Reduction of Maxwell's equations to classical circuit theory Can classical circuit theory based on lumped element models be obtained from Maxwell's equations as a limiting case in an appropriate sense?
If this is the case, what exactly are all the assumptions required to reduce Maxwell's equations to circuit theory? What are some good references with a detailed discussion of this relationship?
 A: Classical circuit theory boils down to Kirchhoff’s laws:
KVL: Kirchhoff's voltage law
A.k.a., conservation of energy.

The algebraic sum of all the potential differences around the loop
  must be equal to zero: $\sum_i V_i = 0.$

This comes from Maxwell's third equation:
$$ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \quad \Longleftrightarrow \quad \oint \mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell} = -\frac{\mathrm{d} \Phi_{\mathbf{B}}}{\mathrm{d}t}, $$
where $\Phi_{\mathbf{B}} = \int\int \mathbf{B}\cdot \mathrm{d}\mathbf{S}$ is the magnetic flux.
Because $\mathbf{E}=-\nabla V$, $V$ being the potential:
$$\oint \mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell} = \sum_{\mathrm{whole\,\,circuit}} V.$$ 
In electrostatics, $\mathrm{d}/\mathrm{d}t = 0$, so 
$$ \oint \mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell} = \sum_{\mathrm{whole\,\,circuit}} V  =0,$$
which is KVL.
Induction
In electrodynamics, $\mathrm{d}/\mathrm{d}t \neq 0$ so:
$$ \oint \mathbf{E}\cdot\mathrm{d}\boldsymbol{\ell} = \sum_{\mathrm{whole\,\,circuit}} V  = -\frac{\mathrm{d} \Phi_{\mathbf{B}}}{\mathrm{d}t},$$
where you would call the net effective voltage $V = \epsilon$ the electromotive force:
$$ \epsilon = -\frac{\mathrm{d} \Phi_{\mathbf{B}}}{\mathrm{d}t}. $$
The minus sign is referred to as Lenz's law.
KCL: Kirchhoff's current law
A.k.a., conservation of charge.

The algebraic sum of ALL the currents entering and leaving a junction
  must be equal to zero as: $\sum I_{\mathrm{in}} = \sum I_{\mathrm{out}}.$

This is essentially equivalent to the continuity equation in electromagnetism. 
From Maxwell's fourth equation:
$$ \nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \frac{\partial \mathbf{E}}{\partial t},$$ and taking its divergence:
$$ \nabla \cdot (\nabla \times \mathbf{B}) = \mu_0 \nabla \cdot\mathbf{J} + \mu_0\epsilon_0\frac{\partial (\nabla \cdot \mathbf{E})}{\partial t},$$
where $\nabla \cdot (\nabla \times \mathbf{B})  = 0$ and $\nabla \cdot \mathbf{E} = \rho/\epsilon_0$ because of Maxwell's first equation.
So:
$$ \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0,$$
which is the continuity equation between the charge density $\rho$ and current $\mathbf{J}$.
In the integral form, this becomes:
$$ \frac{\mathrm{d}Q}{\mathrm{d}t} =  - \int \int \mathbf{J} \cdot \mathrm{d}\mathbf{S},$$ where $Q$ is the total charge in the volume bound by the surface $\mathbf{S}$. This means that a change in charge $\mathrm{d}Q$ requires an inflow of charge $(\mathbf{J}\cdot \mathrm{d}\mathbf{S})\cdot \mathrm{d}t$.
