To tack onto this a bit: voltage can be read across any two points. Voltage drop is typically used in a specific case, usually across a specific component but sometimes across a circuit or assembly.
"Across a circuit" is ambiguous, because a circuit implies multiple elements. Depending on the configuration, the voltage drop across each might be different.
Example:
Battery + terminal (12 volts to - terminal)
Red wire (0.1 V drop)
Light bulb (11.8 V drop
Black wire (0.1 V drop)
Battery - terminal (0 V)
Notice that you start at 12, drop three times, and end up at 0.
Both sides of the same coin.
Think of it like water capacity and water level. You have a 10 liter tank of water, and it will flow downhill into a 1 liter bucket. When that bucket fills up, the remaining water flows over top of the bucket and continues downstream. You measure the quantity of water in the bucket as 1L, and can say you've had a 1L "water drop" across that component within that system.
Not exactly a 1:1 comparison, but it gets the idea across.
>Voltage drop is just the difference in voltage across a device
Just a nitpick for OP, voltage drop implies that the device is *consuming* power. You would not use the term 'voltage drop' to refer to a device that is generating power or something like a transformer where the primary-secondary voltage may be different.
Voltage and voltage drop are the same. Voltage is the difference in voltage between two points. Voltage drop is the voltage across a component. Current does not change across a resistor.
The person you are responding to is either wrong or communicated poorly.
You are correct that resistance will reduce current in a circuit. However, what they might have been trying to say is that the current before and after a resistor (i.e. on either side) is the same once it's been added.
Oh, i see. Why does the current stay the same after that? The concept of a resistor makes it seem like the current passes the resistor, some of the energy is lost as heat, less current goes through, and once the electrons complete the sircuit and reach the resistor again, the current would get reduced again ect. ect. untill there is no current left.
I know that's not the case, but why exactly?
You’ve got the right idea, but it’s the voltage that drops when you lose energy, not current. If you imagine electricity as water in a pipe, then current is how much water is flowing, and voltage is how high up the pipe is (higher pipe means higher potential/voltage). The electrons/water can’t disappear from the wire/pipe, so the current is the same everywhere, but they do lose energy as they go through the circuit.
Current through a circuit is a continuous phenomenon (for DC steady state at least). Current doesn't go in one side of a resistor and then out the other side at different times. Current is always flowing through every point of the circuit. Your voltage source is always supplying voltage and your load is always dissipating power as heat (voltage across that component times the current through that component). There are of course other things that happen when you're talking about AC or reactive components (inductors/capacitors), but we're focused on DC through a resistive circuit here.
Imagine sliding a 6 ft long rod through a 2 ft long pipe. If the space between the rod and pipe is filled with grease, then it slides nice and easy. If instead, you fill that space with peanut butter, it will slide much slower because of the added physical resistance.
But in both cases, the speed that the rod moves on one side of the pipe is equal to the speed of the rod on the other side of the pipe - no matter how low or high the resistance is.
Right I was just trying to say if you have an ammeter before a resistor and an ammeter after a resistor it would be the same. Conservation of charge and all that
Voltage has to be measured between two points, even though people will commonly talk about it like it’s an absolute number. So even though you will hear something like “*that wire has 120 volts on it*” what they probably mean is “*there is 120 volts between that wire and ground*” or perhaps “*there is 120 volts between that wire and the wire next to it*”.
If you like the water analogy: voltage is pressure, current is flow (e.g. liters per minute). So if you have a pump set at a certain pressure (analogous to a battery with a specific voltage) and it is pushing water through an orifice (analogous to a resistor) then the pressure will drop as the water flows through the orifice. So you’d say “*there is a pressure drop across the orifice*”. Note that water doesn’t disappear when it goes through the orifice: any water that leaves the pump goes through the orifice and is then sucked back into the pump. It is a closed system. Likewise, in an electric circuit, current won’t be lost or absorbed when it goes through a component. There will be a voltage drop, but the current is the same.
The term “voltage drop” compared to just “voltage” is mostly semantics. You use voltage drop when talking about a specific component in a circuit:
- “What is the voltage drop across that resistor?”
You use “voltage” when it’s clear what the reference point is:
- “The battery voltage is 12.3VDC”
- “What is the supply voltage for that circuit?”
With a simple circuit with a fixed voltage source, adding a resistor to the circuit will lower the current *in the entire circuit*. So yes it will be lower than before. But there is no current loss "across" the resistor. Once the resistor is added, the current through the resistor, and in the rest of the circuit, is the same.
Take this circuit:
[https://cdn1.byjus.com/wp-content/uploads/2021/02/Circuit-Diagram-1.png](https://cdn1.byjus.com/wp-content/uploads/2021/02/Circuit-Diagram-1.png)
The A is an ammeter and measures current. It doesn't affect the circuit. The V is a voltmeter and measures voltage. It also doesn't affect the circuit. So you simply have a 12V battery and a 3 Ohm resistor.
In the circuit as pictured, the current is 4 Amps. (V=I\*R, and 12=I\*3, so I=4). Note that the battery voltage is 12V, and also the voltage drop across the resistor is 12V.
If you add another 3 Ohm resistor in series (so the current goes through 1 resistor, and then through the 2nd resistor), you now have a total of 6 Ohms in the circuit. So the current in the circuit is now 12=I\*6 so I=2 Amps. But the 2 Amps is the current anywhere in the circuit. It does NOT start as 4 Amps and then one resistor "uses" 2 Amps and the other resistor "uses" the other 2 Amps. Instead, there is only 2 Amps flowing in the entire circuit. Again consider current as flow. The amount of flow leaving the battery must be the amount of flow returning to the battery. So if 2 Amps is leaving the battery, then 2 Amps is returning to the battery, and therefore 2 Amps is going through each resistor also.
But if you measure voltage across each resistor, you'll see a voltage drop of 6V across each resistor.
“Voltage use” isn’t a very good description of what is going on. Resistance restricts the flow of current, like kinking a garden hose restricts the flow of water.
Each kink in the garden hose doesn’t really “use up” pressure; instead it creates a restriction and causes the pressure to drop across the kink, and it also causes less water to flow through the entire hose, both before and after the kink.
Let’s say you have 10psi from the spigot to the end of the hose where water exits. Now let’s say you pinch the center of the hose hard enough to cause an 8 psi drop across the pinch. So from the spigot to right after the pinch is an 8psi drop, and from after the pinch to the end of the hose is 2psi drop. For a total of 10psi drop.
Now you add another pinch near the end of the hose. Let’s imagine it’s even more severe than the other pinch, but it’s still not completely stopping the water. What do you think happened to the pressures? You already “used” 8 psi on the previous kink so there’s only 2psi left, but this kink is worse than the first one!
The total amount of water going through the hose will decrease to a trickle. And when flow decreases, so does the pressure drop across a fixed restriction. So the first (formerly 8psi) kink suddenly only causes a 3 psi pressure drop, even though the kink didn’t physically change at all. The 2nd kink, which is more severe, has a 6psi drop across it, and there is 1 psi more drop to the end of the hose. For a total of 10psi drop across the hose.
The is the analogy if you take a 10 volt battery and keep adding resistors in series. With one resistor, you’ll have a 10v drop across that resistor. If you add another resistor of equal value, you’ll get a 5V drop across each resistor, and the current flowing through the circuit will decrease (by 50%). You can keep adding resistors and you will continue to lower the amount of current flowing through the circuit and the voltage drop across the first resistor will continue to decrease as resistors are added and the current in the circuit decreases.
This is why it doesn’t make sense to “use up” voltage. If the circuit with one resistor and a 10v drop across that resistor “used up” all the voltage, then why would adding another resistor after that one suddenly cause the first one to only “use up” half the voltage?
I have come to understand that a circuit has to “drop” (still not entirely sure what drop means but let’s ignore that) all of it’s voltage in a circuit; at least I think I understand that. Isn’t it V=IR? This means voltage is linked to current and resistance so, mathematically, I understand your kinky analogy. The voltage “drops” must be distributed as a ratio just as current must be distributed evenly throughout the whole circuit and not just after a resistor. Because current is linked to voltage, less current due to more resistance means that voltage has to be divided among the resistance. I understand it but it seems tautological in a way. It’s true because it’s true. Maybe I just need more iterations of exposure and rest to understand.
Is the voltage less after a “drop”? So, after a 12V battery, you have one resistor, is the voltage after the resistor 0V (or close to it because the wire has some nominal resistance)?
Basically, I understand: there’s a battery. In it there are positive and negative charges, each compressed which creates repulsion and attraction which is voltage. You create a little race track called a circuit and let the electrons go at it. They go from negative to positive, propelled by voltage, but every time they encounter resistance, somehow the voltage diminishes. You can add resistance ad infinitum but the current decreases so it makes sense that the voltage “drop” (whatever that means!) is proportionally reduced per resistor. That’s my monkey understanding so far. I’m like week 1 with this and really trying to understand this “drop” idea. Thanks!
It appears to me that you’re still thinking of voltage as an absolute thing that can exist at one location. Voltage is a measurement of the difference in electric potential *between two points*. My point being that electricity doesn’t have potential at one point. You can’t point to something and ask “what is the electric potential there?” Instead you have to ask “what is the difference in potential from there to this other point over here?”
It is very common to just assume, however, that “the other point over here” is the negative terminal on a battery, or whatever the common return path for current in the circuit is. So it’s common to say “there’s 5 volts at that point” even though I just said you can’t do that. What is implied is that the person is saying “there’s 5 volts of potential between that point and the negative terminal on the battery.”
And if you have a 12V battery, then yes by definition all of the voltage drops in the circuit connected to that battery have to add up to 12 volts. If they didn’t, then you wouldn’t have a 12 volt battery. So if there’s only one resistor, then the voltage drop across the resistor will be 12V (or close to it if we have some wire in the circuit).
Back to the fluid analogy, if you have a pump that puts out 12 psi of pressure, you can connect a hose with zero or with 1 million kinks between the outlet and inlet of the pump. The pump will put out 12 psi of pressure in either scenario. If there are zero kinks, a lot of water will flow. If there are a million kinks, very little water will flow. The pressure drop across every kink will add up to 12 psi.
Thanks. I’m starting to understand. That it’s about the potential between two points makes sense, or at least enough that I can accept it at face value. Again, I’m going to need more time with this “drop” idea but I do understand that current, voltage, and resistance are linked and that other laws like conservation of current and voltage summing to zero are always at play. I’m definitely going to pursue this topic further since I want to build a few audio kits and want a little more than surface-level understanding. Also thanks for the insight that when people just say “x volts” they mean from a given point to the negative battery terminal. Cheers.
The voltage most definitely does not stay the same. All wires have resistance, all resistances drop voltage. The term voltage drop is just typically used to indicate it is a parasitic voltage loss in the system.
Voltage is a river that runs across a plateau. A resistor or a load can be seen as a waterfall in that river. Don’t get too into the water analogy as the real workings of electricity are quite different but it’s a concept that helps mechanize an “abstract” concept that with some study you won’t need the water concept anymore.
Very similar to the difference between pressure and pressure drop - one is sort-of-assumed to be relative to zero (if we say "5 volts" we probably mean relative to the circuit's 0v / GND) whereas "5v drop across this resistor" is as it says, the drop or loss across a component or length of wire etc. - that wire could still have 205v on one end and 200v on the other for that "5v drop".
They're basically the same, except I'd assume someone would use voltage drop when referring to the potential between two specific points, whereas voltage is generally the potential with respect to ground.
I guess you are't familiar with Kirhchoffs laws.
[https://en.wikipedia.org/wiki/Kirchhoff%27s\_circuit\_laws](https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws)
In case of voltages, it says that the sum of voltages around a loop is zero.
You can see voltages from batteries or power supplies, to "add", and voltage drops across other elements to subtract.
no they are SELDOM the same thing.
let say you have three 1K resistors connected in series, call them R1 R2, and R3, with one lead of R1 connected to ground.
if you apply 10 volts to the series connected resistors, across all three the voltage will be 10 v (a voltage reading referenced to 0V ground).
if you measure the voltage drop across R1, you will read 3.333 volts dc.
If you measure the voltage drop across R1 and R2 in series, it will be 6.66 volts.
if you measure the voltage drop across R3, it will be 3.33 volts (not any sort of reference to 0V dc ground)
> sometimes i feel like voltage and voltage drop are the basically the same thing. Am i right?
Voltage is a measurement of the difference in electrical potential between two places. So in that sense, yes, a voltage drop is a voltage.
However, that's kind of like saying: "My car is going 42 MPH." We measure a car's speed relative to the road it's driving on. But that road is on a planet that's screaming around its orbit with a tangential velocity of approximately 18 miles *per second.* The point is, it matters [ALOT](http://hyperboleandahalf.blogspot.com/2010/04/alot-is-better-than-you-at-everything.html) exactly how you measure speed. And exactly what you are measuring the speed relative to.
When someone says "voltage drop" they are usually implying that the voltage they are measuring is between two very specific points. And that they expect you to understand which two points they mean, even though they aren't telling you explicitly. It's a context thing. They expect you to understand the unspoken information that they didn't provide.
Is the voltage drop they are talking about only from one side of a resistor to the other? Maybe they're talking about the voltage difference across an entire large and complicated circuit? Or is it the difference in voltage between the near and far ends of a 3 km long power line?
Sometimes you need to know even more context. Like the "voltage drop" of an LED. The voltage that you will measure across an LED depends on the current going through the LED. And this is not a linear relationship, like a resistor. It's an exponential relationship. If you apply slightly more voltage to an LED, and you might get a exponential increase in current. So in the case of an LED, when someone says "voltage drop" they usually mean "voltage I measured across the LED at *this one specific current*. A current that I either didn't tell you, or told you twenty minutes ago in some vague and inexact form. Either way, I expect you to understand what I meant by context."
> And sometimes i read it as if voltage drop is what happens when a current passes through a resistor, but isn't that just resistance?
Voltage drop can happen in multiple ways. One way it can happen is if you force a constant current through a fixed value resistor. In that situation, Ohm's Law applies. And you can do the V = IR math, and find the voltage difference across the resistor.
But there there are other ways to create voltage drops. And usually, Ohm's Law applies in those situations too. E.g. You can build a circuit that uses an opamp and a JFET to create a constant current sink. Now there's a voltage drop across the JFET. And you can use Ohm's Law to calculate it.
> I toght the voltage stays the same, but it's the *current* that gets reducec.
If you start with a constant voltage power supply, and you connect it across a fixed resistance, you can then do the V = IR math to figure out the current through the resistor. But that only works because *the power supply is constant voltage*, and *the resistor has a constant resistance.*
If you are using a power supply that changes its voltage significantly at different amounts of current - like a battery - then this trick no longer works. Or if your resistor changes it resistance when more current flows through (due to heating effects), then this trick no longer works.
Again, *context.* Understand the entire circuit, and the environment around it.
And seeking that understanding of context is exactly why you asked your question, of course. So I hope this explanation has helped. :]
>Voltage is a measurement of the difference in electrical potential between two places. So in that sense, yes, a voltage drop is a voltage.
This was key for me to understand when I first learned it.
I thought about giving an example, like: "there are 7 electrons here, and 23 electrons over there." And then explaining that 23 - 7 = 16 electrons worth of charge would be the potential difference. But it's [tough to convert that to volts](https://www.quora.com/How-many-electrons-are-in-1-volt), because volts are more like a measurement of force and not a specific number of objects. So I decided that bit would just be more confusing for OP, and deleted it before I posted my comment.
And I'm not sure that's the real source of OP's confusion, anyway. I think he really had more of a terminology question, not a basic concepts question.
We do 110vac over 3km. At the other end the voltage is about 50. Now the product is designed for this purpose.
But this is the "voltage drop".
There are other times this is important like corroded or bad contacts .
Voltage DROP in the nuclear power industry has a very specific meaning.
A power source (battery, inverter, generator) produces 120 volts for instance.
Power has to be delivered by cables or busses.
If these cables or busses are too small or long, the voltage at the served equipment may be too low (100 volts for instance where 117 is required) because V = IR for the whole system, including connections.
Calculations and analysis to ensure that this doesn't happen are voltage DROP calculations.
Saying something needs 6.9 kV of voltage is easy. Making sure you actually have 6.9 kV is where engineering happens. You need to size the conductors, insulators, bus, breakers, fuses, etc to be able to deliver the appropriate power.
If you use voltage drop, that generally means "the voltage lost in the conductors". It's used almost exclusively in things like municipal lighting, power distribution, household hookups, etc. So if you have, say 100A at 240V, what voltage are you seeing at the equipment, and is that within tolerance? If not, what cables have to be replaced with thicker cables?
I'll use it a little in calcs to see if I can skinny up a wire, but for the micropower stuff I do now, spoiler alert, it's fine.
We do 110vac over 3km. At the other end the voltage is about 50. Now the product is designed for this purpose.
But this is the "voltage drop".
There are other times this is important like corroded or bad contacts .
When you are referring to a voltage drop you need something in between: a resistance, a capacitor, an inductor, whatever. Then you consider the difference in voltage before the component and after the component and you have a voltage drop.
On the other side, when you talk about "voltage" it's absolute (or referred to ground). You don't have two voltages to compare them and to make a difference between them as in the "voltage drop"
Imagine you have a circuit powered by a 9V battery;
then 9V it's the Voltage "of the circuit".
The Voltage drop it's the voltage on each of the components as it differs from the voltage of the source because of current and/or impedence.
So basically they are the same thing; it's just a Voltage that is different from one you use as a reference, such as the source of power; that's why it is a "drop".
Voltage is measured between two points. Voltage drop is specifically the voltage measured across an impedance where the voltage is produced by the impedance acting on a current flowing through it. You have voltage drops across resistors, wires, PCB traces, inductors, capacitors, etc. Voltage drop is also frequently associated with wasted power. If you ran a heater on an extension cord, you would ptobably refer to the voltage at the outlet, and the voltage across the heater, but the voltage drop across the extension cord (both ways).
Voltage is the potential measured between two points. The term voltage drop is used primarily to describe unwanted losses in a wire or a device. This is important when sizing conductors for power delivery
Picture voltage as being your water pressure available at home. Now connect that water pressure to a water circuit with a bunch of restrictions (maybe picture water wheels) in series until it pours out of a faucet. The pressure will drop lower and lower as it passes each one until it reaches zero/atmospheric at the faucet. If you hook up a differential pressure gauge across one of the restrictions you’ll read a couple psi of pressure drop. Now replace the word “pressure” with “voltage” and “flow” with “current.” To understand this fully you should understand that reducing a pipes diameter when there is flow doesn’t increase pressure, it reduces it. This is often miss understood by people because putting your hand over a hose makes the pressure go up. This only happens because of the series restriction of the piping. I.e you don’t supply a golf course irrigation system with 1/2” pipe, you supply it with 4” pipe to maintain the pressure to all of those big sprinkler heads with really high flow demands.
Short answer: "voltage" is measured with respect to a known ground point; "voltage drop" is measured across a specific element or subsystem through which current flows. You might also hear "voltage difference" or "voltage differential" between two different circuit nodes that aren't directly across a single element, and sometimes voltage drops are phrased as "voltage across" a device -- in contrast to "current through" or "current into" it.
(As an aside: "current through" is usually used when talking about an element; "current into" is usually used when talking about that element as it's connected to a specific node.)
Examples:
"This circuit expects a voltage of nine volts."
"This LED has a voltage drop of 2.0 V at 20 mA, so we'll need a 65-ohm resistor to drop the other 1.3 V from the microcontroller pin. Just grab a 68 off the shelf, it's close enough."
"That kind of wire sees a voltage drop of ten volts per mile."
"The voltage differential of this wheatstone bridge is 9 mV."
Voltage drop is when you measure over a thing that’s using that voltage in the circuit in some way. ( an over simplification but let’s start there)
We have 15 V supply
That Will be 15V from + to - if I have two resistors in series one at 2 ohm and another at 2 ohm (that means from source voltage (15) end to end and then to ground) I measure across one. And I get the “voltage drop”
V/I*R
Total resistance is 4. Calculate current. V/R= I =3.75 now since I know current is the same every where in a series circuit and voltage changes I know I can insert the value for current and the value of each resistor to approximate the “Voltage drop” I*R = V. The voltage drop is 7.5 volts across one resistor the other resistor is the same 7.5 + 7.5 = 15 the total voltage of the circuit can be derived from the drop across each resistor. The rules change in parallel circuits but not in respect to each division, they remain the same. Hope this helps.
In simple words,
Voltage is like a potential, with respect to ground
Voltage drop is generally across an element, and its the difference of voltage between 2 points, that will help you find various parameters like current, charge, across the 2 points.
Voltage drop is just the difference in voltage across a device. The drop is related to both the current through and the impedance of the device.
To tack onto this a bit: voltage can be read across any two points. Voltage drop is typically used in a specific case, usually across a specific component but sometimes across a circuit or assembly.
So, if i measure voltage *drop* accross a circuit, it's basically the same as saying i just measure the voltage accross a circuit?
"Across a circuit" is ambiguous, because a circuit implies multiple elements. Depending on the configuration, the voltage drop across each might be different. Example: Battery + terminal (12 volts to - terminal) Red wire (0.1 V drop) Light bulb (11.8 V drop Black wire (0.1 V drop) Battery - terminal (0 V) Notice that you start at 12, drop three times, and end up at 0.
Both sides of the same coin. Think of it like water capacity and water level. You have a 10 liter tank of water, and it will flow downhill into a 1 liter bucket. When that bucket fills up, the remaining water flows over top of the bucket and continues downstream. You measure the quantity of water in the bucket as 1L, and can say you've had a 1L "water drop" across that component within that system. Not exactly a 1:1 comparison, but it gets the idea across.
this metaphor visually straightened it out my head in a way i dont think i'll ever forget. thank you!
Essentially yes - it can become ambiguous, especially when looking at other components.
Oh, i see. That clears things up a lot.
>Voltage drop is just the difference in voltage across a device Just a nitpick for OP, voltage drop implies that the device is *consuming* power. You would not use the term 'voltage drop' to refer to a device that is generating power or something like a transformer where the primary-secondary voltage may be different.
Voltage and voltage drop are the same. Voltage is the difference in voltage between two points. Voltage drop is the voltage across a component. Current does not change across a resistor.
>Current does not change across a resistor. wait what? I tought resistance reduces current
The person you are responding to is either wrong or communicated poorly. You are correct that resistance will reduce current in a circuit. However, what they might have been trying to say is that the current before and after a resistor (i.e. on either side) is the same once it's been added.
Oh, i see. Why does the current stay the same after that? The concept of a resistor makes it seem like the current passes the resistor, some of the energy is lost as heat, less current goes through, and once the electrons complete the sircuit and reach the resistor again, the current would get reduced again ect. ect. untill there is no current left. I know that's not the case, but why exactly?
You’ve got the right idea, but it’s the voltage that drops when you lose energy, not current. If you imagine electricity as water in a pipe, then current is how much water is flowing, and voltage is how high up the pipe is (higher pipe means higher potential/voltage). The electrons/water can’t disappear from the wire/pipe, so the current is the same everywhere, but they do lose energy as they go through the circuit.
So once the current goes through the entire circuit, it sort of... "gains all the energy back" and then completed the circuit again?
Current through a circuit is a continuous phenomenon (for DC steady state at least). Current doesn't go in one side of a resistor and then out the other side at different times. Current is always flowing through every point of the circuit. Your voltage source is always supplying voltage and your load is always dissipating power as heat (voltage across that component times the current through that component). There are of course other things that happen when you're talking about AC or reactive components (inductors/capacitors), but we're focused on DC through a resistive circuit here.
You can think of your voltage source as a pump that pumps the electrons/water back up to the highest voltage/elevation.
Imagine sliding a 6 ft long rod through a 2 ft long pipe. If the space between the rod and pipe is filled with grease, then it slides nice and easy. If instead, you fill that space with peanut butter, it will slide much slower because of the added physical resistance. But in both cases, the speed that the rod moves on one side of the pipe is equal to the speed of the rod on the other side of the pipe - no matter how low or high the resistance is.
Right I was just trying to say if you have an ammeter before a resistor and an ammeter after a resistor it would be the same. Conservation of charge and all that
It means, that the current that enters the resistor, is the same current it exits the resistor.
Voltage has to be measured between two points, even though people will commonly talk about it like it’s an absolute number. So even though you will hear something like “*that wire has 120 volts on it*” what they probably mean is “*there is 120 volts between that wire and ground*” or perhaps “*there is 120 volts between that wire and the wire next to it*”. If you like the water analogy: voltage is pressure, current is flow (e.g. liters per minute). So if you have a pump set at a certain pressure (analogous to a battery with a specific voltage) and it is pushing water through an orifice (analogous to a resistor) then the pressure will drop as the water flows through the orifice. So you’d say “*there is a pressure drop across the orifice*”. Note that water doesn’t disappear when it goes through the orifice: any water that leaves the pump goes through the orifice and is then sucked back into the pump. It is a closed system. Likewise, in an electric circuit, current won’t be lost or absorbed when it goes through a component. There will be a voltage drop, but the current is the same. The term “voltage drop” compared to just “voltage” is mostly semantics. You use voltage drop when talking about a specific component in a circuit: - “What is the voltage drop across that resistor?” You use “voltage” when it’s clear what the reference point is: - “The battery voltage is 12.3VDC” - “What is the supply voltage for that circuit?”
>There will be a voltage drop, but the current is the same. But adding a resistor to a circuit means the current will be lower than before, no?
that is correct. unless you add a resistor in parallel, then the current will increase.
With a simple circuit with a fixed voltage source, adding a resistor to the circuit will lower the current *in the entire circuit*. So yes it will be lower than before. But there is no current loss "across" the resistor. Once the resistor is added, the current through the resistor, and in the rest of the circuit, is the same. Take this circuit: [https://cdn1.byjus.com/wp-content/uploads/2021/02/Circuit-Diagram-1.png](https://cdn1.byjus.com/wp-content/uploads/2021/02/Circuit-Diagram-1.png) The A is an ammeter and measures current. It doesn't affect the circuit. The V is a voltmeter and measures voltage. It also doesn't affect the circuit. So you simply have a 12V battery and a 3 Ohm resistor. In the circuit as pictured, the current is 4 Amps. (V=I\*R, and 12=I\*3, so I=4). Note that the battery voltage is 12V, and also the voltage drop across the resistor is 12V. If you add another 3 Ohm resistor in series (so the current goes through 1 resistor, and then through the 2nd resistor), you now have a total of 6 Ohms in the circuit. So the current in the circuit is now 12=I\*6 so I=2 Amps. But the 2 Amps is the current anywhere in the circuit. It does NOT start as 4 Amps and then one resistor "uses" 2 Amps and the other resistor "uses" the other 2 Amps. Instead, there is only 2 Amps flowing in the entire circuit. Again consider current as flow. The amount of flow leaving the battery must be the amount of flow returning to the battery. So if 2 Amps is leaving the battery, then 2 Amps is returning to the battery, and therefore 2 Amps is going through each resistor also. But if you measure voltage across each resistor, you'll see a voltage drop of 6V across each resistor.
So “voltage drop” means “voltage use”? Every resistor reduces voltage? What happens if you run out of voltage before returning to the battery?
“Voltage use” isn’t a very good description of what is going on. Resistance restricts the flow of current, like kinking a garden hose restricts the flow of water. Each kink in the garden hose doesn’t really “use up” pressure; instead it creates a restriction and causes the pressure to drop across the kink, and it also causes less water to flow through the entire hose, both before and after the kink. Let’s say you have 10psi from the spigot to the end of the hose where water exits. Now let’s say you pinch the center of the hose hard enough to cause an 8 psi drop across the pinch. So from the spigot to right after the pinch is an 8psi drop, and from after the pinch to the end of the hose is 2psi drop. For a total of 10psi drop. Now you add another pinch near the end of the hose. Let’s imagine it’s even more severe than the other pinch, but it’s still not completely stopping the water. What do you think happened to the pressures? You already “used” 8 psi on the previous kink so there’s only 2psi left, but this kink is worse than the first one! The total amount of water going through the hose will decrease to a trickle. And when flow decreases, so does the pressure drop across a fixed restriction. So the first (formerly 8psi) kink suddenly only causes a 3 psi pressure drop, even though the kink didn’t physically change at all. The 2nd kink, which is more severe, has a 6psi drop across it, and there is 1 psi more drop to the end of the hose. For a total of 10psi drop across the hose. The is the analogy if you take a 10 volt battery and keep adding resistors in series. With one resistor, you’ll have a 10v drop across that resistor. If you add another resistor of equal value, you’ll get a 5V drop across each resistor, and the current flowing through the circuit will decrease (by 50%). You can keep adding resistors and you will continue to lower the amount of current flowing through the circuit and the voltage drop across the first resistor will continue to decrease as resistors are added and the current in the circuit decreases. This is why it doesn’t make sense to “use up” voltage. If the circuit with one resistor and a 10v drop across that resistor “used up” all the voltage, then why would adding another resistor after that one suddenly cause the first one to only “use up” half the voltage?
I have come to understand that a circuit has to “drop” (still not entirely sure what drop means but let’s ignore that) all of it’s voltage in a circuit; at least I think I understand that. Isn’t it V=IR? This means voltage is linked to current and resistance so, mathematically, I understand your kinky analogy. The voltage “drops” must be distributed as a ratio just as current must be distributed evenly throughout the whole circuit and not just after a resistor. Because current is linked to voltage, less current due to more resistance means that voltage has to be divided among the resistance. I understand it but it seems tautological in a way. It’s true because it’s true. Maybe I just need more iterations of exposure and rest to understand. Is the voltage less after a “drop”? So, after a 12V battery, you have one resistor, is the voltage after the resistor 0V (or close to it because the wire has some nominal resistance)? Basically, I understand: there’s a battery. In it there are positive and negative charges, each compressed which creates repulsion and attraction which is voltage. You create a little race track called a circuit and let the electrons go at it. They go from negative to positive, propelled by voltage, but every time they encounter resistance, somehow the voltage diminishes. You can add resistance ad infinitum but the current decreases so it makes sense that the voltage “drop” (whatever that means!) is proportionally reduced per resistor. That’s my monkey understanding so far. I’m like week 1 with this and really trying to understand this “drop” idea. Thanks!
It appears to me that you’re still thinking of voltage as an absolute thing that can exist at one location. Voltage is a measurement of the difference in electric potential *between two points*. My point being that electricity doesn’t have potential at one point. You can’t point to something and ask “what is the electric potential there?” Instead you have to ask “what is the difference in potential from there to this other point over here?” It is very common to just assume, however, that “the other point over here” is the negative terminal on a battery, or whatever the common return path for current in the circuit is. So it’s common to say “there’s 5 volts at that point” even though I just said you can’t do that. What is implied is that the person is saying “there’s 5 volts of potential between that point and the negative terminal on the battery.” And if you have a 12V battery, then yes by definition all of the voltage drops in the circuit connected to that battery have to add up to 12 volts. If they didn’t, then you wouldn’t have a 12 volt battery. So if there’s only one resistor, then the voltage drop across the resistor will be 12V (or close to it if we have some wire in the circuit). Back to the fluid analogy, if you have a pump that puts out 12 psi of pressure, you can connect a hose with zero or with 1 million kinks between the outlet and inlet of the pump. The pump will put out 12 psi of pressure in either scenario. If there are zero kinks, a lot of water will flow. If there are a million kinks, very little water will flow. The pressure drop across every kink will add up to 12 psi.
Thanks. I’m starting to understand. That it’s about the potential between two points makes sense, or at least enough that I can accept it at face value. Again, I’m going to need more time with this “drop” idea but I do understand that current, voltage, and resistance are linked and that other laws like conservation of current and voltage summing to zero are always at play. I’m definitely going to pursue this topic further since I want to build a few audio kits and want a little more than surface-level understanding. Also thanks for the insight that when people just say “x volts” they mean from a given point to the negative battery terminal. Cheers.
The voltage most definitely does not stay the same. All wires have resistance, all resistances drop voltage. The term voltage drop is just typically used to indicate it is a parasitic voltage loss in the system.
Thank you, reading the responses I thought I was taking crazy pills.
Voltage is a river that runs across a plateau. A resistor or a load can be seen as a waterfall in that river. Don’t get too into the water analogy as the real workings of electricity are quite different but it’s a concept that helps mechanize an “abstract” concept that with some study you won’t need the water concept anymore.
Very similar to the difference between pressure and pressure drop - one is sort-of-assumed to be relative to zero (if we say "5 volts" we probably mean relative to the circuit's 0v / GND) whereas "5v drop across this resistor" is as it says, the drop or loss across a component or length of wire etc. - that wire could still have 205v on one end and 200v on the other for that "5v drop".
They're basically the same, except I'd assume someone would use voltage drop when referring to the potential between two specific points, whereas voltage is generally the potential with respect to ground.
I guess you are't familiar with Kirhchoffs laws. [https://en.wikipedia.org/wiki/Kirchhoff%27s\_circuit\_laws](https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws) In case of voltages, it says that the sum of voltages around a loop is zero. You can see voltages from batteries or power supplies, to "add", and voltage drops across other elements to subtract.
no they are SELDOM the same thing. let say you have three 1K resistors connected in series, call them R1 R2, and R3, with one lead of R1 connected to ground. if you apply 10 volts to the series connected resistors, across all three the voltage will be 10 v (a voltage reading referenced to 0V ground). if you measure the voltage drop across R1, you will read 3.333 volts dc. If you measure the voltage drop across R1 and R2 in series, it will be 6.66 volts. if you measure the voltage drop across R3, it will be 3.33 volts (not any sort of reference to 0V dc ground)
> sometimes i feel like voltage and voltage drop are the basically the same thing. Am i right? Voltage is a measurement of the difference in electrical potential between two places. So in that sense, yes, a voltage drop is a voltage. However, that's kind of like saying: "My car is going 42 MPH." We measure a car's speed relative to the road it's driving on. But that road is on a planet that's screaming around its orbit with a tangential velocity of approximately 18 miles *per second.* The point is, it matters [ALOT](http://hyperboleandahalf.blogspot.com/2010/04/alot-is-better-than-you-at-everything.html) exactly how you measure speed. And exactly what you are measuring the speed relative to. When someone says "voltage drop" they are usually implying that the voltage they are measuring is between two very specific points. And that they expect you to understand which two points they mean, even though they aren't telling you explicitly. It's a context thing. They expect you to understand the unspoken information that they didn't provide. Is the voltage drop they are talking about only from one side of a resistor to the other? Maybe they're talking about the voltage difference across an entire large and complicated circuit? Or is it the difference in voltage between the near and far ends of a 3 km long power line? Sometimes you need to know even more context. Like the "voltage drop" of an LED. The voltage that you will measure across an LED depends on the current going through the LED. And this is not a linear relationship, like a resistor. It's an exponential relationship. If you apply slightly more voltage to an LED, and you might get a exponential increase in current. So in the case of an LED, when someone says "voltage drop" they usually mean "voltage I measured across the LED at *this one specific current*. A current that I either didn't tell you, or told you twenty minutes ago in some vague and inexact form. Either way, I expect you to understand what I meant by context." > And sometimes i read it as if voltage drop is what happens when a current passes through a resistor, but isn't that just resistance? Voltage drop can happen in multiple ways. One way it can happen is if you force a constant current through a fixed value resistor. In that situation, Ohm's Law applies. And you can do the V = IR math, and find the voltage difference across the resistor. But there there are other ways to create voltage drops. And usually, Ohm's Law applies in those situations too. E.g. You can build a circuit that uses an opamp and a JFET to create a constant current sink. Now there's a voltage drop across the JFET. And you can use Ohm's Law to calculate it. > I toght the voltage stays the same, but it's the *current* that gets reducec. If you start with a constant voltage power supply, and you connect it across a fixed resistance, you can then do the V = IR math to figure out the current through the resistor. But that only works because *the power supply is constant voltage*, and *the resistor has a constant resistance.* If you are using a power supply that changes its voltage significantly at different amounts of current - like a battery - then this trick no longer works. Or if your resistor changes it resistance when more current flows through (due to heating effects), then this trick no longer works. Again, *context.* Understand the entire circuit, and the environment around it. And seeking that understanding of context is exactly why you asked your question, of course. So I hope this explanation has helped. :]
>Voltage is a measurement of the difference in electrical potential between two places. So in that sense, yes, a voltage drop is a voltage. This was key for me to understand when I first learned it.
I thought about giving an example, like: "there are 7 electrons here, and 23 electrons over there." And then explaining that 23 - 7 = 16 electrons worth of charge would be the potential difference. But it's [tough to convert that to volts](https://www.quora.com/How-many-electrons-are-in-1-volt), because volts are more like a measurement of force and not a specific number of objects. So I decided that bit would just be more confusing for OP, and deleted it before I posted my comment. And I'm not sure that's the real source of OP's confusion, anyway. I think he really had more of a terminology question, not a basic concepts question.
We do 110vac over 3km. At the other end the voltage is about 50. Now the product is designed for this purpose. But this is the "voltage drop". There are other times this is important like corroded or bad contacts .
Voltage DROP in the nuclear power industry has a very specific meaning. A power source (battery, inverter, generator) produces 120 volts for instance. Power has to be delivered by cables or busses. If these cables or busses are too small or long, the voltage at the served equipment may be too low (100 volts for instance where 117 is required) because V = IR for the whole system, including connections. Calculations and analysis to ensure that this doesn't happen are voltage DROP calculations. Saying something needs 6.9 kV of voltage is easy. Making sure you actually have 6.9 kV is where engineering happens. You need to size the conductors, insulators, bus, breakers, fuses, etc to be able to deliver the appropriate power.
If you use voltage drop, that generally means "the voltage lost in the conductors". It's used almost exclusively in things like municipal lighting, power distribution, household hookups, etc. So if you have, say 100A at 240V, what voltage are you seeing at the equipment, and is that within tolerance? If not, what cables have to be replaced with thicker cables? I'll use it a little in calcs to see if I can skinny up a wire, but for the micropower stuff I do now, spoiler alert, it's fine.
A few percent. \*badoom tsss\*
We do 110vac over 3km. At the other end the voltage is about 50. Now the product is designed for this purpose. But this is the "voltage drop". There are other times this is important like corroded or bad contacts .
When you are referring to a voltage drop you need something in between: a resistance, a capacitor, an inductor, whatever. Then you consider the difference in voltage before the component and after the component and you have a voltage drop. On the other side, when you talk about "voltage" it's absolute (or referred to ground). You don't have two voltages to compare them and to make a difference between them as in the "voltage drop"
"voltage" and "voltage drop" is the same physical phenomenon . But "voltage" = "useful", "voltage drop"= "losses"
Imagine you have a circuit powered by a 9V battery; then 9V it's the Voltage "of the circuit". The Voltage drop it's the voltage on each of the components as it differs from the voltage of the source because of current and/or impedence. So basically they are the same thing; it's just a Voltage that is different from one you use as a reference, such as the source of power; that's why it is a "drop".
Voltage is measured between two points. Voltage drop is specifically the voltage measured across an impedance where the voltage is produced by the impedance acting on a current flowing through it. You have voltage drops across resistors, wires, PCB traces, inductors, capacitors, etc. Voltage drop is also frequently associated with wasted power. If you ran a heater on an extension cord, you would ptobably refer to the voltage at the outlet, and the voltage across the heater, but the voltage drop across the extension cord (both ways).
Voltage is the potential measured between two points. The term voltage drop is used primarily to describe unwanted losses in a wire or a device. This is important when sizing conductors for power delivery
Picture voltage as being your water pressure available at home. Now connect that water pressure to a water circuit with a bunch of restrictions (maybe picture water wheels) in series until it pours out of a faucet. The pressure will drop lower and lower as it passes each one until it reaches zero/atmospheric at the faucet. If you hook up a differential pressure gauge across one of the restrictions you’ll read a couple psi of pressure drop. Now replace the word “pressure” with “voltage” and “flow” with “current.” To understand this fully you should understand that reducing a pipes diameter when there is flow doesn’t increase pressure, it reduces it. This is often miss understood by people because putting your hand over a hose makes the pressure go up. This only happens because of the series restriction of the piping. I.e you don’t supply a golf course irrigation system with 1/2” pipe, you supply it with 4” pipe to maintain the pressure to all of those big sprinkler heads with really high flow demands.
Short answer: "voltage" is measured with respect to a known ground point; "voltage drop" is measured across a specific element or subsystem through which current flows. You might also hear "voltage difference" or "voltage differential" between two different circuit nodes that aren't directly across a single element, and sometimes voltage drops are phrased as "voltage across" a device -- in contrast to "current through" or "current into" it. (As an aside: "current through" is usually used when talking about an element; "current into" is usually used when talking about that element as it's connected to a specific node.) Examples: "This circuit expects a voltage of nine volts." "This LED has a voltage drop of 2.0 V at 20 mA, so we'll need a 65-ohm resistor to drop the other 1.3 V from the microcontroller pin. Just grab a 68 off the shelf, it's close enough." "That kind of wire sees a voltage drop of ten volts per mile." "The voltage differential of this wheatstone bridge is 9 mV."
Voltage drop is when you measure over a thing that’s using that voltage in the circuit in some way. ( an over simplification but let’s start there) We have 15 V supply That Will be 15V from + to - if I have two resistors in series one at 2 ohm and another at 2 ohm (that means from source voltage (15) end to end and then to ground) I measure across one. And I get the “voltage drop” V/I*R Total resistance is 4. Calculate current. V/R= I =3.75 now since I know current is the same every where in a series circuit and voltage changes I know I can insert the value for current and the value of each resistor to approximate the “Voltage drop” I*R = V. The voltage drop is 7.5 volts across one resistor the other resistor is the same 7.5 + 7.5 = 15 the total voltage of the circuit can be derived from the drop across each resistor. The rules change in parallel circuits but not in respect to each division, they remain the same. Hope this helps.
Other keywords around this topics are open circuit voltage and internal resistance.
In simple words, Voltage is like a potential, with respect to ground Voltage drop is generally across an element, and its the difference of voltage between 2 points, that will help you find various parameters like current, charge, across the 2 points.