Why is current in a circuit constant if there is a constant electric field? I'm a beginner in E&M, and have just started to learn about current and Ohm's law.
According to this page there is a constant electric field everywhere in a DC circuit pointing against the direction that electrons flow. This is consistent with my textbook and general intuition thus far - a potential difference created by a battery causes charge carriers to flow because of an electric force, which can be represented by having an electric field everywhere in the circuit.
My question is then this - if the charge carriers in a circuit are motivated by an electric force, then how can current be a constant value? I am assuming current to be a measurement of how much charge crosses a cross-sectional area perpendicular to the direction of flow, per time; furthermore, shouldn't the velocity components of the charge carriers in the direction of flow, which is proportional to current, increase because of the electric force that is motivating them in that direction?
In thinking about this question, I was led to a possible explanation related to Ohm's Law - my textbook compares resistance to a frictional force, caused by the accumulation of collisions undergone by a charge carrier (which result in changes in velocity against flow direction). So to answer my initial question, I was thinking that perhaps a reason why current was constant was because in each circuit, resistance provided an effective equal and opposite force to the battery, resulting in a net force of 0 on charge carriers, thus resulting in a constant velocity component in the direction of flow. This explanation also seemed to line up with the concept of 'voltage drops' - take a circuit with a battery and a resistor - the charge carriers before the resistor are motivated by the electric field, then in moving through a bumpy resistor they do work per charge equivalent to the voltage, and thus exit the resistor with their potential gone; the key here being that the resistor provides the exact amount of work needed to counteract the voltage, going with my idea that the resistance in a circuit provides an effective opposite to the electric field at each point (I'm assuming that the resistor affects the entire circuit acting like a plug in a pipe; and, simultaneously, does exist as a physical point in the circuit across which charges do work - potential does change within the resistor, but also the resistor can only let through a constant amount of charge per time, which somehow affects the entire circuit just as the electric field does (?)). Continuing, I would then assume that with a given voltage and given resistance, current changes so that the two exert equal effective forces on charge carriers - this is implying that a resistor would only be able to apply a specific amount of total force (work) at a time, and thus the charge per time must be changed to ensure that the work per charge is equivalent to voltage (this last bit seems the most questionable to me).
Is the idea that resistance provides an equal and opposite force valid reasoning to explain constant current? If not, why is constant current? If yes, is my reasoning above coherent, or what am I missing/misunderstanding?
*Please note: when I refer to 'the direction of flow' above, I am referring to the direction that the charge carriers would flow.
 A: It is a analogic useful way to understand the resistive circuit, in the meaning that the electric expression $V = RI$ has the same form as the mechanical $F = kv$, in an environment where the drag force is proportional to the velocity.
In the case of a conductor, it is important to note that even whithout any applied field, the free electrons have momentum to all directions, and of different magnitudes. But there is no net flow without an E-field. The effect of the E-Field is to increase the fraction of electrons to one direction, and decrease the fraction to the opposite. The scattering with the lattice limits that net flow, and is responsible for the Joule effect of the electrical resistance.
A: Exactly as @Claudio Saspinski says. I was about to write an answer in those same terms when I saw his excellent answer.
Think of the metal as an extraordinarily dense "fluid" in which particles moving (electrons) almost immediately acquire their limit velocity. The faster they want to go, the more resistance they find, in proportion to their own velocity.
$\dot{x}=\textrm{const.}$ is a solution to the equation,
$$
m\ddot{x}=mg-k\dot{x}
$$
A: 
a potential difference created by a battery causes charge carriers to
flow because of an electric force, which can be represented by having
an electric field everywhere in the circuit.

A potential difference created by a battery only causes charge carriers to flow if there is a load connected to the battery terminals. If there is no load connected, there is a potential difference between the battery terminals but no flow of charge carriers.

My question is then this - if the charge carriers in a circuit are
motivated by an electric force, then how can current be a constant
value? I am assuming current to be a measurement of how much charge
crosses a cross-sectional area perpendicular to the direction of flow,
per time; furthermore, shouldn't the velocity components of the charge
carriers in the direction of flow, which is proportional to current,
increase because of the electric force that is motivating them in that
direction?

The force of the electric field causes the charge carriers kinetic energy to flow, but collisions between the charge carriers and the atoms/molecules of electrical resistance in the circuit takes that kinetic energy away and dissipates it as heat. It is a continual process of the electric field giving the charges kinetic energy and the collisions taking the kinetic energy away that keeps the velocity of the charge carriers (i.e., the drift velocity) and thus the current constant. The end result is the electrical potential energy given the charge carriers by the battery is converted is either converted to heat in resistance, or stored as potential energy in the electric fields of capacitance or kinetic energy in the magnetic fields of inductance.

In thinking about this question, I was led to a possible explanation
related to Ohm's Law - my textbook compares resistance to a frictional
force, caused by the accumulation of collisions undergone by a charge
carrier (which result in changes in velocity against flow direction).

And that is indeed a good mechanical analogy. Let's say you push a box at constant velocity on a floor with friction a distance d. The force you apply is analogous to the force of the electric field. The box is analogous to the charge. And the floor friction is analogous to the electrical resistance. The box moving at constant velocity is analogous to constant current. The work you do (Fxd) divided by the mass m is analogous is analogous to the voltage drop (work per unit charge).

Continuing, I would then assume that with a given voltage and given
resistance, current changes so that the two exert equal effective
forces on charge carriers - this is implying that a resistor would
only be able to apply a specific amount of total force (work) at a
time, and thus the charge per time must be changed to ensure that the
work per charge is equivalent to voltage (this last bit seems the most
questionable to me).

Having a little trouble following you. But you should understand that the potential difference, or voltage, between the terminals of a resistor is equal to the work required, per unit charge, or Joules/Coulomb, to move the charge between the terminals of the resistor. Then, since the current in the resistor is the charge per unit time, coulombs per second, moving across any point in the resistor, then the power dissipated in the resistor as heat is the voltage times the current, or Joules/Coulomb x Coulombs/Sec = Joules/sec = watts.

Also, when you said "the potential difference, or voltage, between the
terminals of a resistor is equal to the work required, per unit
charge...to move the charge between the terminals of the resistor" -
to clarify - if this work per charge is a combination of both
resistance and current, and current is charge per time, does this mean
resistance is work per time?

First, what I said is the electrical engineering definition of potential difference, or voltage. In your example it represents the work required by the battery, per unit charge, to move the charge through the resistance.
Second, the definition of voltage is independent of current. There is a voltage across the terminals of a battery when it is not connected to anything, i.e., when there is no current. But the relationship between the current in and the voltage across a resistor of constant resistance is determined by Ohm’s law.
Third, resistance is not work per unit time. Work per unit time is the power dissipated in the resistor and equals $VI$ {Joules/Coul x Coul/sec = watts).
Hope this helps.
A: I think all the answer are missing the point needed to be made . A battery provides an electric field but the arrangement of the electrons in the 'skin' of the conductor creates an opposite electric field which cancels out the battery's electric field which means that the net force is 0 and electrons don't accelerate to the cathode of the battery.However there is an electric potential still from the battery.You may ask how is this possible since didnt I just say that the total electric field is 0?Well yes but the electric field is the first derivative of the voltage so a null electric field means a costant voltage!.
A: In the classical theory, the free electrons are bouncing around at random in thermal equilibrium with the atoms of a conductor. An electric field accelerates them so that the path after each bounce becomes a curve. The average displacement (resulting from this curve) after each bounce determines the “drift” velocity of the free charges.  Continuity of flow requires that the current be the same throughout the length of any single conductor.  If the conductor has a uniform cross-secton and free charge density, then this requires a uniform electric field.  However, maintaining a uniform field requires a gradient in the charge density.  If a uniform wire is connected to the terminals of a battery, there will be an excess of free electrons near the negative terminal and a lower density of free electrons near the positive terminal.  A lower density of free electrons requires a stronger electric field.
