Does Kirchhoff's law apply to displacement currents?
Kirchhoff's Current Law is exact if and only if you include displacement currents, (or displacement currents are not present).
Kirchhoff's Current Law follows immediately from Heaviside's formulation of Ampere's Circuital Law with Maxwell's addition of a displacement current term.
$$\nabla \times \vec{B} = \mu_0(\vec{J} + \epsilon_0\frac{\partial\vec{E}}{\partial t})$$
where \$\vec{J}\$ is the conduction current density, and \$\epsilon_0\frac{\partial\vec{E}}{\partial t}\$ is the displacement current density.
Taking the divergence of both sides (and cancelling out \$\mu_0\$ gives Kirchhoff's Current Law in microscopic form.
$$0 = \nabla \cdot (\vec{J} + \epsilon_0\frac{\partial\vec{E}}{\partial t})$$
In macroscopic form, this says that in any closed region in space, not just in a circuit, and not just with "lumped elements", the sum of all conduction currents and displacement currents into (or out of) that region is exactly 0.
If Maxwell's equations hold, then Kirchhoff's Current Law (with displacement current) holds, without exception, and exactly.
Detailed Discussion of Kirchhoff's law and displacement current point and its implications are found in:
- Eisenberg, R. S. 2019. Kirchhoff's Law can be Exact. arXiv preprint available at https://arxiv.org/abs/1905.13574.
- Eisenberg, B., N. Gold, Z. Song, and H. Huang. 2018. What Current Flows Through a Resistor? arXiv preprint arXiv:1805.04814.