# Question about resistors and voltage

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#### jambo6984

##### Member
I gave a question about rezustors and reducing voltage and it is probably simple but to me im confused, i looked up what resistor to use to reduce 4.2v from a 18650 battery to 3.1v or 3.2v and google told me about 6ohm as i havent completly figured out ohms law i tested 6Ω all the way up to 470Ω with a multimeter and the voltage stayed the same even the flashlight i tore apart to put a different bulb and battery in it has a resistor on it of 10 ohm and google even says its suppose to drop voltage im trying to power a 1 watt led that has a 3.2 voltage drop is what the item description says were i bought it, any clarification and help will be greatly appreciated, is it just my bundle of resistors of every shape and size and even some i have taken out of working appliances ir am i missing something, its probably simple but please someone help ive never felt dumber and here i thought i knew quit a bit in electronics, please answer in lamens terms thank you

Sorry for any mistakes i was typing quick.

I didn't understand most of your post. Resistors drop an amount of voltage based on how much current is flowing through them, You need to know the current. So a resistor is able to drop anywhere between 0V and infinite volts. It depends how much current is flowing through it.

If you want it to drop the same voltage for many different currents, then you need a voltage regulator.

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So if your LED has a voltage drop of 3.2V and you have 4.2V, then you want to drop 1V so the LED doesn't blow out. An LED is a diode and yours has a 3.2V drop. It will drop 3.2V "no matter what amount of current is flowing through it"...until it blows up of course. So you have to decide how much current you want to flow through the LED and a reasonable one. 1W@3.2V means no you can choose more than 312.5mA. Let's pick 150mA.

So if you want to develop a 1V drop in a resistor when 150mA is flowing through it, you need to use R = 1V/150mA = 6.67 Ohms.

I have no idea what you mean when you say you tested 6ohm all the way up to 470ohms.

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Ohhh, I think I see what you are getting confused about.

You are saying that when you hook up 4.2V, the LED, and a resistor in series, the voltage drop across the resistor is always the same no matter how you change the resistance and you don't know why. That's because the LED ALWAYS eats up 3.2V of the 4.2V from the battery so there is only ever 1V remaining and it must appear across the resistor. The battery and the LED "fix" the voltages across their terminals so that remaining 1V must appear across the resistor and however much current will be drawn from the battery to produce make that happen.

The diode is basically turning your battery from a 4.2V battery into a 1V battery. And if you connect a resistor across a 1V battery, guess what? It's always going to drop 1V.

Measure the currents in the circuit AND the voltage across the resistor as you change the resistances and it will make more sense.

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You must learn simple Ohms Law to see about voltage drop, and voltage, current and power calculations. The voltage across a resistor depends on the amount of current in it. You had no current so you had no voltage drop.
To reduce 4.2V to 3.2V if the resistor value is 6 ohms then its current in it must be (4.2V - 3.2V)/6 Ohms= 1/6A or 167mA.

1W in a 3.2V LED is produced with a current that is 1W/3.2V= 1/3.2A or 313mA. A resistor value that will drop 4.2V to 3.2V at 313mA must be (4.2V - 3.2V)/313mA= 3.2 ohms. If you use 10 ohms then the current is (4.2V - 3.2V)/10 ohms= 0.1A or 100mA and the power in the LED is only 3.2V x 100mA= 0.32W.
You should not use its maximum allowed current in an LED without proper cooling so maybe 200mA using a resistor value of 5 ohms will be fine.

The really wierd thing is that im not hooking up the light im just simply using a multumeter a 18650 battery that puts out 4.2 volts and was trying to measure using different resistors to try to get 3.2 volts could it be the battery has circuit protection or does tge voltage never come down with just a resistor i just figured the voltage would show less at some point going up in resistance, it worked in tge factory flashlight it had 3 aaa batteries with a 10 ohm resistor havent took off and measured it yet though but should drop it to about 3.5 right?

The really wierd thing is that im not hooking up the light im just simply using a multumeter a 18650 battery that puts out 4.2 volts and was trying to measure using different resistors to try to get 3.2 volts could it be the battery has circuit protection or does tge voltage never come down with just a resistor i just figured the voltage would show less at some point going up in resistance, it worked in tge factory flashlight it had 3 aaa batteries with a 10 ohm resistor havent took off and measured it yet though but should drop it to about 3.5 right?

No, it is the most obvious thing in the world.This is EXTREMELY basic and literally the first thing someone learns in electronics. If you hook up 4.2V across just a resistor, the voltage dropped across the resistor will be 4.2V. The resistor drops all of the 4.2V because it is the only thing in the circuit. The resistor will limit the current to some current based on the resistance such that 4.2V is dropped across the resistor, because 4.2V must be dropped across the resistor. Where else is the battery voltage going to go if it's not going to drop across the resistor when the resistor is the only other thing in the circuit? All the voltage from the battery MUST be dropped somewhere in the circuit, and if the only thing in the circuit is the resistor, that is where all the voltage will be dropped.

Stare at this diagram for a bit. If the battery is 4.2V across it's terminals, what will be the voltage dropped across the resistor? Hint: The resistor shares the same terminals as the battery in that circuit.

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I know i can spell i just get in a hurry that why i posted if i do mess up its just because im in a hurry, but back to the problem i understand about the diagram and voltage dividers, i was just trying to hook my led up similar to how the old led was hooked up it had 4.5 volts coming from the battery and a resistor on one side of the post of the led going to negative and the positive straight to the battery, is there no way to drop any voltage with a setup like that, because im not having any luck no matter how large of a resistor i use, im just not understanding how the manufacturer did it this way with a 10 ohm resistor and why they did, and i cant always the same voltage all of everyones help is greatly appreciated and i know it seems like a stupid question to everyone here with the knowledge you all know, just trying to understand

I know i can spell i just get in a hurry that why i posted if i do mess up its just because im in a hurry, but back to the problem i understand about the diagram and voltage dividers, i was just trying to hook my led up similar to how the old led was hooked up it had 4.5 volts coming from the battery and a resistor on one side of the post of the led going to negative and the positive straight to the battery, is there no way to drop any voltage with a setup like that, because im not having any luck no matter how large of a resistor i use, im just not understanding how the manufacturer did it this way with a 10 ohm resistor and why they did, and i cant always the same voltage all of everyones help is greatly appreciated and i know it seems like a stupid question to everyone here with the knowledge you all know, just trying to understand

You ARE dropping the voltage in that circuit. The problem is you don't know what you're looking at and you don't understand what the term "voltage drop" actually means. The point of dropping the voltage isn't to make the voltage across the LED less than 3.5V because if you ever did that then the LED would not turn on. If you ever applied more than 3.5V to the LED, it would explode because it would pass more and more current to try to drop the excess voltage across it (whatever voltage is across it minus 3.5V). But since it's voltage drop across a diode is always 3.5V if current is flowing through it, that would take an infinite amount of current and the diode will explode first.

The whole point of dropping the extra voltage from the battery is so the current doesn't try and pass infinite current and explode.

In that circuit, the voltage drops across each components will never change because of the way everything works. What does change is the current flowing through the circuit, and that is dependent on the resistance value you use.

Stop caring about the voltage drop in that circuit. It's confusing you and isn't what you should care about anyways. What you should care about is the CURRENT flowing through the LED. That's why the resistor is there and that's why you choose the resistance value that you do. I have already told you to try measuring the current and you haven't done so. That would clear things up for you if you tried it with a few different resistors. If your next post isn't a question about you measuring the current, I'm not responding.

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I just decided to experiment running the risk of blowing the led and put the 10 ohm resistor on one side of the positive post of the led and hooked it up to the battery and it didnt blow

How can i measure the current in my multimeter i havent figured out how to do that yet

Im sorry if i am frustrating i have read just about everyday all day about electronics and have just about any circuit you could imagine i just havent had much real world experience, my whole ive worked with electronics but never have used resistors or made circuits i have just used existing circuits to do what i need or found a already right existing power supply but really want to learn all of it, and ypur right i was getting stumped in the voltage aspect if it as before thats all i really had to focus on, so to my understanding now the led uses 3.2 and then i meed a resistor to dissipate the other 1 volt, and i need to make sure the current isnt to low or high again thanks for tryung to help me understand i know it must be frustrating trying to teach someone something that seems easy to you thanks again

I just decided to experiment running the risk of blowing the led and put the 10 ohm resistor on one side of the positive post of the led and hooked it up to the battery and it didnt blow
I said before, "If you use 10 ohms then the current is (4.2V - 3.2V)/10 ohms= 0.1A or 100mA and the power in the LED is only 3.2V x 100mA= 0.32W." So the 1W didn't blow when it dissipated only 0.32W.

You do not need to measure current when you have a resistor with the current in it. Simply measure the voltage across the resistor and use Ohm's Law to calculate the current. Above there is 1V across 10 ohms. Then the current is V/R= I.
V is the voltage of 1V, R is the resistance 0f 10 ohms and I= 100mA.

so to my understanding now the led uses 3.2 and then i meed a resistor to dissipate the other 1 volt, and i need to make sure the current isnt to low or high
That is correct. All the voltage from the battery has to go somewhere. If the resistor was not there, the extra voltage would try to drop across the diode. But the diode can't drop more than 3.5V. As a result, it would take an infinite amount of current to drop more than 3.5V across the diode which would destroy it.

At tbe risk of offending everyone who has posted, i think the essential fundamental concept has been overlooked.

Ohm's Law

Resistors follow Ohm's Law. Voltage, resistance and current are related by

V = I×R where

V = volts
R = resistance in ohms
I = current in amps

Lets say we have a 100 ohm resistor with a current of 100mA. What will the voltage across the resistor (i.e., voltage drop) be?

V = 0.1 × 100 = 10 volts - remember, current must be in terms of amps,

We can rearrange the equatuon to calculate the desired quantity. Considering tour circuit of battery, LED and resistor. How can with determine the current through the circuit? Starting with

V = I × R,

we can rearrwnge it:

I = V / R

Lets say we have a 10 ohm resistor and we measure 2 volts across it while the circuit is hooked up and the LED is on.

I = V / R =2 / 10 = 0.2 amps = 200mA

Ohm's Law applies to resistors. The LED is an active device, so Ohm's Law doesn't apply to it.

Power Dissipation

The power dissipated by an element in a circuit is

P = I ×V where

P = power in watts
I = current in amps
V = volts

For the circuit of battery, LED and resistot, the current is the same everywhere in the circuit. In the above example, we calculated the current as 0.2 amps. If we measurethe voltage across the LED and find it to be 3.0 volts, then

P = I × V = 0.2 × 3.0 = 0.6 watts.

The power dissipated by the resistor may be found the same way. Combining the Ohm's Law equation and the power dissipation equation, we can get a couple shortcuts to calculate the power dissipated by the resistor.

P = V^2 / R

P = i^2 × R

Going way back to the beginning...

I gave a question about rezustors and reducing voltage and it is probably simple but to me im confused, i looked up what resistor to use to reduce 4.2v from a 18650 battery to 3.1v or 3.2v and google told me about 6ohm as i havent completly figured out ohms law i tested 6Ω all the way up to 470Ω with a multimeter and the voltage stayed the same even the flashlight i tore apart to put a different bulb and battery in it has a resistor on it of 10 ohm and google even says its suppose to drop voltage im trying to power a 1 watt led that has a 3.2 voltage drop is what the item description says were i bought it, any clarification and help will be greatly appreciated, is it just my bundle of resistors of every shape and size and even some i have taken out of working appliances ir am i missing something, its probably simple but please someone help ive never felt dumber and here i thought i knew quit a bit in electronics, please answer in lamens terms thank you

Sometimes Google can be your new best friend and sometimes not so much. Since the voltage stayed the same maybe we should look at something like how the circuit looked and how your meter figured into things. In the above circuit R2 the 10 MEG resistance simulates the input resistance of my meter. R1 is 6 Ohms hanging off my 4.2 volt battery. Because the input resistance of my meter is so high the current through the circuit will be very low. Actually we get 4.2 Volts / 10000006 = .0000004 Amp so the 6 Ohm resistor is dropping .0000025 volts which we are not even going to see so for all intense purpose V0ut is 4.2 volts. If this was your setup you can see why going from 6 ohms to 470 ohms won't change anything. Now if we have 4.2 volts and we want 3.2 volts we get 4.2 - 3.2 = 1 Volt. Since Google gave you 6 Ohms my read is 1 Volt / 6 Ohms = 166.6 mA?

Ron

Then since Google calculated the 6 ohms resistor it knew that the LED might burn out soon if operated at its absolute maximum power of 1W with a current of 1W/3.2V= 313mA without being cooled with liquid nitrogen or something.

At tbe risk of offending everyone who has posted, i think the essential fundamental concept has been overlooked.

Ohm's Law

Resistors follow Ohm's Law. Voltage, resistance and current are related by

V = I×R where

V = volts
R = resistance in ohms
I = current in amps

Lets say we have a 100 ohm resistor with a current of 100mA. What will the voltage across the resistor (i.e., voltage drop) be?

V = 0.1 × 100 = 10 volts - remember, current must be in terms of amps,

We can rearrange the equatuon to calculate the desired quantity. Considering tour circuit of battery, LED and resistor. How can with determine the current through the circuit? Starting with

V = I × R,

we can rearrwnge it:

I = V / R

Lets say we have a 10 ohm resistor and we measure 2 volts across it while the circuit is hooked up and the LED is on.

I = V / R =2 / 10 = 0.2 amps = 200mA

Ohm's Law applies to resistors. The LED is an active device, so Ohm's Law doesn't apply to it.

Power Dissipation

The power dissipated by an element in a circuit is

P = I ×V where

P = power in watts
I = current in amps
V = volts

For the circuit of battery, LED and resistot, the current is the same everywhere in the circuit. In the above example, we calculated the current as 0.2 amps. If we measurethe voltage across the LED and find it to be 3.0 volts, then

P = I × V = 0.2 × 3.0 = 0.6 watts.

The power dissipated by the resistor may be found the same way. Combining the Ohm's Law equation and the power dissipation equation, we can get a couple shortcuts to calculate the power dissipated by the resistor.

P = V^2 / R

P = i^2 × R

As many good physics books will attest, R=V/I is not Ohm's law. It is the definition of resistance. Ohm's law is an attribute of a material, specifically the resistive linearity. A tungsten wire follows Ohm's law quite well, but a PN junction does not. A material that follows Ohm's law will show a straight line current vs. voltage plot, whereas the opposite is true for non-ohmic materials. In all cases a resistance can be obtained at any current and voltage value because the definition always applies, regardless of the physical attribute of the material.

Ratch

So many words and so little help to the original poster. It's sad that a relatively simple misunderstanding is lost in so many bloated responses that do anything but clarify the original poster's understanding.

Let me try. Typical LEDS have a max current and a nominal current. They also "DROP" some voltage which is different based on the color of the LED

Suppose you have a 5V supply, a 20 mA nominal LED that nominally drops 2.1 V

The series resistor required is <=(5-2.1))/10e-3 = 380 ohms. There may not be a 380 ohm resistor, so pick a standard one less than that.

You do have to size the power dissipation of the resistor as well. P=(I^2) * R, so (10e-3)^2 *380 and you get 0.38 W. This means that a 0.5 W resistor will work.

ASIDE(Multimeter):
Your multimeter current scale inserts a resistor in the circuit. It may significantly alter the real value. An ammeter is placed in series with something and usually one lead has to placed in another jack. There is usually an internal fuse for the ammeter function. the high current ammeter like 10 A range may not be fused.

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