Suppose I have a conductor connected to a battery. The current starts flowing through the conductor.

Why is the current entering the conductor the same as the current leaving the conductor?

Its kinetic energy should be reduced because of collisions inside the conductor and reduced kinetic energy should result in reduced current.

  • $\begingroup$ Doesn't scale make all the difference? Won't a narrower conductor give more resistance, etc… That is to say, everything in your circuit might change the outcome but whether anything will, depends on scale:" ie, its relationship to everything else. $\endgroup$ Commented Nov 23, 2021 at 20:06
  • $\begingroup$ Current != energy $\endgroup$
    – OrangeDog
    Commented Nov 24, 2021 at 11:58

9 Answers 9


If, say, 3 charges enter each second then also 3 charges must leave each second for a steady state current.

  • If more leave than enter each second, then where would the extra leaving charges come from? This is not possible.
  • If fewer leave than enter each second, then some charges are staying behind within the conductor. Over time the net charge in the conductor thus accumulates, increases and increases. This is not impossible, but...

Remember that like charges repel. An ever-increasing negative net charge within the wire will more and more strongly repel new incoming negative charges (electrons e.g.). Soon the net charge is large enough and with a repulsion large enough to balance out the battery voltage. Then no more charges will enter, and the current will stop entirely. Since this does not happen in wires at steady state conditions in working circuits - since we clearly see that the current does not stop flowing - then all charges that enter must also be leaving each second. This is a part of Kirchhoff's current law.

You are correct, though, that charges may lose kinetic energy as heat while flowing through. They will as a result indeed slow down.

  • The next-coming charges will then have to "wait". They will "queue up" behind the slowed-down charges. In a split second all following charges will thus have slowed down to the same (drift) speed.

  • In front of the slowed down charges, we could imagine the leading charges continuing ahead at higher speed. A gap would form behind them and in front of the slowed-down charges. But these leaving charges now do not "feel" the same "push" from behind. So what keeps them flowing at high (drift) speed? Any disturbance that slows them down will bring them down to the same slower speed as the slowed-down charges have reached. Also, the "gap" behind them will be a spot of less negative charge, thus a spot that they are attracted to. Such attraction from behind will also slow them down. Plenty of factors will thus eventually cause any leading faster-moving charges to slow down and match the speed as well.

All this happens in a split second. Close-to instantaneously in most practical purposes. In a split second, all charges move equally fast throughout. The current in all parts of a conductor is then the same. If you turn off the current, add a resistor component and turn on the current again, then in a split second the current will from the same reasoning stabilise at some new, lower steady state current matching this new resistance. So, when steady state has been reached (which happens in a split second in regular conductors and only is delayed when certain components are involved that are specially designed for it, such as capacitors) you will always see the same current at all points.

  • $\begingroup$ Pls explain in detail $\endgroup$ Commented Nov 22, 2021 at 18:17
  • $\begingroup$ I couldn't understand your 2nd paragraph $\endgroup$ Commented Nov 22, 2021 at 18:17
  • 1
    $\begingroup$ @VaibhavTiwari I have rephrased the second paragraph with more detail. The main point is that if more charge enters the conductor than leaves the conductor, then the missing charge must have stayed behind within the wire. Over time, this means that more and more charge accumulates in the wire - this would eventually stop the current from flowing. Since the current does not stop flowing, then we know that this isn't happening. No charge is accumulating, and thus all charge that enters also leaves each second. Let me konw if this is still unclear $\endgroup$
    – Steeven
    Commented Nov 22, 2021 at 18:53
  • 1
    $\begingroup$ "if more charge enters the conductor than leaves the conductor, then the missing charge must have stayed behind within the wire. Over time, this means that more and more charge accumulates in the wire - this would eventually stop the current from flowing." This is what actually happens in capacitors, in DC environments. $\endgroup$
    – nick012000
    Commented Nov 23, 2021 at 0:56
  • $\begingroup$ @nick012000 Yes, indeed. Some components are directly designed for this effect. $\endgroup$
    – Steeven
    Commented Nov 23, 2021 at 8:07

I believe your question is about how a current can remain constant after passing through a conductor. The water analogy introduced by Roger Vadim reminds me of another explanation. A kid asks HC Verma, a renowned Indian physicist, Why does current not decrease on passing through a resistance.

HC Verma's explanation is in some sense a "Proof by contradiction" method. So let's assume 2 things first. Firstly, let's assume that the flow of electrons in a conductor is analogous to the physical flow of water in a pipe. In other words, $I=\frac{q}{t}$ is analogous to the flow of water $f=\frac{V}{t}$ where $q$ is the charge flowing through a cross-section and $V$ is the volume of water flowing through a cross-section. Also assume water is incompressible.

Secondly. let's make the assumption that the current does in fact decrease as it passes through a resistor. This is going to be the assumption we are going to contradict.

Saying that a current decreases after passing due to resistance is sort of like saying that the flow of water passing through a pipe decreases due to friction. Here, the resistance and friction play the role of opposing the flow of electrons and water respectively. But this causes a dilemma. If the initial flow is more than the final flow, this implies that the volume of water entering the pipe per second is more than the volume of water leaving the pipe per second. Where is the water disappearing? Is it getting stuck somewhere in the pipe? Well if that's the case and if chunks of water were to be accumulated within the pipe, then it would cause the pipe to burst... which does not happen in practical scenarios.

What this tells us is that the volume of water entering the pipe per second should be equal to the volume of water exiting the pipe per second. Analogously the amount of charge entering the resistor per second should be equal to the charge exiting the resistor per second. In other words, the initial current and the final current are both equivalents.

This reasoning arises from the conservation of charge which is often expressed by the continuity formula given by Roger Vadim. Incompressible fluids, like the one we used in our analogy, too have a continuity formula given by $A*v=m=$constant, where $A$ is the area, $v$ is velocity and $m$ is mass, which is ultimately the conserved physical quantity.

Note that in the video, HC Verma spoke about the flow of water in terms of velocity, arguing that it remains constant. But this is not strictly true as it depends on the area of cross-section, as shown in the continuity equation. Less area means more velocity. You can picture this easily. Likewise, the velocity of electrons (ie drift velocity) is also inversely proportional to the area of the conductor. But that's a different formula that you will come across later.


I think your question seems to imply that you think that charges at one end of the wire are "fired" through the wire with some initial kinetic energy and and the resistance slows it down.

The electric field inside a wire is constant so at each point in the wire charges experience a force Eq

the resistive forces inside the wire are proportional to the velocity of the charge, normally in the presence of an electric field, charges want to ACCELERATE

When the Electric force equals the resistive force , the charges move at a constant velocity . it is the "terminal velocity" of charges in the wire that makes current density at a point constant.

if the E field is constant throughout the wire then we can say current everywhere is constant

Search up the drude model of conductivity.

The equation of motion of an electron :

$ma = Eq -(m/T) v$

initially the electron in the presence of an electric field IS accelerating , and therefore there IS a changing current density at a point initially. However very quickly the resistive forces equal the applied electric force such that "a" = 0

Meaning $ Eq-m/T v =0$

$v= (qT/m) E$

by definition J = $\rho v $ or $J= nq v$ where n is some number density. Plugging back into our equation

$$J = (nq^{2}T/m)E$$ $$J = \sigma E$$

which means current density J in the steady state is proportional to the electric field at a point. so for a constant electric field, J everywhere will reach some constant value


The charge conservation (often expressed by the continuity equation, $\partial_t\rho + \nabla\cdot\mathbf{j}=0$ means that, the difference between the charge entering the conductor and the charge exiting it, is accumulated as charge within this conductor. So in your scenario, the charge of the conductor should grow to infinity.

What is misleading here is associating the current with instantaneous electron/charge velocity, rather than its average velocity and the quantity of charge. Electrons scatter all the time against the impurities or phonons in the conductor, so their velocity is changing all the time, but on average the number of electrons passing through any cross-section per period of time is the same (unless we have charge accumulation mentioned in the beginning).

A good and obvious analogy here is a flow of water. Imagine a waterfall - the water in the waterfall flows much faster than the water in the river before it reaches the waterfall or after the waterfall. Yet, the quantity of water entering the waterfall and exiting it remains the same: in some parts, it accelerates, in some decelerates, but the average speed is maintained the same.


Electrons are not billiard balls. They are not isolated particles bouncing around in the wire. They interact with one another via electrostatic forces. If you put two electrons in an area, they repel each other.

If you push electrons into a conductor (from a battery) without letting them out, the electrostatic forces will push the electrons apart. This means you will have to put more and more potential into the circuit (stronger battery) to drive more electrons in. This electrostatic force will build until it prevents your battery from putting any more electrons in. Once given a path to a lower potential (such as the other terminal of the battery), they will flow that way naturally because there's less electrostatic force being applied on this side.

The typical analogy for this is water flowing through a hose. Water is incompressible, and the hoses are pretty rigid. If you try to stuff more water in while stopping the water from leaving, the forces pushing back against you quickly grow to insurmountable levels.

Now there is a transient case, right as you connect the battery where this model breaks down. Right as you hook up the battery, electrons do flow into the circuit more than they flow out. However, this is incredibly short. It may take picoseconds (trillionths of a second) for the electrons to rush around and arrive at a stable equilibrium where the electrostatic forces are balanced.

Why doesn't that happen in your conductor? You are learning about ideal conductors. They are simplified to make them easier to learn about. If you are modeling these ultra-fast effects, you will include other properties such as capacitance and inductance to make a conductor that acts like a real conductor. If you're developing modern high-speed networking equipment, you care about these things. But for now, don't worry about them. Just know that the ideal model you are learning about does have its limits, and focus on how the systems behave once they are at "steady state," where the electrostatic forces are in balance and the electrons flowing into a conductor are equal in number to the electrons flowing out. Later, after you learn about capacitors and inductors, you can revisit these non-ideal conductors.


Bead Necklace

This idea of charges going into a resistor moving faster than charges going out makes sense if the pool of incoming charges and outgoing charges aren't directly connected...like a river flowing into a narrow channel, which then drains into the ocean. This would be like charge from a cloud traveling through the air to the ground in a lightning strike.

However, an electronic circuit is not like any of those things. The source and the sink are connected, and this is what changes everything. Instead of imagining the electrons as independent objects, you should think of them as beads on a necklace that all move together. You cannot move one bead without moving all the beads.

Now, voltage is like the force which pushes the beads into motion about the necklace. Current is like the speed at which the beads are moving along the necklace. Resistance would be like putting your hand on the beads to slow them down. At this point, it should be intuitively clear what is happening: you don't just slow down the beads coming after your hand. You actually slow down all the beads, because they are connected. The act of slowing down a bead causes a chain reaction which proceeds all the way round the circuit to the start of your hand.

Because we think linearly, we tend to analyze systems in a reductionist fashion, looking at each component in isolation. One of the most difficult parts of understanding electronic circuits is that everything is happening at once. Something that occurs in one part of the circuit affects what is happening on the other side because of the way effects travel through the current stream. But if you just remember that they are all connected like beads on a necklace, that should clear up at least some of the misconceptions.

Apply Brakes

To put it another way, this question is a bit like asking: "When you apply a brake to a wheel, why doesn't the part of the wheel after the brake slow down to a lower speed than the part before the brake?" Technically, it does, at the microscopic level, but only due to the speed of sound in the wheel material.


You're forgetting that there is a driving force behind the current - in a steady state, the force causing the motion of (e.g.) electrons must be the same as the force that resists their motion. This force is related to the charge. If an individual electron were to slow down, it would get closer to the other like charges, increasing the force that repels them from each other (while ahead, the charge is now more positive, attracting the electron). The electron can't slow down, because there's more electrons pushing from behind, and more "holes" pulling from ahead.

If you want an image that's perhaps easier to grasp, consider an arrow being shot from a bow. It takes a lot of energy to get the arrow to move fast, because the arrow has significant mass. Why doesn't the arrowhead get shot out at incredible speed while leaving the shaft behind? Because they're connected. It's much the same with electrons in a wire.

In contrast, consider making a line of perfectly aligned billiard balls. Striking the first one (at the right angle) will cause the last ball in line to launch with the same energy you gave the first ball - it will not cause all of the balls to launch at correspondingly lower velocity.

If current worked the same way, your picture would be correct - the current would be the highest near the battery, and drop off with distance. But it doesn't, because the balls are connected - they either all move, or none of them do (in the steady state). The current will be lower because of the resistor, but it will have the same value in all the serially connected parts of the circuit.

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    $\begingroup$ Emphasizing the continuously acting electrostatic force is important. It's like pushing a box across the floor -- shouldn't it slow down because its kinetic energy is reduced by the friction? $\endgroup$ Commented Nov 23, 2021 at 23:05

Imagine a long stretch of road, with a speed limit that changes from 80 to 60 in the middle. 5 cars per second enter at 80. How many cars per second leave at 60? Still 5. They're just closer together.

That's actually not like the situation with electrons in the conductor -- they're actually closer together at the entrance, but it shows that differences in speed between entrance and exit have nothing to do with any difference in current.

Any difference in current between the entrance and the exit would be rate at which electrons accumulate inside, but the electrostatic force is so incredibly strong that the number of electrons in the conductor is effectively constant, and always nearly equal to the number of protons.


Maybe this will help. After all the math, it comes down to an offsetting increase in energy compensating for the energy lost to resistance, so the net result is no change at the other end. I think that's what you were asking.

A cell has two terminals – a negative and a positive terminal. The negative terminal has the excess of electrons whereas the positive terminal has a deficiency of electrons. Let us take the positive terminal as A and the electrical potential at A is given by V(A). Similarly, the negative terminal is B and the electrical potential at B is given by V(B). Electric current flows from A to B, and thus V(A) > V (B).

The potential difference between A and B is given by

V = V(A) – V(B) > 0

Mathematically, electric current is defined as the rate of flow of charge through the cross-section of a conductor.

Thus, it is given by I = ∆Q/ ∆t where I is the electric current and ∆Q is the quantity of electric charge flowing through a point in time ∆t.

The potential energy of charge Q at A is Q V(A) and at B, it is Q V(B). So the change in the potential energy is given by

∆Upot = Final potential energy – Initial potential energy

= ∆Q [(V (B) – V (A)] = –∆Q V

= –I V∆t (Since I = ∆Q/ ∆t)

If we take the kinetic energy of the system into account, it would also change if the charges inside the conductor moved without collision. This is to keep the total energy of the system unchanged. Thus, by conservation of total energy, we have:

∆K = –∆Upot

Or ∆K = I V∆t > 0

Thus, in the electric field, if the charges move freely across the conductor, there would be an increase in the kinetic energy as they move.

When the charges collide, the energy gained by them is shared between the atoms. Consequently, the vibration of the atoms increases resulting in the heating up of the conductor. Thus, some amount of energy is dissipated in the form of heat in an actual conductor.


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