# Still trying to wrap my head around Ohm's Law.

#### stormBytes

Joined Jan 26, 2010
43
So after a 27+ response thread which didn't really do much to further my grasp of the concepts, I decided to really tackle the basics.

Example:

Assuming I've got a 12V power supply and I'd like to power an LED with 5V at 25mA. I know that I take the initial voltage (12V), subtract the desired final voltage (5V) and now I know that I need to get rid of 7V while restricting the current to 25mA.

Using Ohms' Law I *know* that I have to divide 7V by the desired current (25mA) which tells me to use a 280Ω resistor. I know that if I plug this resistor into my circuit, that the resistor will [have a voltage] drop [of] 7V and restrict my current to the desired 25mA.

What I don't understand is WHY or HOW this is happening - eg. the actual *mechanism* behind this, and I'd like someone to explain this to me, step by step, on a 4-year-old, 'draw-me-a-picture', level.

I'm trying to wrap my head around this one concept - "voltage drop across a resistor". While probably simple in this example, I don't understand why - in a simple series circuit - 'any' resistor drops 100% of 'any' voltage in 'any' circuit. (so far I've played around with a breadboard, and 5-12V, and this is what I've observed)

Another example, if we forget the LED for a moment, I place a single resistor between the terminals of a battery, the resulting voltage drop across the resistor (I'm assuming that "voltage drop across a resistor" means voltage 'consumed/neutralized' by the resistor) is near 100%, while if I place an LED into that circuit, all of a sudden the 'voltage drop' is 'shared'. Why does it work this way? How can this be calculated/anticipated? What am I missing?

It's baffling! Ughh...

I really appreciate the input I've had from many of the good and knowledgable folks on here, however I want to state in advance that I'm not simply looking for "the answer". Rather, I'd like to understand the fundamental mechanics of the circuit. I say that because I appreciate your time/input and I'd like to be clear on *specifically* the type of response I'm looking for so as not to waste it.

If others might benefit from technical, short, scientific answers to this question, then by all means - post away. I however will not, and as such, I wanted to clarify specifically what type of explanation I'd like to have.

This place is really terrific and I'm glad I came across it. I am certainly doing LOADS of reading on my own and hopefully, these concepts will come together in my brain at some point in the not-so-distant future.

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#### Wendy

Joined Mar 24, 2008
23,122
Assuming I've got a 12V power supply and I'd like to power an LED with 5V at 25mA. I know that I take the initial voltage (12V), subtract the desired final voltage (5V) and now I know that I need to get rid of 7V while restricting the current to 25mA.
This statement doesn't really make sense. If you have a 12VDC power supply then the 5V doesn't make sense. It simply is not involved. If you had 5V then the 12V wouldn't be involved. Given this, the rest of your logic is flawed beyond salvage.

The formula is simple. Don't overcomplicate it.

Voltage = Current Resistance.

You have another issue, LEDs are not linear devices. They drop a constant voltage, which is referred to Vf (forward dropping voltage). I have an article for LEDs.

LEDs, 555s, Flashers, and Light Chasers

The reason the Vf is significant is it is a constant voltage, it doesn't follow Ohm's Law. If a LED drops 2.5V (typical for red) then it is subtracted from the power supply. If you use 12VDC then the number becomes 9.5VDC.

Using Ohm's Law...

9.5VDC ÷ 0.025A = 380Ω

If you want to chop this 9.5V into smaller voltages you can, but the resistance between two resistors will still add up to 380Ω, always, for everything to balance.

I suspect you are wanting to figure out voltage dividers, which also use Ohm's Law, but don't relate to biasing LEDs. You are trying to mix too many things into the mix, while defining it as a simple problem. It is two separate problems, and should be treated as such.

• stormBytes

#### stormBytes

Joined Jan 26, 2010
43
Bill,

Thanks for taking the time to respond to my thread. I've edited the original post since your reply (should probably start using the 'preview post' feature). Perhaps it makes a bit more sense now. My head's a mess! Trying to keep track of all the concepts, figuring out exactly what it is I don't get is probably half the battle. Forgive my incoherences, I'll try to clarify:

This statement doesn't really make sense. If you have a 12VDC power supply then the 5V doesn't make sense. It simply is not involved. If you had 5V then the 12V wouldn't be involved. Given this, the rest of your logic is flawed beyond salvage.
Actually, I'm following a book. In retrospect perhaps not the wisest \$20 I've ever spent. The book introduces resistors in terms of 'reducing voltage' and limiting current. It assumes you're using a 9V battery to power an LED, and the example given is something along the lines of:

"An LED needs around 2.1V and roughly 2.5mA, so to reduce 9V to 2.1V we need to get rid of 6.9V. To accomplish this we divide [the difference] 6.9V by the desired current, 2.5mA and get 276 or a 280Ω resistor value for our circuit"

This is where the voltage reduction (i use the term loosely) issue comes from in my question above.

I don't even know if I'm asking the right question(s). I'm trying to gain a fundamental understanding of resistors, and how these interact with voltage and current. While the latter is relatively simple (I realize that resistors 'restrict' the flow of current), the former just doesn't 'click' in m brain.

So once again -

in a simple series circuit - 'any' resistor drops 100% of 'any' voltage in 'any' circuit. (so far I've played around with a breadboard, and 5-12V, and this is what I've observed). Introduce an LED or another resistor (if that simplifies things) and all of a sudden the voltage drop across R1 is reduced! I don't get it?!

If you've got 5 half-closed valves along a water pipe, the pressure-drop that occurs after the FIRST valve simply does NOT vary with the number of valves downstream. Also, the first half-closed valve does NOT reduce the pressure to absolute zero! While any resistor in any circuit I've built so far (voltages: 3-15V, resistors used: 100Ω - 10kΩ) have 'neutralized/reduced/consumed/whatever' 100% of the voltage in the circuit. So basically, as I've said, it'd be nice to wrap my head around this because after 3 days of reading/posting/youtube I'm frankly at wits end.

I suspect you are wanting to figure out voltage dividers, which also use Ohm's Law, but don't relate to biasing LEDs. You are trying to mix too many things into the mix, while defining it as a simple problem. It is two separate problems, and should be treated as such.
Bill - I'm about as close to 'figuring out voltage dividers' as I am to setting up a lawn chair on the moon and sipping cappuccinos. While I've learned a great deal in the course of breaking my head over this one issue, I've yet to have it all come together. Right now I'm just looking at a series of random dots.

#### Potato Pudding

Joined Jun 11, 2010
688
Resistance controls the ratio of Voltage to Current.

You have 12 Volts across 100 Ohms and it will be 120milliAmps.

You have learned the math but you still have a disconnect to the theory.

There is a reason why you can divide the Voltage by a Current to get the value of resistance that fits.

The water analogies are ok but try to think of it in terms of what is really happening.

Resistance is resistive conduction. It takes power to get current through a resistor. The more resistance the more power it takes. Current through a resistor bleeds off heat.

Because of the power requirement the resistance sets up E=IR, aka Ohms law.

You can change the voltage and the resistor will still conduct. It will just change the current.

#### eblc1388

Joined Nov 28, 2008
1,542
Resistor does not drop current or voltage automatically. The concept of whatever value of resistance drop whatever voltage IS Wrong.

So there isn't a particular value of resistance which will drop 9V into 5V, generally speaking.

The correct way to understand the idea is that "a voltage develops across a resistor and its magnitude depends upon the current flowing through it and the resistance value"

Our Mr. Ohm said that if the resistor is a linear one, then v is proportional to the current flowing and the proportional constant is defined as the resistance.

So, any resistor, regardless of its resistance, will not drop voltage if there is no current flowing.

#### wes

Joined Aug 24, 2007
242
Ok so from my understanding of your questions you want to know why is it that the voltage drop across 1 resistor or led in a circuit with a voltage source always equals 0. correct.

but why is it when you introduce another resistor or led (so 2 now ) the voltage drop across any one is not 0, correct

So from that, the only thing I can think of to explain your issue is it's the way you are measuring the voltage. Since the battery or voltage soure alway's has a positive side and a negative side and you can think of one as plus 12 volt's and the other as 0 volts. so when measuring the battery your volt meter would get positive 12 volt and if you switch it give's -12 volts.

So when you now place a resistor into the circuit and measure the voltage between the resistor terminal instead of the battery you get a lower voltage then the battery voltage, let's say 6 volts, so your voltage drop is 6 volts. Again if you measured between the battery terminals , even with the resistor in the circuit, the voltage would be 12volts. now add a second resistor or led of the same rating and now you measure the voltage across the second resistor and what you should find is the voltage is zero. Because the voltage drop across the second resistor or led was again 6 volts and since there were 2 resisitors or led, the total voltage drop is 12 volts, exactly the same as the applied battery voltage of 12 volt's.

I believe this called Kirchhoff's Voltage Law
I think this will help to explain alot

If I am wrong then please someone correct me. I don't want to think I know something and be totally wrong about, lol

#### wes

Joined Aug 24, 2007
242
actually I am wrong on one part the 2nd resistor would also read 6 volt's, half the applied voltage since there are 2. the only time you would read zero is if you measured between the 2nd resistor and the negative terminal of the battery , same would apply if you measure between the positive and the 1st resistor.

#### stormBytes

Joined Jan 26, 2010
43
Resistor does not drop current or voltage automatically. The concept of whatever value of resistance drop whatever voltage IS Wrong.
What can I say? Perhaps you'd care to engage my DVM in debate. Based on the measurements I took, in a circuit that had nothing more then a fixed resistor across the terminals of a power supply (for all intents and purposes), the voltage across the resistor was just about 100% (again, for all intents and purposes) of the supply, regardless of what voltage the supply was set to. The voltage between the output-leg of the resistor and the positive terminal of the supply was just about zero.

The book said, you're starting out with a 9V battery, you need to get rid of 6.1 V. So 9V - 6.1V = 1.9V which is what's needed to power (and not fry) the LED. Then you want to have 25mA of current. So you divide 6.1/0.025 and you get 276 or 280Ω.

RThe correct way to understand the idea is that "a voltage develops across a resistor and its magnitude depends upon the current flowing through it and the resistance value"
WHY does the magnitude of voltage that develops across a resistor 'depend' on the current flow or the resistor value?? Voltage is, as I understand, the force of electrons repelling each other. A 12V potential difference (voltage) between 2 points means that there's '12V' of 'repulsion difference' - meaning that point-N is 12V 'more negative' then point-P. As I understand it, point-N is either more negative or NOT. What difference does it make how many electrons are actually 'flowing' between these points? Voltage, as I understand is a measure of "electromotive [POTENTIAL] force' -

By all means, correct me where (not if!) I'm wrong Our Mr. Ohm said that if the resistor is a linear one, then v is proportional to the current flowing and the proportional constant is defined as the resistance. So, any resistor, regardless of its resistance, will not drop voltage if there is no current flowing.
I don't understand why a resistor should affect Voltage at all!? Again - Voltage, as I understand it is a measure of 'potential' difference. Who cares if the electrons are at the resistor, an inch away, 2 inches, a yard... The potential difference between point-N & point-P should, in (my) the idiot's mind remain unchanged.

#### wes

Joined Aug 24, 2007
242
The only time you would read 0 or close on the second resistor terminals is if the first resistor had a very high ohm and the second was a low ohm otherwise if they are equal then the voltage would be equal across each one

sorry if I am making this confusing, lol

#### stormBytes

Joined Jan 26, 2010
43
Okay, this warranted a second-reply. If any n00b ever reads this and has a problem understanding Ohm's law, then READ THE FIRST LINKED ARTICLE quoted above! That article is the best thing since sliced bread! And while it doesn't explain away everything (I'm not a natural-born physicist in case you haven't gotten that by now) it really helps put a lot of things in perspective.

I'm going to read the other two linked resources (printed them off..) but that first one's a 5-star deal.

Again, thanks for sharing!

#### Jony130

Joined Feb 17, 2009
5,439
in a simple series circuit - 'any' resistor drops 100% of 'any' voltage in 'any' circuit. (so far I've played around with a breadboard, and 5-12V, and this is what I've observed). Introduce an LED or another resistor (if that simplifies things) and all of a sudden the voltage drop across R1 is reduced! I don't get it?!
For this simple circuit with only one resistor R1 all voltage apply by voltage source Vin will be present "across" resistor R1. Because in this circuit we have only one resistor.
So from Ohm's law
I = Vin/R1
For example if Vin = 10V and R1 = 100Ω then current is equal:
I = 10V/100Ω = 0.1A = 100mA

And know if we add another resistor R2 is series with R1 we will have different situation. Total resistance of a circuit is now lager (Rtot=R1+R2)
So current is smaller, for example if R1=R2=100Ω then current in the circuit is equal
I = 10V/ (200Ω) = 0.05A = 50mA
So if current is smaller voltage drops on the resistor also will be smaller.
V1 = I * R1 = 50mA * 100Ω = 5V
V2 = I * R2 = 50mA *100Ω = 5V
But the sum of all voltage drop across resistors must be equal Vin
Vin = V1 + V2

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#### stormBytes

Joined Jan 26, 2010
43
Ok so from my understanding of your questions you want to know why is it that the voltage drop across 1 resistor or led in a circuit with a voltage source always equals 0. correct.
Forget the LED, that's too advanced for me Right now I'm keeping this really *really* simple. One battery/source, one resistor going across. That's it.

Since the battery or voltage soure alway's has a positive side and a negative side and you can think of one as plus 12 volt's and the other as 0 volts. so when measuring the battery your volt meter would get positive 12 volt and if you switch it give's -12 volts.
Good so far...

So when you now place a resistor into the circuit and measure the voltage between the resistor terminal instead of the battery you get a lower voltage then the battery voltage, let's say 6 volts, so your voltage drop is 6 volts.
No, if I got 6V I'd be happy as a school boy! When I measure the voltage between the output-terminal of the resistor (assuming the input-terminal is connected to the negative/ground terminal of the battery) and the positive terminal of the battery/source, i get next to zero Volts! That's what I'm not getting! (considering the water analogy)

Again if you measured between the battery terminals , even with the resistor in the circuit, the voltage would be 12volts.
Correct. The measurement would then be 'potential difference' between the two terminals of the battery, completely avoiding the resistor.

now add a second resistor or led of the same rating and now you measure the voltage across the second resistor and what you should find is the voltage is zero. Because the voltage drop across the second resistor or led was again 6 volts and since there were 2 resisitors or led, the total voltage drop is 12 volts, exactly the same as the applied battery voltage of 12 volt's.
Again if this was the case, everything would make sense (assuming all components are being connected in series). R1 eats up the first 6V, R2 eats more up, Led1 eats up the rest. We're all happy & dandy!

Problem is -

When I connect the FIRST resistor across the terminals (nothing else in the circuit at this point), the voltage (between R1's output terminal and the remaining battery terminal) is ZERO, and the voltage ACROSS R1 is 100% (or nearly) of the battery output. Now if I connected R2 in series with R1, the voltage across BOTH (connected in series to one and other) will equal whatever the battery's spitting out, and the voltage across each INDIVIDUAL resistor will vary with that resistor's value. If I then add an LED to the circuit (again in series), the voltage is re-divided AGAIN! Like that piece of cheese that keeps getting re-divided depending on how many mice are vying for it.

I believe this called Kirchhoff's Voltage Law
If it is, I'd certainly love to give that Ruski a piece of my mind! ughhhhhh!

In the meantime, thanks for posting the resource. I'll have a look as soon as I'm done reading through another 200 pages of everything-and-the-kitchen-sink! Last edited:

#### stormBytes

Joined Jan 26, 2010
43
For this simple circuit with only one resistor R1 all voltage apply by voltage source Vin will be present "across" resistor R1.

View attachment 22319

Because in this circuit we have only one resistor.
So from Ohm's law
I = Vin/R1
For example if Vin = 10V and R1 = 100Ω then current is equal:
I = 10V/100Ω = 0.1A = 100mA

And know if we add another resistor R2 is series with R1 we will have different situation.

View attachment 22320

Total resistance of a circuit is now lager (Rtot=R1+R2)
So current is smaller, for example if R1=R2=100Ω then current in the circuit is equal
I = 10V/ (200Ω) = 0.05A = 50mA
So if current is smaller voltage drops on the resistor also will be smaller.
V1 = I * R1 = 50mA * 100Ω = 5V
V2 = I * R2 = 50mA *100Ω = 5V
But the sum of all voltage drop across resistors must be equal Vin
Vin = V1 + V2
Yes I realize this, and I get the math. I'm clueless about 'mechanism' - WHY is this happening? What's going on (on a 4-year-old's level) with the electrons that 'accounts' for this law/phenomenon?

#### wes

Joined Aug 24, 2007
242
okay well I guess I will give my 2 sense on why I think the voltage across the resistor (just one resistor in the circuit) is 12 volts and not 0 like when you measure between the resistor terminal and the battery terminal.

If you think of the resistor as a road block and the wire as the highway.
then if the circuit was nothing but wire then the voltage drop across it would be 12v , so you would read 0 volts. The reason because the electrons are never encountering much resistance except the wire but that is small, so they are like cars going from terminal a (+12 volts) which has a excess of electrons constantly being created by the battery to terminal b (0 volts) where electrons are being absorbed. so If you measured between one point on the wire and another then voltage would read 0 because there is no charge build up between those two points of wire. the reason the voltage reads 12 volts across the battery is because the battery has a constant charge build up on one side and absorbing it on the other, so the voltage reads 12 volts. When the battery dies and the charge is equaled out on the otherside of the battery , you would read 0 volt's because there is no excess charge build up on terminal a.

so now on to resistors and why you read 12 volt's across them.
using this you can see that the resistor is like a road block. whether you have a small resistance or huge, it is still 12 volt's because one side will always have a higher buildup of charge on one side then the other. you would think that it would be a bigger voltage with a larger resistance but since this is just one, you will get 12 volt''s because the batterys is always creating a buildup of charge on one end.

So now on to two resistors and why you would read equal voltage across two equal resistors. Since resistor 1 is say 1 ohm the voltage across it would be 6 volts and since resistor 2 is also equal, the voltage across it would 6 volts aswell. but why is that when resistor 1 is say twice that of resistor 2 , so 2 ohm that the voltage read across it would be 8 volts instead of 6 volts and resistor 2 voltage is 4 volts. well back to the roadblock idea. resistor 1 causes a bigger roadblock and so more charge is being built up on the side connected to the battery +12volt and the reason for 4 volts on resistor 2 is because it is like a smaller roadblock, so there is less voltage buildup on the side connected to rsistor 1. Now you might ask why is that resistor 2 only shows 4 volts instead of 6 volts like it did when resistor 1 was only 1 ohm. well I like to think of it as it's because each electron has a charge and since resistor 1 is now letting through less current and resistor 2 has stayed the same, there is even less less charge buildup on resistor 2. But if we raise resistor 2 up to 2 ohm s then again we will get 6 volts because the higher resistance is causing a bigger charge build up at resistor 2 and resistor 1 drops back to 6 volts again because , this might sound wierd.
but I believe it is because since resistor 2 is now causing a bigger buildup of charge on the side connected to resistor 1. this means when you measure acrosss resistor 1, resistor 2 now causes more voltage to be subtracted from the resistor 1 reading because the buildup of charge on it's terminal connected to resistor 1.

Whew, lol

So does that make any sense to anyone, or am just crazy and this is totally wrong. If I am then Please tell me and Please tell me why.

#### stormBytes

Joined Jan 26, 2010
43
As day-4 of this question dawns on me, I'm starting to think that I'm looking at this all wrong. Perhaps "Voltage" is only marginally relevant? That is, relevant only as it applies to the potential for current. That is, Voltage is a measurement of the 'potential' for current flow. But the *point* is current. Without current you don't get any work done!

I mean.. This is a total copout. It does not answer my original question, even if this approach is more practically-correct. Even so, it leaves many questions unanswered. Like for instance if I have my circuit, with one resistor in it, and I choose to add a second load, how would I know how much voltage is available (and therefore current) to the second load in the circuit?

I think my brain is officially offline. Time to get some shut eye.

Thanks to all those who've indulged, humored and tolerated (and continue to do so..)

#### Jony130

Joined Feb 17, 2009
5,439
In series circuit, the current will be the same through all of the series components.
So if we put another resistor in to the circuit we change the total resistance of a circuit. So, more resistance mean less current, less current means less voltage drop across individual resistors, it's simply Ohm's law.
Try to think of it, in terms of water analogy, where current is represent by water flow, voltage by a pressure and resistor by pipe with a constricted region.
The water pump produce constant pressure (voltage) to try to force water through the pipe.
So if we have a water pump + valve + constricted region.
And now if we open the valve to it's 50% then for sure there will be pressure drop across 50% open valve, so hydraulic resistor don't get 100% of a water pressure produce by water pump, becaues of a pressure drop on a valve resistance.

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#### Jony130

Joined Feb 17, 2009
5,439
Like for instance if I have my circuit, with one resistor in it, and I choose to add a second load, how would I know how much voltage is available (and therefore current) to the second load in the circuit?
For two resistors V2 = I * R2
I = Vin/( R1 + R2 )
V2 = Vin/( R1 + R2 ) * R2 = Vin * R2/(R1+R2)

#### Wendy

Joined Mar 24, 2008
23,122
Right now I'm keeping this really *really* simple. One battery/source, one resistor going across. That's it.
The Ohm's Law formula does it all, and it is simple.

E=IR, where E=Voltage, I=Current, and R=Resistance.

This simple formula is the basis for all else. Learn to use it and you are there.

If I have a 9V battery, and a 10KΩ resistor, then I will have 0.9ma. Always.

If you have two resistors in series they add, and you deal with the total voltage, and the total resistance, which determines the total current.

I'm not sure where you are getting these books or the other web sites (as for the relay), but you need to be somewhat selective as too your sources. The local text book at this site isn't perfect, but it is pretty good.