How does the current remain the same in a circuit? [duplicate]

Question

I understand when we say current, we mean charge (protons/electrons) passing past a point per second. And the charges have energy due to the e.m.f. of the power supply.

Now tell me, if a lamp has resistance and you hook it in the circuit, how will the current stay the same? The charges obviously lose energy in the lamp and so become SLOWER, which should mean current decreases, right?

[Edit] All answers explained a bit of everything, so it was hard to choose one. If YOU are looking for an answer, please check the others too, in case the accepted one doesn't answer your question.

Answer 1

The charges obviously lose energy in the lamp and so become SLOWER

The charges lose potential energy, not kinetic energy. Since they're not slowing down, it's not a problem.

Imagine that I let my car drift down a mountain. The engine isn't running, but I'm in control with the brake pedal. And I'm going to make sure the car always moves at exactly 15 mph.

The car starts with some energy at the top of the mountain. As it moves, it pushes energy into the brakes. Why doesn't the car slow down? Because it's gaining exactly the same amount of energy by dropping down the mountain. The speed and kinetic energy don't change because the energy sent into the brakes is identical to the change in potential energy by moving lower down the mountain. Or you could say the force from the brakes is balanced by the force from gravity.

If the slope is steep, I press harder on the brakes and they get hotter. If the slope is gentle, I release the brakes and they cool off a bit.

In the circuit, when the charges move through the resistor, the energy lost to the resistor is balanced by the energy from the electric field. The resistor is trying to slow down the charge, but the field in the resistor is pushing them along so they keep their speed.

At the "end" of the circuit, the charges have minimal potential energy (like the car at the bottom of the mountain). The voltage source does work on the charges and moves them to the part of the circuit with the greatest potential energy.

Answer 2

The circuit in its whole will settle into an equilibrium state. If you were to shoot a very short pulse of electrons through your lamp, then indeed, the current would decrease right after the lamp.

But if you have a constant flow, then, as another user illustrated, you'd have electrons bunch up right before the lamp, which limits the output, but ALSO the input! The "traffic jam" works its way backwards, all the way to the source, and eventually slows down all electrons from source to lamp.

Answer 3

The question

The charges obviously lose energy in the lamp and so become SLOWER, which should mean current decreases, right?

shows that you have a misconception about the motion of the conduction electrons.

If you were correct then to maintain the same current around a circuit by a miracle more conduction electrons would need to contribute to the conduction process as you went around the circuit.
If this did not happen then the conduction electrons would move slower and slower and . . . . . eventually stop?

In fact what happens is that the electrons gain kinetic energy (and lose electric potential energy) between collision with the lattice (bound) ions from the electric field in the wire and then lose that extra kinetic energy to the lattice ions upon collision with them.
The net effect is that the temperature of the material increases as the internal kinetic energy of the material has increased (the lattice ions vibrate more) and the conduction electrons move along the conductor with a constant average speed.

Answer 4

Ok , so I think that confusion lies in the concept of current. When we apply potential difference through a circuit having resistance, a single electron does not move from one end to the other. What actually happens is that it replaces electron next to it which in turn replaces the electron next to it and this happens till the last electron reaches the other terminal. The resistance from beginning itself shows its effect on the movement of charge and not when the charge passes through it. What u say is that a charge when passing through a resistance should lose energy and hence current shoould slow down is partially correct. Yes, resistance does slow down the movement of electron but that effect is observed in overall circuit and not just in the path where resistance is connected.

Answer 5

Here is why the current stays the same going through the bulb:

Think of current as water flowing downstream in a river. The water comes to a dam, and flows over the top, and then falls all the way down to the bottom of the dam, then resumes flowing downstream. Every gallon of water that was flowing in the river will go over the top of the dam. Every gallon of water that flows over the top will fall to the bottom. Every gallon that falls to the bottom resumes flowing downstream. Therefore, every single gallon of water that was originally flowing downstream before coming to the dam will wind up flowing downstream again after leaving the dam behind.

We say that water is conserved. One gallon in, one gallon out. Current through a circuit behaves the same way: One ampere in, one ampere out.

Answer 6

Your question is very nice, I have concluded that your question is how does the current remains the same when it supplies the power to the lamp for glowing it up.

There is an experimental law called Ohm's Law, it's mathematical statement is $$ V = I R$$

Now, let's consider this circuit

If we consider that electrons flow from negative terminal to positive terminal and ignore the Quantum Mechanical aspects then for wire and bulb in this circuit we can write Ohm's Law like this $$ V_{wire} = I_{wire} R_{wire}$$ $$V_{bulb}= I_{bulb} R_{bulb}$$ .

we usually use the conducting wires whose resistance is small so from Ohm's law we can at once see that at any two points the voltage is smaller, but the resistance of the bulb is obviously much higher than the wire, therefore, the voltage between the two ends of the bulb is quite high. You should see that I have not used the fact here that current is constant, I have just used the experimental fact that voltage is directly proportional to resistance.

The relation of drift velocity with current is $$ I \propto v_d $$ and I want you to understand that when the voltage will be high the electrons will have more energy and therefore will have high drift velocity, higher the drift velocity higher will be the current. Therefore, the current in the bulb will be higher than the current in the wire.

So, we have found that the current will be higher in the bulb and the electrons with high drift velocity collides with the atoms of the bulb, passing some of their energy to them and this energy is released as photons, now our actually traveling electrons have less energy (as they lost some energy during their collision with atoms inside the bulb material) so we can say that their drift velocity has reduced so the current and hence qualitatively we can say the current has now reached the same magnitude as it wass in the wire.

If anything is unclear you're welcome to ask it in comments.

Answer 7

@Farcher answer, particularly the last paragraph, sums it up perfectly. The positive work done by the electric field on the charge giving the charge kinetic energy equals the negative work done by the lattice structure that takes away the kinetic energy of the charge increasing the internal energy of the structure. Ultimately, the energy is dissipated as light and heat to the surroundings (a.k.a resistance heating).

A mechanical analog is pushing an object at constant velocity on a surface with friction. The positive work done in pushing the box between two points exactly equals the negative friction work for a net work of zero and no change in the kinetic energy (velocity) of the box. The result is an increase in the temperature at the interface and eventual heat transfer to the surroundings.

Although not exact, you can think of the external force as analogous to the electric field force, the box analogous to the charge, the velocity analogous to current, and the surface with friction analogous to electrical resistance.

Hope this helps.

Answer 8

When you connect the terminals of a battery to the terminals of a light bulb, physically it means you are establishing an electrical field inside the conductor, in direction from the positive side to the negative one. How the conductor has free charges, the electrical field will force them to move from the terminal of higher charge concentration towards the other one of lower concentration, in order to achieve equilibrium (lowest energy).

Metals are elements that their valence electrons are not very well connected to their atomic nuclei and, besides that, form substances that the valence electrons can be shared among several atoms. Because this we can use metals as conductors, and the shared electrons are the necessary free charges. So, electrons have mass and charge, and then the force created by the electrical field will act over each electron in a way that obeys the Newton’s second law: $$F = m_e.a$$ But the poor electron collides every time because it is inside a substance (metal). In each collision it loses kinetic energy to the metal structure. As we know, kinetic energy loss means decreasing electron speed, and the energy is transformed to structure vibration (heat), and these are the principia of resistance and Joule effect. The light bulb, for example, uses these effects to heat extremely its filament in order to produce light via black body radiation.

The average speed of an electron inside the conductor is kept about the order of millimeters per second, what travels quickly is the information of the electrical field, about the light speed, and what increases when current value rises is the number of electrons passing by a point in a linear metal wire per unit of time, and this ratio remains constant everywhere in the circuit in a macroscopic measuring.

Answer 9

There's a lot of overly complicated answers here. I will keep it as simple as possible, and try to address your points. Current is the movement of Electrons. Protons don't move. A power supply does not "produce current" as such. Current is caused by Voltage (which is a difference in electrical potential.) The best analogy is water pressure. Voltage is the water pressure, the resistor is your tap or faucet, current is the flow of water. If your tap is partially on and left constant, then the flow of water is determined by the "Voltage". The higher the water pressure, the higher the flow. Alternatively, with the pressure constant, you can vary the "resistance" with the tap. Open the tap more = lower resistance = higher water flow. So the formula is Current = Voltage/Resistance.

There can be some confusion when we talk about a circuit being a "10Amp Circuit" but that is the maximum current, not a constant.

Lets consider a simple example using a Car Battery, which is typically referred to as 12 Volts. (Its actually closer to 13, but we like even numbers) It is always at 12V, even with no current flowing. Once a circuit is completed, current begins to flow. For the purposes of discussion we can ignore the resistance of the wires and focus on the light bulbs. Something like a brake-light would have resistance of about 12ohms so the current for one light would be around 1amps. Whereas a headlight has lower resistance, around 3ohms, so the current would be around 4amps. That's the tricky part to understand. A brighter, more powerful light, actually has LOWER resistance. Again, I'll go back to the water example, only lets make the water hot, and say you're trying to heat something up by running hot water over it. The faster the water, the more heat is put into it.

Hope that helps