This is from chapter 13 of "Engineering Electromagnetics" by Hyat, third edition.

**THE LAWS OF CIRCUIT THEORY**To indicate how Maxwell’s equations, the potential definitions, and the concepts of resistance, capacitance, and inductance combine to produce the common expressions of circuit analysis, consider the configuration shown in Fig. 13.1 Between the two points 0 and 1 an external electric field is applied. These terminals are very close together, and a sinusoidal electric field may be assumed. Perhaps we could visualize a microscopic transistor oscillator, a rotating machine the size of a pinhead (complete with prime mover), or even a cooperative flea moving a mouthful of charge alternately toward point 1 and point 0. Whatever the nature of the source, it continues to produce an electric field between these two terminals that is independent of any currents that may consequently flow. Between points 2 and 3 there is a region of lossy material of cross-sectional area \$S_R\$, length \$d_R\$, and conductivity \$\sigma\$. At points 4 and 5 are located two capacitor plates of area \$S_c\$, with separation \$d_c\$ and dielectric of permittivity \$\epsilon\$. These several points are connected as shown by a filamentary perfect conductor of negligible cross section. Between points 6 and 8 the filament is wound into a helix of N turns having a very fine pitch.

It is probably obvious that we are now going to develop the familiar circuit equation

(1) \$V_{10} = IR + L_{ind}\frac{{dI}}{{dt}} + \frac{1}{C}\int\limits_{ - \infty }^t {Idt}\$

from Maxwell’s equations. As we do so, it is interesting to watch how each of these terms arises and what assumptions we have to make in the process. Our beginning point is the integral from of Faraday’s law,

(2) \$\oint {E \cdot dL = } - \frac{\partial }{{\partial t}}\int\limits_S {B \cdot dS}\$

We shall see that the right side of this equation provides us with only one term in (1), that involving the inductance. The other three terms all arise from the closed line integral.

Let us consider the surface integral on the right side of (2). Since the configuration of the circuit does not change with time, the partial derivative may be replaced with the ordinary derivative. Also, the filament between points 6 and 8, an N-turn helix, serves to produce a much larger magnetic field within the helix than in any other region along the filament. If we assume that a total magnetic flux \$\Phi \$ links all N turns, the surface integral becomes \$ - \frac{{d\Phi }}{{dt}}\$, or \$ - L_{ind}\left( {\frac{{dI}}{{dt}}} \right)\$ by the definition of inductance, and

\$\oint {E \cdot dL = - L_{ind}\frac{{dI}}{{dt}}} \$

where I is the filamentary current in each turn of the helix.

The closed line integral is taken along the filament, directly between the capacitor plates and points 0 and 1, as indicated by the dashed line. The contribution from the perfectly conducting filament is zero, because tangential E must be zero there; this includes the helix, surprising as that may be. We therefore have

\$\oint {E \cdot dL = \int\limits_0^1 {} } + \int\limits_2^3 {} + \int\limits_4^5 {} \$

The first integral on the right is the negative of the voltage between points 1 and 0,

\$\int\limits_0^1 {E \cdot dL = - V_{10} }\$

This integral is a function only of the external source and does not depend on the configuration shown in figure 13.1. The path is directly between the adjacent terminals, and since we are more used to considering an external source as a voltage than as an electric field intensity, we usually call \$V_{10} \$ an applied voltage.

The second integral is taken across the lossy material, and we apply Ohm’s law in point form and the definition of total resistance R,

\$\int\limits_2^3 {E \cdot dL = } \int\limits_2^3 {\frac{J}{\sigma } \cdot dL = } \int\limits_2^3 {\frac{{JdL}}{\sigma } = } \frac{{Jd_R }}{\sigma } = \frac{{Id_R }}{{\sigma S_R }} = IR\$

The same total current I is assumed, and this justified only when two conditions are met. There can be no displacement currents flowing from one point of the filament to another (such as from point 3 to 8 ), because we require the continuity of conduction plus displacement current density to be satisfied by conduction current alone. In other language, we are assuming that stray capacitances are neglected. Also the dimension of the filamentary path must be small compared to a wavelength. This will be applied in the final section of this chapter, but our study of wave motion should indicate the complete reversal which may occur in afield over a distance of one-half wavelength. Here we wish to avoid radiation, but later in the chapter it will provide our main item of interest.

The third integral is evaluated across the region between the capacitor plates where conduction current is zero but displacement current is equal to the current I, as we assumed earlier. Here we may represent the integral by

\$\int\limits_4^5 {E \cdot dL = } \int\limits_4^5 {\frac{D}{\varepsilon } \cdot dL = } \frac{{Dd_C }}{\varepsilon } = \frac{{Qd_C }}{{\varepsilon S_C }} = \frac{Q}{C}\$

or

\$\int\limits_4^5 {E \cdot dL = } \frac{1}{C}\int\limits_{ - \infty }^t {Idt} \$

where we assume that there is no charge on the capacitor at \$t = - \infty \$.

Combining these results, we have

\$ - V_{10} + IR + \frac{1}{C}\int\limits_{ - \infty }^t {Idt} = - L_{ind}\frac{{dI}}{{dt}}\$

or

\$V_{10} = IR + L_{ind}\frac{{dI}}{{dt}} + \frac{1}{C}\int\limits_{ - \infty }^t {Idt}\$

which is the familiar equation for an RLC series circuit that we hoped we would obtain.