. It's electrons rushing along through the wire and through the bulb filament. They are constantly knocking into atoms along the way and making everything jiggle. Subatomic jiggling is heat at a macroscopic level. But the electrons don't stop. They come back out the other terminal. This is why you have to have a complete circuit. Otherwise the electrons would all just pile up at the end of the wire. (In fact they do this and that's how we make capacitors.)

#### dcbingaman

Joined Jun 30, 2021
1,065
This is a semantic issue. 'Voltage' is just how much energy it takes to move a specific amount of charge from one place to another. The specific amount of charge is defined as 1 Coulomb which is equivalent to about 6.25e18 electrons. The energy is measured in Joules and 1 Volt is just 1 Joule/Coulomb of displaced charge. So take the 28V battery you are referring to. Another way to look at this: If you are charging the battery, then is will take 28 Joules of energy to move 1C of charge 'uphill' into the battery. The opposite is also true, any load placed across the battery will provide a path for current to flow, it will flow from the positive terminal of the battery to the negative terminal of the battery (conventional flow) or other direction (electron flow) and start discharging the battery. With each 1C of charge moved, 28 Joules of energy will have been 'used' or 'moved' from one location to another. I don't like the word 'used' because energy is never really 'used' it is simply transferred from one place or form to another place or form. If the load is 'resistive' then that energy will be dissipated as heat into the ambient air surrounding the resistive element. Of in this case a lamp will expend the 28J of energy as both heat and light. If it is a motor then a certain amount will be converted into mechanical work along with heat. If it is a capacitor, then the energy will be transferred and stored in the capacitor for later use. In all these cases energy is being transferred from one location and form to another location and or form. The Volt unit is just how much energy it takes to move 1C of charge from one location to another. This is why energy is transferred using higher voltages for the Electrical power grid. You get more Joules of energy transferred per C with higher voltages which results in higher transfer efficiencies.

#### dcbingaman

Joined Jun 30, 2021
1,065
The Bulb has what is known as "Resistance".
There has to be something in the Circuit to limit how much Current can flow
or there would be "infinite" Current, and maybe even a Fire or Explosion.

All this is governed by "Ohms-Law",
( that is until You start getting into more advanced concepts ).

Ohms Law is simply a set of Math-Equations.
.
.
.
View attachment 274490.
View attachment 274491
.
View attachment 274492.
.
.
I like this notation:

You could apply it to other fields for remembering various formulas, like F=ma, W=Fd, and just about any form of x=yz format.

#### Ya’akov

Joined Jan 27, 2019
8,512
I like this notation:

View attachment 274498

You could apply it to other fields for remembering various formulas, like F=ma, W=Fd, and just about any form of x=yz format.
You didn’t mention the coolest feature of this, simply cover the unknown to know the formula to find it.

#### Ya’akov

Joined Jan 27, 2019
8,512
So you‘ve gotten a lot of good explanations and my purpose in adding another is not to challenge any of them but to try to turn the elephant parts into the whole elephant¹.

Discussions about electric current and its propagation through an electrical circuit that aren’t mostly mathematical are fraught. The reason being that electricity is a field effect and fields are not mechanical things but we tend to imagine the behavior of systems as mechanical.

Lord Kelvin (William Thomson), a famously brilliant and productive scientist, had a lot of trouble with the electrical theories of people like (James Clerk) Maxwell whose equations unifying the electric and magnetic forces into electromagnetism departed from the mechanical. He said:
I can never satisfy myself until I can make a mechanical model of a thing. If I can make a mechanical model, I can understand it. As long as I cannot make a mechanical model all the way through I cannot understand.
He didn’t recognize the incredibly brilliant insight of Maxwell’s equations, now the foundation for our understanding of electricity (that is, electromagnetism, which cannot be divided and be correct) as fields and waves. When Maxwell submitted his work for publication in 1864 Kelvin wrote to him in late 1865, saying:
I am sorry for being slow, I‘ve read most of it and it seems entirely suitable for publication
What seems like praise for Maxwells work is a startling understatement. If Kelvin had actually understood it he’d have been amazed at the power of the equations. Maxwell’s Equations, with the refinement of others, like Poynting, and built on the foundations laid by Farady, Helmholz, Ohm, Volta, Coulomb, Gauss, and many others², literally changed the world.

But it required an extreme paradigm shift to comprehend. It appeared to involve the very hard to swallow idea of action at a distance. For those embedded in a mechanical worldview, this just didn’t work because electromagnetism does not survive the analogies we use with great practical effect. If you take the various analogies, each useful in its domain, they don not add up to electromagnetism, only Maxwell‘s equations have that power, which was the tremendous insight of Maxwell in working them out.

So you are not to blame for your confusion, it is inherent in the use of analogies, no matter the topic or the analogy concerning it. But, you have to recognize these things are analogies and where they go off the rails. The very best way to understand this is to start with various laws and theorems that help to analyze circuits from (George Simon) Ohm on the relationship of voltage, current, and reissitance, (Gustav) Kirchhoff on the current and voltage in a circuit, (Léon Charles) Thévenin on equivalent resistive circuits, (Edward Lawry) Norton on the equivalent reactive circuits, and others³.

These tools provide a way to understand, analyze, and predict the behavior of electrical circuits—but they do not describe why this behavior is the case, and they cannot be used to work that out independent of the unified theory of electromagnetism embodied in the work of Maxwell.

So, to fully understand the nature of what is happening in an electrical circuit you must to use the calculus that describes it in terms of fields and waves. You can’t use a mechanical model because it is not mechanical, and you can’t use the math that describes what to expect in a circuit but doesn’t account for the E-field (electric) and B-field (magnetic).

In particular, while if may be the case you are aware that a current flowing through a conductor creates a magnetic field at right angles to the current flow, and that a magnetic field cutting across a conductor does the inverse, these are not just interesting and useful facts, they are fundamental to what is happening in the circuit under examination and if you don’t account for the fields, you don’t have a description of what is going actually going on, nor can you tell when your analogy is breaking down and becoming a lie instead of the truth.

This is not to say the laws and theorems of network analysis are not incredibly accurate in their domains. When working with DC circuits, or AC at low frequencies, you would never even need to have a clue about the fields. But as soon as you try to engineer a motor you begin to involve yourself in this mysterious magnetic world of the B-field, and when you increase the frequency of the AC you find you need to know about the E-field and how it works.

So, all of this is to say that the models (analogies) embodied in the simplified analysis of DC circuits are very useful but they aren’t the elephant and until someone who can see the whole thing, like Maxwell and his collaborators, can describe how the parts are connected together, you are just grabbing on to a trunk, or a leg, or tail and your conclusions will quickly fail to explain things once you go past that.

If you are inclined to math, or even if you aren’t but can understand the explanations and ignore the derivations (as I have to do, sadly), I can strongly recommend the excellent lectures of Professor Walter Lewin of MIT, all available on YouTube. He teaches physics to undergraduates and his lectures on electromagnetism (actually the full MIT course!) are lucid and insightful. It is well worth the watching if you are interested in getting more insight, and if you have the ability to work out the calculus, even better.

I know this is long but I hope it might have provided some insight into why you need to know much more to answer very reasonable questions like the one you pose. If you only have hold of the tail, you can expect to miss out on everything else.

*This is a reference to the fable of the blind wisemen who, upon feeling different parts of an ee plant conclude it is variously like a snake, a tree, a rope, etc. The point being none were wrong but without seeing the whole elephant a wrong conclusion can easily be drawn.

2. You might note that five of the six mentioned have given the names to basic units in the SI system for describing electromagnetism. This is no accident.

3. The descriptions in this list are very simplified for brevity’s sake, the details are important, though so don’t take my short labels as revelatory of the nature of these things.

[EDITED: for typos and minor wording changes for clarity]

#### DickCappels

Joined Aug 21, 2008
10,104
I was pleased to see him drawing a cloud of electrons rather than drawing the out-dated Bohr model.

#### Ya’akov

Joined Jan 27, 2019
8,512
I was pleased to see him drawing a cloud of electrons rather than drawing the out-dated Bohr model.
His lectures are very up to date and rigorous. Worth watching even if you don’t “need” them

#### MisterBill2

Joined Jan 23, 2018
16,550
That "out-dated Bohr Model" and the associated "Lewis Dot System" are convenient tools for balancing chemical reactions, although they do not represent the physical reality. My high school chemistry teacher explained that atoms are closer to a cotton ball with jelly beans poked into it, and the cotton and beans representing probability clouds.
Of course the analogy struggles when comparing electrical field to cotton balls. Useful but not accurate.

#### Tonyr1084

Joined Sep 24, 2015
7,549
Two meters showing 28.00VDC and 0.00VDC. Why? Because on the left upper meter you're reading across the battery. The lower central meter is reading the same wire. There should be 0.00VDC there because you're not reading across a power source.

Regardless of AC or DC, you're just reading a short as far as the meter is concerned. Don't confuse volts with amps (current). My dad always referred to "House Current of 110 VAC". That was a common way of referring to the voltage at the power plug in your room. (or 220 in other countries). It's the wrong way to think of it. Voltage is equivalent to pressure. How much voltage the lamp can withstand is like a balloon. It can stand its rated pressure. Exceed the pressure and the balloon will pop. Exceed the voltage by enough and you can burn the lamp out.

Current is the movement of electrons through the lamp. If you were to test for current you'd read some amperage related to the voltage and the resistance of the bulb. For giggles let's assume the lamp when fully lit at 28VDC the filament resistance is 22.4Ω. Current is equal to voltage divided by resistance. In this example, 28 ÷ 22.4 = 1.25 amps. The wattage of the lamp is equal to the voltage times the amperage. 28V x 1.25A = 35W. The lamp is dissipating 35 watts of light and heat energy. The current in this theoretical lamp is 1.25 amps.

Using Ohm's Law we can calculate the filament resistance if we know the voltage and the wattage.
The way I'm accustomed to doing it is to calc one part then calc the other. My way is:
35W ÷ 28VDC = 1.25A
Knowing 28VDC and 1.25A you calc the resistance:
28VDC ÷ 1.25A = 22.4Ω

For my example I decided that the lamp was going to be a 35 watt bulb at 28 volts. You must know two factors in order to figure out all other numbers. Like I said this is the way I go about it. There are other short cut ways but I've never gotten used to using them. I should. They're easier I'm sure.

#### EdgarW

Joined Mar 6, 2017
2
Ya'akov; Thank you for being so kind to share your time and knowledge, PLEASE be assured it Is greatly appreciated by most, if not all.

I have spent numerous hours on Lewin’s YouTube lectures, the lectures appear to be very well done; however, in no way suggesting I clearly understood.

Water running down a hill was a confusing moraaaasssssss, until gravity was discovered, then, it became a simple explanation.

Initially, I created a single bulb circuit and measured the 24 volts to the bulb; then,

added a 2nd identical bulb and found the 1st bulb now passed on 12v to the 2nd bulb; then,

added a 3rd identical bulb and found the 1st bulb passed on 16v to the 2nd bulb and 8v to the 3rd.

Then, out of total frustration, disconnected the circuit from Ground, the result was all bulbs passed on 24 volts; that was the end for me – done!

Again, thank you for sharing your time and knowledge that WHAT electricity does is one thing, but HOW and WHY electricity does what it does is another world. Hopefully one day the Gravity of electricity will be discovered and simplify this mess that very few really comprehend.

#### crutschow

Joined Mar 14, 2008
33,348
Hopefully one day the Gravity of electricity will be discovered and simplify this mess that very few really comprehend.
The theory of electricity is well developed and fully explains the how and why of its operation.
There is no "mess" and I very highly doubt there is a "Gravity" of electricity to yet be discovered.

#### Jon Chandler

Joined Jun 12, 2008
852
Maybe the gravity of electricity explains why some things don't work when I turn them over!

#### MisterBill2

Joined Jan 23, 2018
16,550
We do have a term "EMF", which stands for Electro Motive Force, that was very common, and seemed to be the equivalent to voltage. Perhaps not similar to gravity, as it seems to work in all directions..
The frustrating experiment described in post #15 is why study should come before experimentation, applicable i low voltage work and vitally important when experimenting with mains power because of no battery being available. Basic circuit theory would explain exactly what happened. With no current flow there is no voltage drop.

#### MrChips

Joined Oct 2, 2009
29,841
TS needs to start thinking about current and how to apply Ohm’s Law.
I will show an example later.

#### MrChips

Joined Oct 2, 2009
29,841
Resistors don't stop voltages.
Resistors resist current flow.

We use Ohm's Law to calculate current.

I = V / R

In this example the total resistance of resistances in series is the sum of individual resistances.
Rtotal = R1 + R2 = 40Ω + 20Ω = 60Ω
The current through each resistor is calculated using Ohm's Law,

I = V / R = 120V / 60Ω = 2A

The current flowing in each resistor is the same.
Working backwards, we can calculate the voltage across each resistor.
V = I x R

V1 = I x R1 = 2A x 40Ω = 80V
V2 = I x R2 = 2A x 20Ω = 40V

Note that V1 + V2 = VT, i.e. 80V + 40V = 120V
There is no voltage loss. There is no stopping of voltage.
Some people would say "R1 drops 80V and R2 drops 40V" or "the voltage drop across R1 is 80V".
I prefer to say "the voltage across R1 is 80V and the voltage across R2 is 40V" and avoid using the word "drop".

Reference: https://ecampusontario.pressbooks.pub/multimeters101/chapter/voltage-drops-in-series-circuits/