# I don't understand how I am "calculating for the voltage drop"

• BeautifulLight
In summary: Voltage drop = voltage across resistor." So, in this case resistor will only change the current in the circuit.
BeautifulLight
Assume I have connected a single light-emitting diode (2V, 30mA) and resistor to a standard 9V battery. Ohm's Law is used to calculate the value for the resistor. The observed relationship between voltage, current, and resistance is as follows, V=IR.

I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V.

So, 7=.03R, R=233ΩWhat's the problem? I understand everything I just did, and yet I am oblivious to the fact that when I solved for value R, I was also solving for a 7V drop across the resistor.

Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else...

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If you put your 233 ohm resistor across a 9 volt battery, it drops 9 volts. Voltage drop = voltage across resistor.
Resistor "see" 9V across his terminals because you connected a 9V battery directly across a single resistor.
So in this case resistor will only change the current in the circuit. Because how can a resistor change the battery voltage?

We have a different situation when we add another resistor in series

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BeautifulLight said:
Assume I have connected a single light-emitting diode (2V, 30mA) and resistor to a standard 9V battery. Ohm's Law is used to calculate the value for the resistor. The observed relationship between voltage, current, and resistance is as follows, V=IR.

I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V.

So, 7=.03R, R=233ΩWhat's the problem? I understand everything I just did, and yet I am oblivious to the fact that when I solved for value R, I was also solving for a 7V drop across the resistor.

Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else...

Your first calculation is correct. Start with the datasheet for the LED to see what Vf is at your desired current (brightness). Then the rest of the battery voltage needs to drop across the current-limiting resistor. Different color LEDs and different efficiency LEDs have different Vf values, so you pretty much need to always check the datasheet to find the right Vf to use for the calculation.EDIT -- Nice post by Jony!

I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V.

So, 7=.03R, R=233ΩWhat's the problem? I understand everything I just did, and yet I am oblivious to the fact that when I solved for value R, I was also solving for a 7V drop across the resistor.

Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else...

Sounds to me like an algebra problem not an electronic one.
Remember beginning algebra - one equation per unknown?

If A = B X C, you must know two things to figure out the third one.
Ditto for V = I X R .
I am oblivious to the fact that when I solved for value R, I was also solving for a 7V drop across the resistor.
No, you weren't solving for 7 volts - you figured out earlier that 7 was the voltage across resistor. And you correctly wrote 7 in for V. You got 7 by subtracting 2V diode drop from 9V battery supply per Kirchoff.

Then you wrote in .03 for I. You got .03 from LED datasheet.

So you were left with one equation and one unknown, which is solvable. And you solved it.

Keep it simple. Talk your way through these things - talking stimulates different parts of the brain than does just that "inner voice"..

I thought voltage drop was strictly the result of a component actively using voltage. Remember this?

"I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V."

So when you told me there was a 7V drop across the resistor, I immediately asked myself how in the world a resistor "uses" 7V and hence the last part of my initial post,

"Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else..."***There is a difference between voltage drop as a result of a component actively using voltage VS voltage drop by simply shorting out the leads of a 9V -of course the drop is going to be 9V, that's the EMF of the battery. In my case, it was easiest for me to think of the 7V as "left-over" voltage ...having absolutely nothing to do with the value of the resistor.I have one additional relevant question,

Why should a light-emitting diode have a minimum current rating such as 30mA? As long as there is enough voltage supplied -so electrons can jump from cathode to anode (or vise versa), then there should be no reason why electrons shouldn't flow even if there's only 0.00000000000001 amps supplied.

Is the 30mA some kind of threshold for the naked eye?

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BeautifulLight said:
I thought voltage drop was strictly the result of a component actively using voltage. Remember this?

"I intuitively understand there will be a voltage drop across the diode because it takes a certain amount of energy (in joules) for electrons to "jump" from cathode to anode, and voltage is defined as joules/coulomb. Therefore, the potential difference will now be 9-2=7V."

So when you told me there was a 7V drop across the resistor, I immediately asked myself how in the world a resistor "uses" 7V and hence the last part of my initial post,

"Would it be better to disregard the diode and just use a single resistor to help me understand? So, 9=I(233). -I randomly chose that value for R. I'm not sure how you figure voltage drop using only 9=I(233). From that, I can only derive current, nothing else..."***There is a difference between voltage drop as a result of a component actively using voltage VS voltage drop by simply shorting out the leads of a 9V -of course the drop is going to be 9V, that's the EMF of the battery. In my case, it was easiest for me to think of the 7V as "left-over" voltage ...having absolutely nothing to do with the value of the resistor.I have one additional relevant question,

Why should a light-emitting diode have a minimum current rating such as 30mA? As long as there is enough voltage supplied -so electrons can jump from cathode to anode (or vise versa), then there should be no reason why electrons shouldn't flow even if there's only 0.00000000000001 amps supplied.

Is the 30mA some kind of threshold for the naked eye?

The 30mA is the value for that LED to give good brightness, apparently. That's actually pretty high for modern LEDs, BTW. It's more common to see 5-10mA on datasheets currently.

As you learn more about basic electronics, you will stop using terms like "voltage used up" and so on. Just thing about voltage drops due to PN junctions, resistors, and so on. Later you will learn about reactive elements like inductors and capacitors, which can have AC voltage drops across them that vary with frequency...

BeautifulLight said:
I thought voltage drop was strictly the result of a component actively using voltage.

But the resistor IS actively using the voltage. It is a little bitty heater. How do you think BIG electric heaters work? Do you think that they are not "actively" using the voltage that provides the current that creates the heat that they radiate?

phinds said:
But the resistor IS actively using the voltage. It is a little bitty heater. How do you think BIG electric heaters work? Do you think that they are not "actively" using the voltage that provides the current that creates the heat that they radiate?

My chemistry teacher told us there was such things as insulators and conductors. I think those terms are for people who don't understand that you can pass electric current across any substance granted you supply enough voltage. I believe whatever the resistor is composed of has a threshold voltage; that is the required voltage of a substance for current to pass through it (assuming normal conditions like temp. etc). So yes, you are correct in saying that the resistor IS actively using voltage. I do, however, think that it is irrelevant here because it's so insignificant and it is better to think of that 7V as "left-over" voltage.

I may be very wrong, but I'm letting you know how I see it.

Edit: For instance, maybe the resistor does require a tenth of a volt for current to pass through it, but that figure is so insignificant compared to the 7V that has been left-over. If that figure would be more significant, then we would have to start dropping voltages on every resistor as we did with the 2V drop on the light-emitting diode.

I thought about this^ over dinner and it's (partially) incorrect, but mostly! That 233Ω resistor IS using all 7V granted there's .03 amperes of current. I forgot about Ohm's Law...

I say partially because when I read your post, it made it seem like or at least to me, that all 233Ω resistors require 7V. That isn't true. You can double your power (14V, .06 amperes) and it still works out, hence 14=(.06)(233) or 14=14.

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BeautifulLight said:
My chemistry teacher told us there was such things as insulators and conductors. I think those terms are for people who don't understand that you can pass electric current across any substance granted you supply enough voltage. I believe whatever the resistor is composed of has a threshold voltage; that is the required voltage of a substance for current to pass through it (assuming normal conditions like temp. etc). So yes, you are correct in saying that the resistor IS actively using voltage. I do, however, think that it is irrelevant here because it's so insignificant and it is better to think of that 7V as "left-over" voltage.
The terms insulator and conductor are very useful for people that understand electronics and electricity because they classify solids and how and when you can/should use them. A true resistor is a linear device, and has no such thing as a threshold voltage. It will obey Ohm's law down to the point where you can no longer reliably measure current. The only threshold effect you could see would be due to the interface between the resistor material and its contacts to whatever metal is used in its terminal, but that would be indicative of a poor design. Indeed, resistor bulk is coupled to metal using what is called an "ohmic" contact just to avoid any such thresholding effect.

BeautifulLight said:
I may be very wrong, but I'm letting you know how I see it.

Edit: For instance, maybe the resistor does require a tenth of a volt for current to pass through it, but that figure is so insignificant compared to the 7V that has been left-over. If that figure would be more significant, then we would have to start dropping voltages on every resistor as we did with the 2V drop on the light-emitting diode.

Yes, you're very wrong so you've come to the right place! :) The idea that a resistor would require a 10th of a volt for current to pass through it but 7V would be "left over" makes no sense and indicates you don't understand how a resistor works. It's rough, but people have found the water analogy useful to get started in electronics. Imagine a garden hose connect to a faucet on the side of your house. Basically, the water pressure is the voltage, the flow of water through the hose is the current and the resistance analogous to the inverse of the diameter of the hose.

V = IR, or water pressure = flow*(1/diameter).

Think about it, if you constrict the hose with constant pressure the flow increases (you use this to blast suds off your car, for instance). If you imagined a fat hose with ten times the diameter the water would just trickle out, since the volume of water (or the "power") would be constant.

If you don't like the water analogy, ignore it (plenty of people hate it because it can be misleading at times)... in that case remember that for resistor circuits V = IR. Manipulate this equation to see how volts, amps, and ohms are related when it comes to flow and power. As you can see in the equation, the idea of "left over" volts makes no sense. All the voltage is dropped across the total resistance.

Thank you analogdesign. You caught me while I was editing my post (#8). Here

I thought about this over dinner and it's (partially) incorrect, but mostly! That 233Ω resistor IS using all 7V granted there's .03 amperes of current. I forgot about Ohm's Law...

I say partially because when I read your post, it made it seem like or at least to me, that all 233Ω resistors require 7V. That isn't true. You can double your power (14V, .06 amperes) and it still works out, hence 14=(.06)(233) or 14=14.

The only time I like to see water analogies is with "valves" (diodes/triodes) because you can take it literally Oh, and I don't mean to knock the whole insulator/conductor thing; I'm just one of those 0.0000000001 isn't 0 kinna guys.

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Yes

I was in the middle of drawing a pic when you replied ...

seeing a circuit often helps visualising the problem

Dave

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How did you make that schematic? I don't see anything underneath "reply." Is that another program you're using and you're just hitting the paperclip icon to copy it?

so we know there is 9V across the resistor and LED ... ie. the voltage measured between A and C = 9V

the datasheet tells us that there will be a 2V drop across the LED ... between B and C
that means there must be a 7V drop across the resistor ... between A and B
BECAUSE all the voltage drops must add up to the total voltage applied to the circuit

Dave

BeautifulLight said:
The only time I like to see water analogies is with "valves" (diodes/triodes) because you can take it literally Oh, and I don't mean to knock the whole insulator/conductor thing; I'm just one of those 0.0000000001 isn't 0 kinna guys.

Ha! I'm an engineer so I'm a "0.0000000001 = 0" kind of guy when we're talking about amps, but I'm a "0.0000000001 isn't 0" kind of guy when we're talking about nanoamps. It's all about context.

BeautifulLight said:
How did you make that schematic? I don't see anything underneath "reply." Is that another program you're using and you're just hitting the paperclip icon to copy it?

I just drew it quickly in MS paint

If you click on the Go Advanced button below the text box, it gives you the option to manage attachments

so some things you can now work out

here's a problem for you to solve ...

in a series DC circuit, the current is measured the same everywhere, therefore there is 30mA through the LED, through the resistor and through the wires coming from and to the battery

Using Ohms Law --- R = V/I

knowing that the voltage drop across the LED is 2V and the current is 30mA ---
What is the resistance of the LED ?

knowing that the voltage drop across the resistor is 7V and the current is 30mA ---
What is the resistance of the resistor ?

Dave

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BeautifulLight said:
I do, however, think that it is irrelevant here because it's so insignificant and it is better to think of that 7V as "left-over" voltage.

I'd suggest that you not try to become an Electrical Engineer.

phinds said:
I'd suggest that you not try to become an Electrical Engineer.

awww don't be too harsh

At least he came to the right place to learn the error of his ways and is starting to get the hang of it

Dave

phinds said:
I'd suggest that you not try to become an Electrical Engineer.

Am I supposed to be embarrassed for attempting to come up with a logical solution rather than have something be spoon fed to me & then pretend I'm smart? Right, because that takes a lot of thought... You must be one of those guys who got their BA in Chemistry w/out ever thinking about the "everyday" stuff around them. Rofl, yeah what an achievement.

Resistance of R1 was already calculated (233Ω). Total resistance of system is 300Ω, hence 9=.03(300), so if you subtract the two, you get the resistance of the L.E.D. which is around 67Ω.

Internal resistance was ignored, but most 9V put out 750mA, so that gives a value of 12Ω. Can be ignored here.

"Shut up & calculate." -Feynman. Always wondered who/what he was talking about. I guess I know now. Please refrain from posting in my threads, especially when you're going to be that discouraging. And by the way, George Ohm is dead so I don't know who I'm talking to, but it sure isn't the guy who came up with Ohm's Law, so you're in the same boat as I am. Now am I supposed to tell you to find another field of study?

don't get too upset mate

read the PM I sent you

is there anything else concerning this that you are unsure of ?

cheers
Dave

Internal resistance was ignored, but most 9V put out 750mA, so that gives a value of 12Ω. Can be ignored here.

Yes we usually ignore the internal resistance of the battery/power supply

You do understand that it is the battery's internal resistance that limits the current from it ?
which is what stops an infinite current flowing under short circuit conditions

Dave

Mr Light - these concepts become second nature after you work a few dozen exercises solving circuits by laws of Kirchoff and Ohm.
I learn in the sequence "What" then "Why".

I suggest you practice until the concepts of loop current, node voltage, mesh currents seem as natural as riding a bicycle.

We all started out awkward.

old jim

BeautifulLight said:
Resistance of R1 was already calculated (233Ω). Total resistance of system is 300Ω, hence 9=.03(300), so if you subtract the two, you get the resistance of the L.E.D. which is around 67Ω.

It is true that the ratio of volts to current is resistance but an LED does not exhibit the same resistance as the current varies - it's non-Ohmic because it's not a piece of metal. If you try to use that value of resistance you (correctly) calculated anywhere else, you will fall over. The voltage drop across an LED is more or less the same over a range of currents - so that Voltage is what you have to use in calculations. In this case, Resistance of a non-Ohmic component is like a wet bar of soap in the bath - you can't latch onto it.

sophiecentaur said:
It is true that the ratio of volts to current is resistance but an LED does not exhibit the same resistance as the current varies - it's non-Ohmic because it's not a piece of metal. If you try to use that value of resistance you (correctly) calculated anywhere else, you will fall over. The voltage drop across an LED is more or less the same over a range of currents - so that Voltage is what you have to use in calculations. In this case, Resistance of a non-Ohmic component is like a wet bar of soap in the bath - you can't latch onto it.

What happens when I pull the resistor leaving only the L.E.D. across the 9V battery? I want to calculate both amperage & resistance, but it's a little confusing. I know there's a 2V drop across the L.E.D., but I also know at those same points there'll be a 9V drop across the terminals of the battery.

BeautifulLight said:
What happens when I pull the resistor leaving only the L.E.D. across the 9V battery? I want to calculate both amperage & resistance, but it's a little confusing. I know there's a 2V drop across the L.E.D., but I also know at those same points there'll be a 9V drop across the terminals of the battery.

It's confusing because an LED is an active device and doesn't follow Ohm's Law.

If you pull the resistor leaving only the LED you will burn out the LED because the source resistance of the battery is probably small. The 2V LED will have 9 V across it, so the battery current will get really high and kill the LED.

The LED will either fail closed in which case the battery will be drained by sourcing current to ground, or it will fail open, in which case the current will drop to zero because there won't be a closed circuit. Whether it fails open or closed depends on what fails first, the semiconductor junction or the bondwires/case connectin.

BeautifulLight said:
What happens when I pull the resistor leaving only the L.E.D. across the 9V battery? I want to calculate both amperage & resistance, but it's a little confusing. I know there's a 2V drop across the L.E.D., but I also know at those same points there'll be a 9V drop across the terminals of the battery.

if LED always "eats" 2 V on every current, other 7 V will have to be dissipated on internal resistance. current will rise until IxR_internal=7 V. on battery terminals you will actually measure 2 V, same as on diode terminals.

this will be a large current which will destroy your LED in short time, LED will not be able to cool itself.

your battery schematics is "9 V source + internal resistance". battery resistance is serial resistor which can not be removed.

BeautifulLight said:
What happens when I pull the resistor leaving only the L.E.D. across the 9V battery? I want to calculate both amperage & resistance, but it's a little confusing. I know there's a 2V drop across the L.E.D., but I also know at those same points there'll be a 9V drop across the terminals of the battery.

If you remember, I wrote that the volts drop across an LED is constant over a range of currents. That range doesn't include the current that would flow when it's put across a 9V battery. I(t will melt, of course.

analogdesign said:
It's confusing because an LED is an active device and doesn't follow Ohm's Law.
.
Afaiaa, the term "active" refers to devices that use an external power source in order to work - e.g. amplifiers. An LED is just Non-Ohmic, with a particular V/I characteristic.

sophiecentaur said:
Afaiaa, the term "active" refers to devices that use an external power source in order to work - e.g. amplifiers. An LED is just Non-Ohmic, with a particular V/I characteristic.

A lot of people agree with you, but it is not standard. What would you call a diode-connect transistor? An active device connected in a passive configuration?

analogdesign said:
A lot of people agree with you, but it is not standard. What would you call a diode-connect transistor? An active device connected in a passive configuration?

I would not call an electric drill that's used to hammer in a nail, using its handle, a 'power tool' and I would not call a transistor which does not 'control' the flow in one terminal by the signal on another terminal, an active device. It's what it is doing that counts, surely; not the name on the side.

Can you quote me a credible example of the use of the word 'active' that doesn't involve a power supply plus some element of control? I am sure that there will be places where all sorts of jargon are used but that wouldn't convince me. Many so-called Engineers can be extremely sloppy when they get talking amongst themselves.

sophiecentaur said:
I would not call an electric drill that's used to hammer in a nail, using its handle, a 'power tool' and I would not call a transistor which does not 'control' the flow in one terminal by the signal on another terminal, an active device. It's what it is doing that counts, surely; not the name on the side.

Can you quote me a credible example of the use of the word 'active' that doesn't involve a power supply plus some element of control? I am sure that there will be places where all sorts of jargon are used but that wouldn't convince me. Many so-called Engineers can be extremely sloppy when they get talking amongst themselves.

I get it. You're a pedant. I'm a sloppy engineer. Fine. I'm sorry you didn't sleep well last night.

It is interesting in light of your signature how wrapped up you get in classifications.

The meaning of language is its use. We so-called engineers are too concerned making circuits that function as intended to worry too much about these kinds of classifications.

To the OP: I was wrong. An LED is a passive device, not an active one. You may hear diodes referred to in conversation as active devices but don't let that fool you. They are identifying themselves as sloppy engineers so avoid such people. Hopefully this clarifies your understanding of the circuit.

analogdesign said:
I get it. You're a pedant. I'm a sloppy engineer. Fine. I'm sorry you didn't sleep well last night.
I like to think that 'pedants' will be responsible for checking the aeroplane that I plan to fly in on Friday.
I can also be as sloppy as the next man, when conversing with colleagues - but, when I am trying to help someone to learn something that is new to them, I think I owe it to them to be as accurate as possible. That's only fair to them, I feel. Langauge is, as you say, defined by its use. But if terms are used inappropriately and their meanings are not defined and held in common, what happens to the communication? I would never carry on this way on a Ham Radio or Amateur Constructors' Forum but I thought PF had some standards.
Did you have any evidence about the term 'active' or are you too cross to give me any?

sophiecentaur said:
Did you have any evidence about the term 'active' or are you too cross to give me any?

I'm not cross. The issue here is you're not "more accurate" because there is no objective definition by a generally accepted body clearly defining the terms "passive" and "active" with respect to electronics. You pick a definition you like (often for sound reasons, then act like the definition came down from on high and was etched on stone thousands of years ago).

Many people, yourself included, subscribe to the definition that an active device has a power supply and a method of control.

Other people define active as a device that requires a source of power to operate. Often diodes require that. Sometimes they don't.

You have to set up an LED with a constant current to drive it. You can't just let the signal power it as a resistor or capacitor can (unless you've got a very unusual signal). To some sloppy people, that is considered an active device.

As I mentioned, I agree with you. Describing a diode as passive is probably more descriptive. But being so rigid in your thinking will just turn people off electronics as a hobby. Hell, it's turning me off from helping answer questions as a hobby.

Here are some examples of people describing diodes as active. The first one is a university. The second is wikipedia. The third is a tutorial from a company. The last is an MS thesis from MIT.

http://www.ami.ac.uk/courses/topics/0133_itc/
http://en.wikipedia.org/wiki/Electronic_component
http://sound.westhost.com/beginners.htm
http://dspace.mit.edu/bitstream/handle/1721.1/47505/40295442.pdf <-- calls PIN diodes active elements

http://sound.westhost.com/beginners.htm

Actually that one lists them as passive ... but a special case

Passive: Capable of operating without an external power source.
Typical passive components are resistors, capacitors, inductors and diodes (although the latter are a special case).

personally I have been told and understood all semiconductor devices to be active

Passive was for resistors, capacitors, inductors

Dave

davenn said:
http://sound.westhost.com/beginners.htm

Actually that one lists them as passive ... but a special case

personally I have been told and understood all semiconductor devices to be active

Passive was for resistors, capacitors, inductors

Dave

The old 'categorising' problem has reared its head again. If you try to put your finger on what exactly is 'active' then you need a definition and not just an arbitrary classification. If 'active' implies control of some parameter, by another (i.e. there has to be a source of power and an output which depends upon the value of an input quantity ) then where do you draw the line between a mixer diode and a full blown amplifier? If you are going to include all semiconductors in the category 'active' then the terms 'active' and 'non-linear' mean the same thing - which means the term is redundant.

If 'active' is taken to mean that a power supply (of some sort) is involved then, at least, there is less possibility of confusion. An 'active' loudspeaker system would not be expected to contain merely some diodes, for added distortion; we would assume there is a power supply somewhere, other than just the audio power form the amplifier.
otoh, a mixer diode (used as part of a larger circuit, of course) could be said to use the Power from a local oscillator to provide a frequency shifted version of an input signal. In that context, it could be active. But so could a piece of resistance wire where the current through it can be controlled by its temperature. It satisfies the condition of control of one quantity by another so it would have to b e an 'active' device.

I think the total voltage drop in the circuit is = to the supply voltage in a simple dc series circuit like this. The led has a certain resistance, as does the passive resistor. The voltage drop across each will be equal to V=I*R.

The current that is flowing in the circuit will be equal to I = V/R (where R is the total resistance in the circuit) The voltage drop across each resistance, is according to the resistance of of that part of the circuit, and the current flow will be the same in all parts of the curcuit. Maybe neglible in most circuitry is the fact that the wires or other conductors like circuit board traces also have some resistance to the current flow.

<h2>1. What is voltage drop?</h2><p>Voltage drop is the decrease in voltage that occurs when electrical current flows through a resistance in a circuit. It is measured in volts and can affect the performance and efficiency of electrical devices.</p><h2>2. How is voltage drop calculated?</h2><p>Voltage drop is calculated using Ohm's Law, which states that voltage drop is equal to the current flowing through the circuit multiplied by the resistance of the circuit. This can be represented by the formula V=IR, where V is voltage, I is current, and R is resistance.</p><h2>3. What factors can affect voltage drop?</h2><p>Several factors can affect voltage drop, including the length and thickness of the wire, the type of material the wire is made of, and the amount of current flowing through the circuit. Higher resistance and longer wire lengths can result in a larger voltage drop.</p><h2>4. Why is voltage drop important to consider?</h2><p>Voltage drop is important to consider because it can impact the performance and efficiency of electrical devices. If the voltage drop is too high, it can cause devices to malfunction or not work at all. It can also lead to overheating and potential safety hazards.</p><h2>5. How can voltage drop be reduced?</h2><p>Voltage drop can be reduced by using thicker wires with lower resistance, minimizing the length of the wire, and reducing the amount of current flowing through the circuit. Properly sizing wires and using high-quality materials can also help to reduce voltage drop.</p>

## 1. What is voltage drop?

Voltage drop is the decrease in voltage that occurs when electrical current flows through a resistance in a circuit. It is measured in volts and can affect the performance and efficiency of electrical devices.

## 2. How is voltage drop calculated?

Voltage drop is calculated using Ohm's Law, which states that voltage drop is equal to the current flowing through the circuit multiplied by the resistance of the circuit. This can be represented by the formula V=IR, where V is voltage, I is current, and R is resistance.

## 3. What factors can affect voltage drop?

Several factors can affect voltage drop, including the length and thickness of the wire, the type of material the wire is made of, and the amount of current flowing through the circuit. Higher resistance and longer wire lengths can result in a larger voltage drop.

## 4. Why is voltage drop important to consider?

Voltage drop is important to consider because it can impact the performance and efficiency of electrical devices. If the voltage drop is too high, it can cause devices to malfunction or not work at all. It can also lead to overheating and potential safety hazards.

## 5. How can voltage drop be reduced?

Voltage drop can be reduced by using thicker wires with lower resistance, minimizing the length of the wire, and reducing the amount of current flowing through the circuit. Properly sizing wires and using high-quality materials can also help to reduce voltage drop.

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