2
$\begingroup$

Okay so consider a circuit with a cell and a resistor. Lets say that the circuit has a switch which is closed at time $t=0$. The following question are based on the state of charges in the circuit BEFORE steady state is achieved.

QUESTIONS:

  1. Current flows through the cell gains some potential energy and then looses this potential across the resistor. But isn't the drift speed of electrons really low, if so why is the potential drop established across the resistor instantaneously?

  2. An electron goes through the battery and gains some potential. What will the potential of a charge away from the negative terminal (this charge has not crossed the cell yet) and somewhere close to the resistor be?

  3. Do the charges rearrange themselves in order to generate a uniform electric field in the circuit?

$\endgroup$
  • $\begingroup$ 3.) Yes. The field across the cross-section of the wire is constant. This is achieved by a combination of rearranged surface charges and charges elsewhere, such as on capacitor plates. Here is a very rich simulation by Bruce Sherwood that demonstrates this. Warning: there's a lot to absorb in that simulation. $\endgroup$ – garyp Nov 20 at 20:07
2
$\begingroup$

1) Current flows through the cell gains some potential energy and then looses this potential across the resistor. But isn't the drift speed of electrons really low, if so why is the potential drop established across the resistor instantaneously ?

The drift speed isn't relevant here. How do you know the potential drop is established instantaneously? It isn't. Immediately after the circuit is connected, the current and the voltage drop across the resistor will be very small. Both will increase until the circuit is at steady-state.

2)An electron goes through the battery and gains some potential. What will the potential of a charge away from the negative terminal (this charge has not crossed the cell yet) and somewhere close to the resistor be?

This depends on how long it's been since closing the circuit (and where the circuit was closed) and the inductance of the current.

When the circuit is closed, you will have some charges at a high potential and some charges at a low potential adjacent to each other. But the unequal potential means there is an electric field. This field (and a path for mobile charges) accelerates some of the charges. This acceleration changes the field at other points in the circuit until everything stabilizes at steady-state.

Because you probably completed the circuit somewhere outside the battery, the battery's terminals were still maintaining a potential difference.

3)Do the charges rearrange themselves in order to generate a uniform electric field in the circuit?

Not uniform. The product of the field strength and the distance along it will equal the difference in voltage between the beginning and end points. A very low resistance wire (with a tiny voltage drop) will have a very small electric field. The resistor with a large voltage drop will have a stronger electric field.

Hyperphsics voltage and electric fields

What confuses me is how do the charges 'know' how to rearrange so as to create a voltage drop equal to the battery's voltage?

How does a skydiver "know" to fall at terminal velocity? It's because if they're falling at some other speed, there is a net force that accelerates it.

Imagine a circuit with a voltage source and a resistor, and the current around the circuit (and therefore the voltage drop across the resistor) are less than steady-state.

If you sum the voltage around the circuit, it is not zero. That means a charge can gain energy just by going around the circuit. As they do, that energy gain goes into increasing the KE and therefore their drift velocity. Increased drift velocity, increased current.

As the current increases, more charges gather at the boundary with the resistor. Eventually there are so many charges building up there that the electric field inside create a voltage drop equal to the voltage source. When that happens, there's no more energy to speed up the charges and the current stops increasing.

$\endgroup$
  • $\begingroup$ The electric field across every component of the circuit will have a potential drop which when added will be equal to the potential difference created by the battery. What confuses me is how do the charges 'know' how to rearrange so as to create a voltage drop equal to the battery's voltage? $\endgroup$ – Aditya Ahuja 2 days ago
  • $\begingroup$ Added a bit to the answer. $\endgroup$ – BowlOfRed 2 days ago
  • $\begingroup$ so the Loop rule can only be applied at steady state? If so then I'm case of charging of a capacitor why do we use the Loop rule before the steady state is achieved ? $\endgroup$ – Aditya Ahuja yesterday
0
$\begingroup$

1) The chemicals in the battery keep the electrons at a potential whether there is an external circuit or not. The electrons are waiting at a high potential and when a switch is thrown they are allowed to flow. Depending on how many electrons are needed the battery may droop its potential as electrons have to come from deeper within the battery on average for larger loads (more internal resistance). For an ideal battery we use 0 internal resistance.

In the extreme you can also consider very small capacitance and very small inductance (called parasitic) that also effect current flow in addition to the resistor, these cause an undershoot or overshoot if you were to use a fast oscilloscope.

When the current is 1 amp there are many many more electrons in the wire and resistor involved in the flow, thus drift or average velocity is slow compared to the rate of current.

The potential remains in the battery and as well on the resistor, even after the electrons are flowing, the field (or potential) is a force that is present instantly.

$\endgroup$
  • $\begingroup$ So initially the components have a non-uniform potential drop across them and ohm's law cannot be applied right ? $\endgroup$ – Aditya Ahuja Dec 6 at 14:05

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.