There are a number of questions here.

When you mention [math]12V\ 1A[/math], I will assume that you mean a [math]12V[/math] voltage source that can supply [math]1A[/math]. This means that while the voltage is (more or less) fixed, the current is not. The current does not have to be [math]1A[/math]. It could be less, but it can be maximum [math]1A[/math]. The current will depend on the resistance in the circuit.

You have a device or a chip that requires [math]5V[/math] with a max. current of [math]0.5A[/math]. This means that we need to drop the voltage from [math]12V[/math] to [math]5V[/math]. If you’re using a series resistor, the remainder of the voltage will be dropped across your device. The value of the resistor would be: [math]\frac{12V-5V}{0.5A}=14\Omega[/math].

There are a number of problems with this approach, however.

The first problem is that we’re wasting a lot of power. Concretely, we’re dissipating [math]7V*0.5A=3.5W[/math] to get[math] 5V*0.5A=2.5W[/math] of power. That’s only [math]\frac{2.5W}{2.5W+3.5W}\approx42\%[/math] efficient.

The second problem is that your average resistor can only dissipate about [math]\frac{1}{4}W[/math] of power — assuming the ambient temperature is low enough. We’re trying to dissipate more than 10 times as much power. That’s going to require a fairly beefy resistor. Not to the extent that it will be very hard to find, but it’s going to be more expensive than a standard resistor, which in turn will make the other options even more attractive.

The third problem is that the voltage will not be vary stable. Your load will likely not be as stable as the [math]10\Omega[/math] resistor I used to model this example. This can result in voltage spikes and drops and thus make the power supply unreliable, and it could even destroy your device.

The fourth problem is that your voltage source is unlikely to be a stable voltage source in the first place. If you’re starting from a [math]12V[/math] voltage source, my guess would be that it’s a [math]12V[/math] lead-acid battery (although most [math]12V[/math] batteries can deliver much more than [math]1A[/math] of current). The voltage of this battery will vary from [math]12.6V[/math] when fully charged to about [math]7V[/math] when fully discharged (although you shouldn’t discharge it this much). The voltage source’s voltage may vary, and thus the voltage drop across the load will vary as well:

When the battery voltage drops, we want the value of our series resistor to be lower as well, so that the voltage drop across the load can remain more or less constant. Fortunately, such device does exist: it’s called a voltage regulator:

This will make sure that the output voltage is a more or less stable [math]5V[/math] ([math]4.8V[/math]-[math]5.2V[/math] at room temperature). The most well-known (linear) voltage regulator is probably the [math]7805[/math]. This device would solve three out of four problems. It can handle the power dissipation, depending on the ambient temperature, but you will likely need a heatsink. The voltage will be stable regardless of the input voltage (as long as it’s within range of what the [math]7805[/math] can handle, which is [math]7V-25V[/math]) and the load.

However, we’re still struggling with the heat and the power loss. We really need a more efficient solution. Fortunately, such solution does exist: a switching DC-DC power supply, more specifically a DC-DC step-down buck converter. It has all the advantages of the linear voltage regulator, but it will generally be [math]>90\%[/math] efficient.

Given that you want to step down the voltage from 12V to 5V, which is extremely common, as this is what all cigarette socket to usb converters do, you can find these readily available, both as a bare-bones circuit and as a cigarette plug adapter. They will generally look like this:

and you can find them on eBay or AliExpress for less than 1$. You simply solder wires from your 12V voltage source to the board and you get 5V out of the USB port.

The cigarette plug adapters will look like this:

Given the low current requirements, any circuit/adapter will suffice.

In conclusion, given the low cost and availability of step-down buck converters that will do the job efficiently and effectively under a variety of circumstances, it is by far the best option and therefore should be preferred over all others options.

Look, the short answer is no.
However, the concept is a little tricky, so let me give it a shot.
Let’s start with an analogy, it might help:

I’ve tried to depict here a pipe section, where water is flowing. Do you think that water can escape the pipe, other than from the left or right sections (where I put the arrows)? It should be intuitively apparent that the answer is no. There is no water increase or decrease in this system, because all that’s happening, is happening inside the pipe, which is sealed, unless at its boundaries.

Imagine you now have a pipe-loop, filled with water. However you make the water move inside it, since the pipe is sealed, no water can escape, right?

That is, even if the water is flowing through the pipe, its mass must be constant at any time. We refer to such a fundamental observation stating a law of mass conservation. We don’t expect in such a system, which we call closed, any mass to enter or exit if not through the gates which we indicated with the in and out arrows, when we consider any (ideal, we’re not physically cutting the loop) section of it.

This is quite powerful, ponder it for a minute. [..] Done? Good :)

So imagine now that you have a piston, and you push the water on on the left border, towards the right of the diagram. Well, if we stick to our conservation law, then two things must happen.

1. The volume left by the water I’m pushing, must be immediately filled by new water coming from ‘somewhere else’ (but this is a loop, right? So…)
2. The exact same volume of water I’m pushing, must also leave from the other side, since there’s no other place where water can enter or leave, and we stated that the mass of water within the pipe cannot increase.
   (given this fact, and that we have a loop, I would say that what leaves from the right side is exactly the water that’s coming in…)

So, if you look at those grey rectangles I’ve depicted, they represent the volume of water I’m moving from the left to the right. They must be equal, for conservation to hold.

If we consider those two grey volumes as generated over one second, we give them the special name of flow.

We also say: the flow of water through this pipe is constant. As you see, this comes from the fact that no water can enter or leave the system if not through those sections. I cannot compress more water in that section volume, and I cannot have less water either.

However, there is something changing between the left and the right sides, isn’t there? What changes is the ratio between the horizontal and vertical dimensions of those rectangles. On the right, I’ve got ‘a lot’ of water travelling a small distance. On the left, I’ve got ‘far less’ water, travelling a larger distance.

But the product of these two dimensions stays constant. And this product is what we call ‘the water flow’.

You can now draw the parallel to the electrical case.

The pipe becomes an electrical wire, with different resistance (the section of the pipe).

Water becomes charge.

Water flow becomes current.

As in the analogy, the amount of charge within a circuit is constant. It cannot come from anywhere. If it moves from one section of any part of a loop, it has to be replaced by the same amount and this same amount must leave other places, like in a domino or… In a pipe loop.

Therefore, as in the water analogy, the flow of charge, we call current, must be constant throughout. If you take as a ‘section’ of your circuit, just a resistor, this applies in the same way. No charge can accumulate inside the resistor, so any charge that gets in from one side, must get out on the other side.

The ‘difference’ between current flow in the wire (larger section pipe) and the resistor (smaller section), is that in the wire there’s more charge, moving slower, whereas in the resistor there is less charge, which then has to move faster.
You know what makes charge move faster? A higher potential difference, aka Voltage drop. This means, to have the same current flow through a wire and a resistor, the voltage drop in the resistor must be higher than that of a wire.

[math]V = R \cdot I[/math]

Voltage drop is result of current flowing thru resistance and is calculated using Ohm law:

Consider electricity (wires) as pipes and flowing water. In pipe water that get’s in must get out. Water cannot just evaporate.
Same for electricity, current which enters circuit from battery must be identical to current existing circuit, returning back into battery.

All these is based on most fundamental law - conservation of energy.

For example this circuit:

Current thru all three resistors is constant - I.
Sum of voltage drops on each resistor must be Vs (12V).

But could be said that voltage drop does not affect current, current affects voltage drop.

In a way, there is. The higher the resistance, the lower the current flow.

But the reason you don't have a current drop across a resistor is because the same electrons that are entering the resistor are also existing the resistor. If you place an ammeter before and after the resistor you should get the same amperage.

If you ran water through a garden hose, the same amount of water comes out of the hose as goes in. The pressure changes depending on where you take the measurement, but the flow of water doesn't change throughout the branch.

So, measuring current is really measuring the number of electrons that pass a point over 1 second. The same electrons pass through the resistor's entrance as the exit, therefore flow is the same no matter where it's measured.

It is tempting to imagine a resistor works by reducing either current or voltage, but that’s not the most useful mental model of how a resistor works.

A better way is to think of a resistor ([math]R[/math]) as creating or forcing a relationship between current ([math]i[/math]) and voltage ([math]v[/math]), as captured by Ohm’s Law, [math]v = i R[/math].

You can of course write Ohm’s Law other ways, [math]i = v/R[/math] or [math]R = v/i[/math].

Ohm’s Law represents the relationship between three quantities, which makes it awkward to focus on only two, as in, “does a resistor reduce current?” It is more useful to think about resistance with phrases that include all three quantities, like this,

If you are given a voltage, increasing resistance makes current go down.
If you are given a current, increasing resistance makes voltage go up.

in the atomic level, there are only two components to be considered. the positive charge particle (proton) and the negative charge electron in this configuration…

F is the electric force acting on the particles to accelerate towards one another. the electric force is called the coulomb force which is also the unit of charge.

voltage is the energy generated by the coulomb force exerted in the particles. if there is an equal amount of charge in the particles, they are said to be in an equilibrium state or neutrally charged and no motion of charge is present. when there is unequal amount of charge in the configuration, the particles will exert a force of attraction. the particular energy generated by these moving charges is expressed as the potential energy with a unit of volt.

1 volt is equal to 1 coulomb of force when 1 ampere of charges create unequal net (negative) charge between the particles.

for me, the most intuitive analogy is to imagine the electric field that binds the proton and the electron as an elastic substance or a coil spring.

let's say the wall to which the one end of the spring is attached is the unmoving massive positive proton and the square box is the lighter moving negative electron. when a coulomb force is applied in either direction as attraction or repulsion in the box, a compression or stretching force is generated in the spring. that pressure in the spring or the electric field is what we call the voltage. its oscillating motion in alternating current is the electrical energy that does work in the electrical load.

*Voltage Drop and Resistors:*

Adding resistors to a circuit affects voltage drop in the following ways:

1. *Voltage Division*: Resistors divide the total voltage into proportional parts, depending on their values.

2. *Voltage Drop*: Each resistor reduces the voltage by an amount proportional to its resistance and current flowing through it (V=IR).

*Current Limiting with Resistors:*

Resistors are primarily used for current limiting purposes due to:

1. *Reducing Current*: Resistors reduce current flow by converting excess energy into heat.

2. *Protecting Components*: Resistors prevent damage to sensitive components from excessive current.

3. *Voltage Regulation*: Resistors help regulate voltage by limiting current fluctuations.

*Key Factors:*

1. *Resistance Value*: Higher resistance values reduce current and increase voltage drop.

2. *Current Flow*: Increased current flow increases voltage drop across the resistor.

3. *Power Rating*: Resistors must be able to handle the power dissipated (P=V^2/R).

*Types of Resistors:*

1. *Fixed Resistors*: Provide constant resistance.

2. *Variable Resistors*: Allow adjustable resistance.

3. *Power Resistors*: Designed for high-power applications.

*Applications:*

1. *Voltage Regulation*

2. *Current Limiting*

3. *Signal Attenuation*

4. *Impedance Matching*

5. *Heat Sink Applications*

*Theoretical Framework:*

1. *Ohm's Law* (V=IR)

2. *Kirchhoff's Laws*

3. *Thevenin's Theorem*

4. *Norton's Theorem*

*Expert Guidance:*

Qaisar Hafiz

Ex IES, Managing Director - Engineers Zone E-Learning P Ltd

B.Tech. Hons. IIT Roorkee

5 Times IES qualified, AIR 2

9873000903/9873664427

Would you like more information on:

1. Resistor networks?

2. Voltage division?

3. Current limiting techniques?

4. Power electronics?

5. Circuit design principles?

Or explore:

1. Resistor materials and properties?

2. Thermal management in electronics?

3. Electromagnetic compatibility (EMC)?

4. Circuit simulation software?

5. Electronic design automation (EDA)?

Farmer Jack has three tonnes of hay and one complete, serial circuit containing three horses.

Three tonnes of hay shall be delivered by Farmer Jack's voltage wagon at an interval of ‘per week'. In one week:

Horse #1 eats a half tonne of hay.

Horse #2 eats one and a half tonnes of hay.

Horse #3 eats one ton of hay.

Each delivery of high-voltage hay serves mutually-exclusive purpose.

Voltage wagon hay drop arrives at Horse #1, who sees three tonnes of hay arriving, but only gets a half tonne of hay.

Tally: Half-tonne of hay dropped, 2.5 tonnes of high voltage hay remain.

Jack's wagon arrived at Horse #2, where horse realizes he accepts only 1.5 tonnes of hay, only to realize one tonne of hay eludes him.

Tally: Two tonnes of hay dropped, 1 ton of high voltage hay remains.

Its eyes wide open, hay wagon departs drooling Horse #2 and proceeds along to Horse #3, who sees one tonne of hay and receives one tonne of hay.

Farmer Jack dematerializes like Donkey Kong Mario and starts the cycle again the following week.

The resistor serve function like each horse, using only a specific amount of potential from the three tonne hayload.

The high voltage hay across the horse resistor will not be what drops but rather what passes by to the next horse.

High voltage hay is not something farmer Jack wants falling on him.

• Coming up Next, we substitute “week" with “microsecond,” as Farmer Jack gets a faster hay wagon … and new feed.

When a wire overheats from too much current running through it, what is happening at the atomic level?

This is easiest to understand if you first have a look at what’s happening in a conductor with no current. All the conducting electrons still exist in the conductor - they’re just some of the electrons associated with the conductor atoms. So with no current flowing are these electrons stationary? No, they’re not. Every particle in a material has kinetic energy because the material contains heat (assuming that the conductor is not at absolute zero degrees kelvin), and heat is nothing more than kinetic energy in the atomic particles.

Now the atoms just vibrate with this energy because they’re held in position, but the conducting electrons are free to move and so move they do. In fact you can work out their average speed simply from the temperature, and it turns out that at room temperature for example every electron is travelling at around at an average of over 100km per second.

So why isn’t there a huge current caused by these high speed electrons? The answer is that it’s because the direction that these electrons are moving in is random. Each electron is moving around bouncing off the atoms in a random path. And every time an electron bounces off an atom then there is an energy exchange between the two. If the electron leaves more slowly than it arrived then it has given energy to the atom, and if it leaves faster than it arrived then it has received energy. This means that the electrons and atoms are in a constant thermal equilibrium. Their individual kinetic/thermal energies vary all the time, but their continuous interaction shares out the heat energy evenly.

OK, so now consider what happens when a current flows. This happens when an electric field is formed along the conductor, causing each conducting electron to experience a force and so accelerate along the conductor. But this movement is just a small amount of additional kinetic energy gained between each collision with an atom, on top of the kinetic energy the electron already has. The only difference is that this movement is a very slight bias so that the random electron paths now all creep in one direction. The individual electrons of course carry on with their thermal paths, just with slightly more kinetic energy being gained continuously from the electric field causing the current.

But kinetic energy is kinetic energy - the current is just a small increase in the average speed of the electrons. And as the electrons are in thermal equilibrium with the atoms then they just share this additional energy with these atoms, increasing the temperature of the conductor as a whole. Indeed if it wasn't for this continuous energy sharing then the electrons would just keep accelerating indefinitely due to the electric field.

So in reality the kinetic energy in the electrons due to the current is nothing more than additional thermal energy continuously being given to the electrons by the electric field. The stronger the electric field then the more rapidly they acquire this additional thermal energy. And this thermal energy shared with the rest of the conductor increases the temperature of the conductor. The higher the current then the faster the additional heat is being added and so the higher the temperature of the conductor. At some point the wire will overheat, depending on how quickly the wire is able to lose heat to the environment.

Electric current as far as electrons are concerned is nothing more than additional heat being acquired from an electric field.

That’s not exactly what happens. In a basic DC circuit with a given resistance for the load, if the voltage drops, so does the current. This is what Ohm’s Law says:

V = IR,

where
V is voltage in Volts
I is current in Amperes (Amps)
and
R is resistance in Ohms.

This relationship can be written in two other forms, each isolating one of the 3 variables:

V / R = I,

and

V / I = R.

As an exercise, try using the second of these (V / R = I) with any arbitrary value of V and R to obtain a current I.
Next, using the same value for R, try a lower value for V. You’ll see that I is smaller as well.

In order for current to increase when voltage drops, the circuit resistance R must decrease by more than the voltage.

If you got a “voltage drop “ across a resistor in a DC circuit, doesn’t it mean that the rate flow of the electrons, at the other end of the resistor would decrease per second?

Yes, and it will decrease at both ends.

If you get a pressure drop in a pipe across a partly open valve, will you have more water coming in than leaving, or will the flow rate of water coming out match the flow rate of water coming in?

Electrons are particles with a charge of -1. If more flowed into a resistor than left, they would build up, raising the voltage. Otherwise, you’d have to violate both conservation of mass and conservation of charge.

This answer for who really understand voltage and resistance.

To simplify this answer we will use one electron

We will deal with real moving of electrons not Conventional current notation

when an electron go from battery from negative terminal to positive terminal it has energy to do work. right?

when connect resistance to the battery and form a circuit.

Again an electron go from battery from negative terminal to positive terminal we now know that electron has energy to do work when go through resistance it lost all of its energy which make voltage drop.

which means that all voltage lost across that resistance.

point before resistance an electron has energy and point after it that electron has no energy.

voltage means potential difference between two points, so point before resistance is high potential and point after it is low potential.

let’s give another example. we will deal with two resistors.

when an electron go from negative terminal of battery to positive terminal and go through first resistor it will lost some of its energy and continue to second resistor to lost remaining energy.

quantity of lost energy in two resistor dependent on resistors values.

SO FOCUS HERE.
voltage is potential difference between two points.
point before first resistance has least potential and point after first resistance has greater potential.
point before second resistance has the same potential after first resistance and point after second resistance has greatest potential.

Nothing less Nothing more

Actually there is a lot of loss.. just not how you perceive it!

A resistor tries to prevent current flow through it. So to FORCE current to flow through it, one needs an electric potential source. This potential source pushes the current, through the resistor (which essentially throws a tantrum, get all hot and bothered!) This causes the resistor to heat up and this heat dissipates in the air/surrounding medium causing loss of energy. Meanwhile the potential source (assuming it is a battery) finally gets tired (drained) and the current flow drops - eventually going to zero.

So: yes there is a loss - but not really of charge (or current), rather loss of energy which changes its state from electical to thermal and eventually goes into the air (or the medium surrounding the resistor).

There’s another point to remember: Electrical conduction through a resistor works as follows: The potential source pushes electrons from one end. Now since there can’t be any excess electrons inside the resistor, an equal number of electrons are forced to pop out from the other end. This process is obviously not easy and creates heat .

A good analogy is as follows: imagine you are boarding an overcrowded bus: The bus has a finite capacity. In order for you to get on the bus (say from the rear gate), someone from the front gate needs to get off! Now you get on the bus and some one behind you gets on the bus! Which means one more person from the front needs to get off! This process keeps repeating till you get off and by the time you do - you feel tired since a lot of energy was expended in pushing people ahead of you!!!

Hope this helps!

Current constantly changes in an inductor

Self inductance

resistance is opposition to current flow

By putting a resistor in you are limiting current

.

There's resistance

Inductance and

Capacitance

Resistance is linear

In it of itself it's not changing anything

An inductor is changing things

• Since V=I×R

From the below given diagram you can observe that,

• If R increases then I decreases (as R varies inversely with I)
• Whereas voltage is remaining constant.

Image source : Google.

Joule per Coulomb.

The voltage is the pushing force. It is also a potential. How much a hill is tilted. The more tilt, the faster the ball will roll.

When in doubt, consult the hyperphysics website:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html#c3

In an atomistic sense, it translates to the sum of energy levels contained within several states occupied by the particle-wave. Or, in plain English, how much work is needed to go from energy state A to energy state B.

from STM: The Details

A transmittance scheme through tunneling from two different potential wells. Note the change in workfunction(phi) from the Fermi('sea') level.

By definition, electric potential exists with respect to another point, often coveniently (but sometimes arbitrarily) chosen as 0.

Voltage drop where? Across the load? Across the wires? Are the wires ideal or real? The question doesn't refer to any device in particular, so it can be answered differently depending on what you refer to. Let's consider a resistor, an inductor, an open circuit, and a short circuit.

• In a typical resistor (or a device/equipment modeled as a resistor, such as cables/conductors), the answer is no, if the current is zero, the voltage of the resistor will also be zero. This is Ohm's law. In the following picture, even if [math]v' \ne 0[/math], the resistor's current and voltage will both be zero, due to the open circuit and Ohm's law, respectively.

Another example: In the following picture, an ideal short circuit (0 volts in the short circuit) has occurred in parallel with the resistor, so from Kirchhoff's voltage law the resistor's voltage is also zero, and from Ohm's law the resistor current is also zero, so all the current flows through the short circuit.

• In an inductor/solenoid/coil/winding, there can be instantaneous voltage even if the instantaneous current is zero, because the induced EMF in the inductor depends on the change of magnetic flux (or current), not on magnetic flux (or current) itself. This is Faraday's law. In the following picture it is shown the voltage and current, against time, of an inductor operating in sinusoidal steady-state, where you can see that when [math]i = 0[/math], the voltage is actually maximum in magnitude.

If instead of instantaneous voltage and current you meant RMS voltage and RMS current, then for a constant-inductance inductor operating in sinusoidal steady-state, the answer is no. If the RMS value of the sinusoidal instantaneous voltage across the inductor is zero, so will the RMS value of the sinusoidal instantaneous current through the inductor. This is Ohm's law generalized to the frequency domain.

• In the open-circuit terminals of a source (battery/generator), the answer is yes, there is voltage and the current is zero; however, this shouldn't be called voltage drop as you said, but an EMF (voltages can be classified as EMF or as voltage drops). The present voltage is due to the source, not to Ohm's law (in an ideal source, its voltage is independent of its current, so it doesn't obey Ohm's law). In the following picture, the current is zero due to the open circuit, and the output voltage is equal to the source's voltage, which is non-zero.

A real-life example would be the wall outlets in homes: ignoring transmission line theory, when there's nothing connected to the outlet, there's no current, yet the voltage is always there, available.

Some people may explain this by saying something like “applying Ohm's law, where [math]R = \infty[/math] (open circuit) and [math]v[/math] is the source's voltage, you get [math]i= \lim_{R \to \infty} v/R = 0[/math]”, but I find it inappropriate to use Ohm's law in an infinite resistance (open circuit), although it gives the correct answer, because if you look at the situation from the following point of view, you get a contradiction: from Ohm's law, we know [math]v \propto i[/math], so how come in an infinite resistance you have [math]i=0[/math] but the voltage can be non-zero (i.e. there is a voltage across it)?; it is contradictory because if the current is zero then the voltage must also be zero, yet it isn't. An ideal open circuit isn't a resistor, so you shouldn't apply Ohm's law to it. Another way to look at why you shouldn't apply Ohm's law in an open circuit: since [math]R = \infty[/math] and [math]i = 0[/math], the voltage is [math]v = R i = \infty \times 0[/math], which is an undefined mathematical operation.

• In an ideal wire, the answer is no, the voltage across it will always be zero, whether there's current flowing through it or not. This is by definition. In the following circuit, a shortcircuit has occurred (or perhaps it's just a wire carrying current), and since the wire is assumed ideal, it has no voltage drop, yet current it has non-zero current flowing through it.

Some people may explain this by saying “applying Ohm's law, where [math]R=0[/math] (short circuit), you get [math]v = R i = 0 i = 0[/math], with [math]i[/math] equal to any value”, but I find it innapropiate to use Ohm's law in an ideal wire (shortcircuit), although it gives the correct answer, because if you look at the situation from the following point of view, you get a contradiction: from Ohm's law, we know [math]v \propto i[/math], so how come in an ideal wire you have [math]v=0[/math] but the current can be non-zero (i.e. current is flowing through it)?; it is contradictory because if the voltage is zero then the current must also be zero, yet it isn't. An ideal short circuit isn't a resistor, so you shouldn't apply Ohm's law to it.

It moves through resistors with effort. That's what resistance means -- specifically, resistance to the flow of electrons.

A resistor is not a barrier to current flow; a capacitor is, in a DC circuit. A resistor does not store up energy, as an inductor does. A resistor just impedes current flow.

In so doing, a voltage differential builds up from one terminal of a resistor to the other. Within the resistor, the energy consumed in overcoming the resistance shows up as heat.

In a low-value resistor, a given current will create some amount of heat. In a high-value resistor, the same amount of current can only be produced if a higher voltage is applied. That current, equal to the current in the small resistor, will produce significantly more heat. In fact, the heat will be exactly proportional to the values of the two resistors.

Think of electric current like water flowing through a pipe. If you would stuff some balled-up steel wire inside the pipe, the water would have a more difficult time moving through the pipe. The wire would provide resistance, which would impede the flow of water.

You could then get the same rate of water flow through the pipe, if you would increase the pressure of the water. The stream would have to do more work to get through the obstruction, but with a sufficient increase in pressure you would get the same output; or by analogy, with a higher voltage applied when you put a resistor into a circuit, the same current can be achieved.

Why does the voltage drop occur when the current passes through a resistor? What does it actually mean?

The voltage drop takes place between one end of the resistor and the other. If there is no connection through the resistor there is no voltage drop because there is no current flow.

When electrons flow work is done. Heat is produced. The rate of energy used is calculated from the voltage drop and the current or resistance.

The movement of electrons is easy in conductors and very difficult to impossible in insulators. In between are semi-conductors, which are pure materials to which very small and carefully controlled impurities have been added. Their conductivity can be altered by applying external influence, light or extra, special, connections for example. All conductors have resistance. Some special materials become super-conductors, with really zero resistance, and temperatures near to absolute zero.

Resistors are made of non metallic materials, metals that are not good conductors, or alloys of metals that are not good conductors.

‘Resistance’ (Ohms) is a measure of the ease of movement of electrons through materials. It relates to the atomic and crystalline structure of materials.

In order for current to double

Voltage is doubled first

Current then would double

Voltage went up so drop went up

Cant stay the same

Google these:

“Electron theory of current flow” and “water pipe analogy of electric circuits”.

That fact that you have asked this question tells me you don’t have any concept of what is going on in an electric circuit. You need to have some sort of mental concept of what is going on in a circuit to be able to work with it.

In the Navy they taught “Electron theory of current flow” in a circuit. It provides you with a mental picture of current flowing in a conductor. The other explanation you might what to look at on Google or you tube is “water pipe analogy of electric circuits”.

Now to answer your question.

E = I X R as E increases so does I, R is usually a fixed value of resistance.

E volts = I (current in AMPs) X R resistance in ohms

E volts this is a Force, think of it like pressure from a pump. I is current, think of it as a flow of electrons, like water. R is resistance to the flow of electrons. Think of it like a crimp in a water pipe or a kink in a hose. In a circuit Voltage is usually a constant within the limits of the supply. Think of a pump that puts out a certain amount of pressure. If you increase the Pressure you will get more current. And a greater voltage, pressure drop across the resistance.

Ohm's law states that the current flowing through a conductor is directly proportional to the voltage applied across it, provided the temperature and the physical properties of the conductor remain constant:

[math]I=V/R[/math]

where I is the current, V is the voltage, and R is the resistance of the conductor. In this case, the conductor is our resistor, where the same relationship holds. It

For a reason, I think Ohm's law sums it up well. The experimental proof is straight-forward:

1. Ohm's law can be experimentally proven by measuring the current and voltage across a conductor and showing that they are directly proportional.
2. This can be done using a variety of electrical measurement devices, such as
1. ammeters,
2. voltmeters, and
3. ohmmeters.

To verify theoretically, you could analyze circuits using Kirchhoff’s laws, which are based on the fundamental physics principles:

• conservation of charge
• conservation of energy

These laws are Kirchhoff’s voltage law and Kirchhoff’s current law.

• Using these allow you to analyze and solve more complex circuits, measure the results physically, and compare theoretical and physical results.
• if Kirchhoff’s laws hold true, and Ohm’s law can be used to verify Kirchhoff’s laws, you have your theoretical proof.

Worthy of consideration: Ohm’s, like many physics-based derivations, describes an ideal relationship between current, voltage and resistance. For real-world experiments, other factors may be needed to quantify internal or external confounding variables, as mentioned above.

The relationship between voltage and current in a transformer is based on its turns ratio, which is the ratio of the number of turns on the primary winding, to the number of turns on the secondary winding.

As you can see from these relationships, when the secondary voltage is lower than the primary voltage (i.e., a step-down transformer), the secondary current will be higher than the primary current by the same ratio. Likewise, when the secondary voltage is higher than the primary voltage (i.e., a step-up transformer), the secondary current will be higher than the primary current, again by the same ratio.

You can understand this relationship intuitively by considering a lossless transformer, where the power flowing into the primary winding (P1 = V1 • I1), is equal to the power flowing out of the secondary winding (P2 = V2 • I2). Since (in a lossless transformer) no power is lost in the transformer itself, P1 = P2, so that V1 • I1 = V2 • I2., or V1/V2 = I2/I1, as outlined above, which illustrates the intrinsic inverse relationship between primary and secondary voltage and current in a transformer.

The current does not change as a result of passing through the resistor. The current is always the same on both sides (or both ends, if you prefer) of the resistor.

A resistor, in conjunction with the applied voltage, defines the current in the circuit (on both sides of the resistor).

If there is more than one connection on each side of the resistor, the sum of the currents in the conductors on one side equals the sum of the currents in the conductors on the other side. What goes in has to come out!

Assuming that the circuit is composed of passive components…

If you insert a resistor into a branch of a circuit, the current in that branch is reduced by its presence. Both sides of the resistor.

On the other hand, there will be a voltage drop across the resistor.

Resistors impede the flow of current. They ‘resist’ it. You can alter current in a circuit by altering the value of a resistor, or applied voltage. Current is not a property, it is a condition, the result of other factors. The value of a current, defined in amperes, is the result of calculation. When one volt is connected to a load (resistor) of one ohm, the current that passes is defined as one ampere. That is the way it is. It cannot be changed. This is Ohm’s Law. It is universal.

‘Passive components’, above, are resistors, capacitors and inductors in any combination. (Wire, or connections, are assumed to have zero resistance and inductance.) My exclusion relates to non-linear components, amplifying devices and things like motors and cells. The ‘constant current’ circuit device is very useful and maintains, through the use of ‘active’ devices, a steady current in spite of changing voltage.

There's no current drop across a resistor because current is the term used to indicate the effective rate of flow of charged particles[math]^1[/math] (usually electrons) through the resistor, and this rate is the same at both ends of the resistor. There IS a drop in potential energy across the resistor; the potential energy associated with charged particles exiting the resistor is less than that of the charged particles entering it.

For example, consider a 1Ω resistor connected across the terminals of a 1V battery. Ohm's Law tells us that 1A of current is the resulting flow rate (1A = 1 coulomb per second, where a coulomb is 1.62 x 10[math]^{19}[/math] electrons). The potential energy drop of each electron is 1V, and this energy is lost in the form of heat.

[math]^1[/math] Current is really a measure of the flow of energy, which is not exactly the same as the rate of flow of discrete particles, but the difference is unimportant in a discussion of the basic differences of current and voltage in a resistor.