How can KVL (Kirchhoff's Voltage Law) & KCL (Kirchhoff's Current law) be derived from Maxwell equations in lumped circuits? (Lumped network: if $d$ is the largest dimension of the network and $\lambda$ is the wavelength of the signal, a lumped network satisfies the condition $d \ll \lambda$ where $\lambda=\frac{c}{f}$)


1) KVL & KCL works for lumped circuits (not for every circuits). Trying to derive KVL and KCL from Maxwell equations without using lumped assumption (Lumped assumption: assuming that the circuit is lumped) is wrong. Actually 4 Maxwell equations + lumped assumption $\implies$ 2 Circuits laws (KVL & KCL)

2) The other thing we can assume is that there are no incoming energy from outside, such as external magnetic field.

  • 3
    $\begingroup$ Sorry would you elaborate more on what KVL and KCL stand for? $\endgroup$
    – Ali
    Commented Mar 7, 2014 at 16:40
  • $\begingroup$ @Ali Kirchhoff laws for Current and Voltage $\endgroup$
    – Nephente
    Commented Mar 7, 2014 at 16:40
  • $\begingroup$ I see. For the current it's just conservation of electric charge. For the voltage: in the absence of time varying magnetic field, $\nabla \times \vec E = 0 \Rightarrow \vec E = -\nabla V \Rightarrow \oint \vec E . \vec {dl} =0$ $\endgroup$
    – Ali
    Commented Mar 7, 2014 at 16:48
  • $\begingroup$ @Ali $\nabla \times E = -dB/dt$ $\endgroup$
    – rza
    Commented Mar 7, 2014 at 22:49
  • 1
    $\begingroup$ @rza : That's why she specified "in the absence of a time varying magnetic field". $\endgroup$
    – Colin K
    Commented Mar 9, 2014 at 0:28

1 Answer 1


$\def\vE{{\vec{E}}}$ $\def\vD{{\vec{D}}}$ $\def\vB{{\vec{B}}}$ $\def\vJ{{\vec{J}}}$ $\def\vr{{\vec{r}}}$ $\def\vA{{\vec{A}}}$ $\def\vH{{\vec{H}}}$ $\def\ddt{\frac{d}{dt}}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$ $\def\rmC{{\mathrm{C}}}$ $\def\rmM{{\mathrm{M}}}$ $\def\ph{{\varphi}}$ $\def\eps{{\varepsilon}}$

Faraday's law in integral form: $$ \oint_{\partial A} \vE \cdot d\vr + \ddt\int_A \vB\cdot d\vA = 0 $$ Thereby, $A$ is some surface and $\partial A$ its boundary. The boundary can be split into partial paths $\bigcup_k C_k = \partial A$ and the integral becomes $$ \sum_{k} \int_{C_k}\vE\cdot d \vr + \ddt\int_A \vB\cdot d\vA = 0 $$ You can define the voltage drops $V_k := \int_{C_k}\vE\cdot d\vr$ and the induced voltage $V_i$. This way you get Kirchhoff's Voltage law $$ \sum_{k} V_k + V_i = 0. $$ Ampere's law reads $$ \oint_{\partial A} \vH\cdot d\vr = \int_A \vJ\cdot d \vA + \int_A \dot\vD\cdot d\vA. $$ If you choose a closed surface $A$ then the boundary $\partial A$ is empty, the left-hand integral is zero and the equation for a closed surface becomes $$ 0 = \oint_A \vJ\cdot d \vA + \oint_A \dot\vD\cdot d\vA $$ We dissect the surface into partial surfaces $A_k$ of conductor cross-sections and a partial boundary of an insulator $A_i$. The corresponding integrals $I_k:=\int_{A_k}\vJ \cdot d\vA$ are the currents through these conductors. Furthermore, we have the charge transfer currents $I_{Ck}:=\int_{A_k}\dot\vD \cdot d\vA$ for the conductors and for the insulator $I_{i} := \int_{A_i}\dot \vD \cdot d\vA$. This gives you Kirchhoff's current law $$ \sum_{k} I_k + \sum_{k} I_{Ck} + I_{i} = 0. $$

If one includes induction voltages and transfer charge currents in KVL and KCL these laws directly represent the laws of Faraday and Ampere. These quantities can be modeled through parasitic inductance and capacitance in the network diagram.

The fields $\vE$, $\vJ$, $\vD$, $\vH$, $\vB$ of the exact solution of Maxwell's equations satisfy Faraday's and Ampere's law for every piecewise smooth surface $A$.

Network modeling can be interpreted as discretization of Maxwell's equations. Through the finite selection of path integrals and surface integrals the number of degrees of freedom and the number of equations are reduced from infinite for the vector fields to the finite number of voltage and current variables and the finite number of corresponding loop and cut equations.

A discretization can give the exact results if the V-I-relations of the finite elements give exact field solutions (as boundary value problems). This is (almost) possible with DC-circuits.

For higher frequency applications one needs to refine the discretization. A good guidance for the choice of the discretization width is the wavelength of the considered electromagnetic waves in the circuit. In practice that means that one needs to include more parasitic elements in the network model for higher frequencies.

Slowly I hopefully understand what you mean by the `integrated principle'. I think it is helpful to see how the behaviorial relation of a network element is derived from Maxwell's equations and embedded into the network theory.

Note, in this answer, I have just derived the V-I-relation of a resistor from Maxwell's equations. The V-I-relation for a capacitor can be derived in a similar way. For the inductor you need the induced voltage as defined above.

In a wide range of applications the fields can be approximated as quasi-stationary. With this approximation it turns out that for many basic structures the path integral over the field strength, i.e., the voltage drop is directly related to the cross-section integral over the current density, i.e., the current. This fact and the above equations for certain path-integrals over the field strength (KVL) and certain surface integrals over the current density (KCL) are exploited in network theory.

Note, that the splitting of voltages into the path integrals $V_k:=\int_{C_k} \vE\cdot d \vr$ and the induction voltage is only one way of interpretation which has its draw-backs.

For an instance with a time-variable magnetic field the spatial separation suggested by the formula $\sum_{k} V_k + V_i = 0$ does not really exist.

There is another approach where the field strength $\vE$ is split into a Coulomb-part and a magnetic part. Pityingly, I do not know the reference anymore and I learned this stuff more than 10 years ago. But, if you are really interested I can try to retrieve it (this is not so easy and will take its time).

The idea is to apply the well-known technique of magnetic vector potentials $\vB=\rot \vA$ which solve the divergence equation $\div\vB=0$. As a gauge condition $\div\vA=0$ is used in this setup.

Faraday's law in differential form then reads $$ \rot(\vE+\dot\vA) = 0 $$ which ensures in a simply connected domain the existence of a potential $\ph$ for the vector field $\vE+\dot\vA$, i.e., $$ \begin{array}{rcl} \vE + \dot\vA &=& -\grad\ph\\ \vE &=& \underbrace{-\grad\ph}_{\vE_\rmC} \underbrace{-\dot\vA}_{\vE_\rmM} \end{array} $$ In this formula $\vE_\rmC$ is the Coulomb-part of the electrical field strength and $\vE_\rmM$ is the magnetical part of the field strength.

Kirchhoffs Voltage law then holds unrestricted for Voltage drops defined with the Coulomb-part of the field strength $V_k := \int_{C_k} \vE_\rmC \cdot d\vr$.

For an example, in the case of constant permitivity $\eps$ the Coulomb part is the cause of space charges: $$\begin{array}{rl} \rho &= \div \vD\\ & = \div \eps (\vE_\rmC - \dot\vA)\\ & = \eps \div \vE_\rmC - \eps\frac{\partial}{\partial t}\underbrace{\div\vA}_{=0}\\ &= \eps \div \vE_\rmC \end{array}$$ One can then derive the V-I-relations for the network elements based on the Coulomb part and the magnetic part of the voltages and the currents.

This approach respects that the Coulomb-part and the magnetical part coexist everywhere in the circuit.

But, I have not seen this approach in practical usage yet.

  • $\begingroup$ @Tobias May I ask the justification for neglecting the integral on the LHS of the Ampere's law in the fourth equation? Why does the surface being closed imply that the line integral of H-field along the boundary of this surface to be zero? $\endgroup$
    – Vesnog
    Commented May 6, 2015 at 20:16
  • $\begingroup$ @Vesnog The boundary of a closed surface is empty. $\endgroup$
    – Tobias
    Commented May 6, 2015 at 21:29
  • $\begingroup$ I would agree with Tobias's answer,but there is a fundamental flaw when it comes to the discussion of derivation of KVL.The nature of the induced electric field from a time varying magnetic field is irrotational.Hence it cannot be expressed as gradient of some potential.So I have serious doubts on the terms of induction voltages. $\endgroup$ Commented Oct 18, 2016 at 13:39
  • $\begingroup$ Actually the derivation for quasi-static fields is much more complicated than this. For a complete derivation see Robert M. Fano, Lan Jen Chu and Richard B. Adler, Electromagnetic Fields, Energy, and Forces and this paper. $\endgroup$ Commented Oct 18, 2016 at 15:42
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    $\begingroup$ @SambeetPanigrahi I assume you meant that the electric field induced by a time-varying magnetic field is not irrotational. True. That is the reason for the decomposition of the electric field into a Coloumb component $\vec{E}_{\rm C}$ and a magnetic component $\vec{E}_{\rm M}$. Please, check their definition in the text above. Note also that the induction voltage $V_i$ is defined for the pre-defined path $\partial A$ via Stokes theorem. It is not the gradient of some potential in the text above. $\endgroup$
    – Tobias
    Commented Oct 18, 2016 at 16:38

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