Ohm's Law Intuition When we derive Ohm's Law using the Drude Model, we assume at one point of time that $E=V/L$, when is fact, $E=dV/dL$, unless $E$ is constant, in which case the assumption $E=V/L$ is true. But I don't understand why the electric field in a conductor must be constant as current flows. Is there a convincing explanation that is perhaps related to the way atoms behave and orient themselves?
Also, if the assumption $V=E\cdot L$ makes sense, I can understand why Ohm's Law should work for a homogeneous electric circuit. However, I don't understand why it should work for a heterogeneous circuit - perhaps one with two different resistors connected in series. And please don't use the traffic jam analogy. Surely there must a more theoretical way to explain this (using Classical Physics).
 A: An easy way to prove Ohm's law for electric fields that aren't constant is to first assume that the electric field is approximately constant over short lengths, just like $E=dV/dL$ suggests. Using that, you can derive Ohm's law for short lengths of material, $dV=IdR$. We'll assume that "current in = current out", which is true at steady-state. This allows us to integrate this equation (since current is a constant relative to both dV and dR), and you get regular Ohm's law $V=IR$. This is equivalent to saying that small resistances combine in series to form a net resistance for a material, for which Ohm's law also holds. This is regardless of how complex the geometry is that makes up the resistor.
A: In Ohm's Law, we take $E$ as constant for simplicity. When $E$ changes, the Ohm's Law varies to $\vec{J}=\sigma \vec{E}$ (where $\sigma = \frac{1}{\rho}$), according to Kirchhoff.
A: A well-designed resistor has a consistent, constant cross-sectional area, so any given cross-section looks like any other, and the electric field $E$ is the same in all, giving a linearly varying voltage as one travels the path.  (More theoretically, this happens to be the solution given by Laplace's equation, which turns out to apply to the potential in a material with constant electrical conductivity $\sigma$.)  If there were a pinched, narrower cross-section along the way, the electric field would have to be higher there to support the same current as in the un-pinched sections.
For two resistors $R_1$ and $R_2$ in series across a voltage source $V$, the above argument works for each resistor, but you can't conclude that the electric fields in the two parts are equal!  Instead, the current $I$ in both resistors is (must be) the same, and the voltages $V_1$ and $V_2$ across the two parts divide such that their sum is the applied voltage $V$:
$$ V_1 = I R_1 = \frac{V}{1+\frac{R_2}{R_1}}  \quad V_2 = I R_2 = \frac{V}{1+\frac{R_1}{R_2}}  $$
A: The more detailed way to solve for the current with a given applied voltage is to compute a self-consistent solution. A key point in Ohm's law is that the current is the steady state current, and so the current has to be divergence-less, so that the charge distribution is steady. Transient currents are much, much more complicated.
To compute a current in a given location you need some sort of physical mechanism (like Drude model) which depends on the voltage landscape near that position. However the self consistency requirement comes in here: the current distribution must be divergence-free, and so you need to adjust the voltage landscape until that is the case. There is a natural feedback here where charge wants to pile into areas where the voltage is too low; once the charge goes there, the voltage increases (at a rate depending on capacitance) and the charge distribution becomes more stable. Once the charge distribution is totally stable, that is the steady state.
In the special case of a rectangular material geometry with uniform end-contacts and uniform resistivity, this process does end up giving a uniform electric field.
A: 
When we derive Ohm's Law using the Drude Model, we assume at one point of time that E=V/L, when is fact, E=dV/dL, unless E is constant, in which case the assumption E=V/L is true. But I don't understand why the electric field in a conductor must be constant as current flows.

Generally, the electric field in a conductor does not have to be constant (in time or space) as current flows.  If the time variation of the electric field is large enough, it can cause rapid flux variation which in turn can cause the electric field can become a function of space.  In this case, Ohm's law in the traditional sense ($V=IR$) breaks down and transmission line theory must be applied.  Let's back up for a second and see how we can stay away from this general case, and remain in the limit where voltage is well-defined.
Starting with Faraday's law, we know that:
$\nabla \times E = -\frac{\partial B}{\partial t}$
Which can be re-written as:
$\oint E \cdot dl = -\int{\frac{\partial B}{\partial t}}\cdot dA$
So if the right side of these equations is negligible, we know that the sum of changes in voltage along any closed path (for example, in a circuit consisting of a battery with two resistors in series) must equal zero.  In addition, note that $\nabla \times E = 0$ is also required for us to be able to write $E = - \nabla V$.  If $\nabla \times E$ is nonzero, then we can't define this relationship between electric fields and voltage this way because $\nabla \times \nabla V$ must always be zero (this is just a vector calculus identity).
Now, even with this constraint, E doesn't has to be constant in space all throughout the circuit.  Fundamentally, it is constant within a medium like a resistor because that medium is uniform (I'll get to this shortly).  

Also, if the assumption V=E/L makes sense, I can understand why Ohm's Law should work for a homogeneous electric circuit. However, I don't understand why it should work for a heterogeneous circuit - perhaps one with two different resistors connected in series. And please don't use the traffic jam analogy. Surely there must a more theoretical way to explain this (using Classical Physics).

I'll give you the classical physics picture that will explain why E = constant within a uniform resistor and explain why Ohm's law holds for a series combination.  An insightful way of deriving Ohm's law is to start with the Lorentz equation and apply it to the bulk electrons within the resistive medium (not a single electron, but the bulk, assuming that they move collectively):
$F = n m \frac{dv}{dt} =  n q (E + v \times B)$
The electron is accelerated by the electric field and for our ideal circuit a magnetic field doesn't exist so we'll drop the second term (this is sometimes called the Hall term and gives rise to the Hall effect).  Note that $n$ in the above expression is the bulk electron density.  Now, we know that this bulk movement can't continue to accelerate indefinitely - electrons will give up their energy as they undergo collisions within the medium.  We can model this by adding a loss term to the above equation.  Electrons have a momentum of $n m v$ and they lose this momentum at some rate; let's call that $\nu$.  This gives:
$F = n m \frac{dv}{dt} =  n q E  - n m v \nu$
Note that $\nu$ has units of inverse seconds.  It's just a definition at this point.  We'll need to calculate $\nu$ by thinking about how long it takes the average electron to lose a momentum equal to $m v$ (by our own definition of the problem).  How this value is calculated is beyond the scope of this post and depends on the medium.  For some materials, we only need to take into account individual coloumb collisions and ask how long it takes for an average electron to lose momentum equal to $m v$.  For other materials, different effects may come into play that can cause the electron to lose energy.  In any case, however, we can lump all of that Physics into $\nu$, and the latter may be a function of temperature, mass, charge, etc.  Then we can ask the question, given the above force equation for the bulk electrons, what is the steady state velocity?  To find that we'll set $\frac{dv}{dt} = 0$ and rearrange terms:
$\frac{n q}{m \nu} E = n v$
If we multiply both sides by the charge $q$ and recognize that current density is defined as $J\equiv n q v$ we arrive at:
$\frac{n q^2}{m \nu} E = J$
That explains why, for many materials, this equation holds.  All of the physics is contained in $\nu$, which is lumped with some other constants and called the conductivity.
$\sigma \equiv \frac{n q^2}{m \nu}$
This definition for the entire extent of the resistor makes sense only if the density of charge carriers, n, are constant and have the same mass all throughout the resistor.  Then we can say that $\sigma$ is constant in space, which implies that $E$ and $J$ are constant in space.  The underlying physical picture has to do with electron energy loss; if that loss occurs uniformly then $\sigma$ is uniform in space. So now if you have two different materials in series, you can apply this equation to each material independently.  One thing you must recognize however is that the current through material 1 ($I_1 = J_1 A_1$) must equal current for material 2 ($I_2 = J_2 A_2$) because of the charge continuity equation (here $A_1$ and $A_2$ are the cross-sectional areas of the resistors).  If the $I$'s aren't equal, then one or more of the resistors will build up with charge.  As long as there's an electric field across the medium, electrons will accelerate and undergo collisions; the net result is that they will continue to flow across the medium and not physically build up. So, given this condition, we get:
$E_1 \sigma_1 A_1 = I_1 = I$
$E_2 \sigma_2 A_2 = I_2 = I$
From here on out, we're just using definitions.  If the resistor lengths are $l_1$ and $l_2$ we can multiple and divide by the length:
$E_1 l_1 \sigma_1 A_1 / l_1 = I$
$E_2 l_2 \sigma_2 A_2 / l_2 = I$
Using $V = E l$ and $R\equiv \sigma_1 A_1 / l_1$, we get:
$V_1 / R_1 = I$
$V_2 / R_2 = I$
Using what we derived earlier:
$\int{ E \cdot dl} = V - V_1 - V_2 = 0$
We arrive at:
$V = V_1 + V_2 = I (R_1 + R_2) = I R_{eff}$
Hope that helps!
