I know that resistors are used to limit the value of the current flowing in a circuit. But why is the value of current the same before it enters the resistor and after it leaves it? I do understand that it creates a congestion for current flow. But why do they have the same values?

enter image description here

In the above circuit, the value of current flowing Between \$B\$ and \$C\$ is 0.5 Ampere, and the value of the current flowing between \$C\$ and \$D\$ is also 0.5 Ampere. Why and how is it that they are equal? Shouldn't the current flowing between \$B\$ and \$C\$ be greater than that flowing between \$C\$ and \$D\$?

Thanks in advance.

  • 8
    \$\begingroup\$ One electron goes in, one electron comes out. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 27 '16 at 12:08
  • \$\begingroup\$ Electricity takes about 6 ns to travel a metre so, in the case of a small resistor that's about 60 ps. Given that there will always be thermal noise, and, taking it to the limit, the current on one side will hardly ever be the same as the current on the other side. \$\endgroup\$ – Andy aka Aug 27 '16 at 16:05
  • \$\begingroup\$ A lot has been added here. Just bear in mind that galvanically connected conduction band electrons have very long range effects because of the incredible magnitudes of electric field repulsion (almost unimaginably high.) Everything is local, true. But any slight shift in position immediately is passed along at near speeds of light. In equilibrium, which non-reactive circuits essentially are, loop current is everywhere the same as it must be. You can get local effects from reactive components, which temporarily store small bits of energy for a moment. But that's just a detail. \$\endgroup\$ – jonk Aug 27 '16 at 19:39

Current is the flow of charge.

  • If charge goes in one end of the resistor it has to come out the other otherwise the resistor would get charged (and it doesn't).
  • You can think of current as like a bicycle chain. It goes round in a loop. What leaves the battery on one terminal must return to the other.


simulate this circuit – Schematic created using CircuitLab

Figure 1. The water circuit analogy.

A simple water circuit analogy might help. In Figure 1b the battery is represented by the pump. The resistor is represented by a restriction in the pipework - in this case a radiator panel.

It should be clear that the water current is flowing in a loop. What goes into the radiator at the top comes out at the bottom and returns to the pump. Similarly in the electrical circuit the current that goes in at the top of the resistor comes out at the bottom and returns to the battery.

So water current is analogous to electrical current. Water pressure is analogous to electrical voltage.

Don't push the water analogy too far!

[From OP's comment:] \$ I = nAeV_d \$ [where n is the number of charge carriers free to move per cubic meter, A is the cross-sectional area, e is the electron charge and \$ V_d \$ is the drift velocity].

So Bart said it's due the drift velocity. n , A and e are constants, so \$V_d \$ is the one which is changing. So The value of the current is directly proportional to \$ V_d \$. Since they are subjected to same electric field \$V_d \$ is constant so "I" must be the same throughout the circuit. So I guess resistance is directly proportional to temperature, and drift velocity is directly proportional to temperature, \$ V_d \$ depends on resistance. Am I right or wrong?

First of all, apologies for thinking you were a beginner and pitching my answer a bit low. It was the PixelPaint edited image in your OP ...

I think you've got this mostly right. \$ V_d \$ will depend on electron mobility which will vary with the conductivity of the material. Watch out for materials with negative temperature coefficients too.

|improve this answer|||||
  • \$\begingroup\$ Unfortunately the water analogy doesn't explain why the electron flow at the end is the same as at the beginning. It is not like the ingoing electrons are pushing the others out of the resistor, which is what happens in a water flow. The incoming water molecules exert a pressure on the ones already inside. All electrons in the resistor, and the connecting wires are subject to the same electric field, which causes them to drift in the same direction on average. \$\endgroup\$ – Bart Aug 27 '16 at 13:11
  • \$\begingroup\$ Thanks, Bart. The answer is pitched to suit the level of the question. \$\endgroup\$ – Transistor Aug 27 '16 at 13:33
  • 2
    \$\begingroup\$ I like to think of conductors as tubes filled with ping pong balls. Push a ball into one end of the tube, and one will pop out the other. And this will happen at the speed of sound through the ping pong balls, even if the actual speed the ping pong balls are being pushed is very slow. Sure, electrons are bouncing all over the place, but the aggregate charge in the conduction band ultimately averages out to virtually the same effect. Heck - the electrons even move slowly, but the "compression wave" (current) propagates at the speed of light in that conductor. \$\endgroup\$ – metacollin Aug 27 '16 at 16:46
  • 1
    \$\begingroup\$ The electron does know about the resistor before it has reached it. But in the same way the ball at one end of the tube 'knows' there are other balls in front of it. If you push on the ball, for it to move, it must push on the adjacent ball, and that ball in turn must push on yet another ball further down...on and on. If the tube narrows and makes it harder to push the balls, then that resistance is felt by all the balls, regardless of where they are. But there is no sorcery happening there. You're assuming the resistor has a localized effect, which is incorrect. \$\endgroup\$ – metacollin Aug 27 '16 at 16:49
  • 2
    \$\begingroup\$ Basically, a brake pad on a bicycle wheel can slow down the entire loop of the wheel despite only creating resistance to the rotation via one small section of the wheel. Why would it be any different for electrical resistance? A braking bike wheel is no more mysterious than current in this context. :) \$\endgroup\$ – metacollin Aug 27 '16 at 17:03

Your intuition makes you think "something" must be lost, and you're right. But you must think in "Energy", some energy is lost indeed, current is constant, but voltage drops in the resistor, thus V*I gives you the ammount of energy lost per second. Beware electrons are particles, and matter cannot vanish, nor can be created, by the way it is the background for the "Kirchoff's law of current". Regards.

|improve this answer|||||

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.