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My electromagnetics textbook has a picture of an $RC$ circuit with a current source.

screenshot of textbook

As you can see, there are three current density vectors, $J_i$, $J_c$, and $J_d$. The book labels them as follows:

$J_i =$ impressed (source) electric current density

$J_c =$ conduction electric current density

$J_d =$ displacement electric current density = $\frac{\partial D}{ \partial t}$, where $D$ is electric flux density

It then uses this figure to explain the following Maxwell equation:

$\text{curl }H = J_i + J_c + J_d$, where $H$ is magnetic field intensity

What I don't understand is... Isn't $J_i$ the same as $J_c$? I'm getting this from the idea that if you do a Kirchhoff Current Law evaluation at the top left node (between the source and resistor), then $J_i$ is going in and $J_c$ is going out, thereby making them equal? ... So wouldn't this mean that $J_i$ and $J_c$ are the same current? And if they're the same current, then doesn't that mean that the Maxwell equation is counting it twice?

I'm looking for a solid explanation of why $J_i$ is different than $J_c$, despite Kirchhoff's Current Law.

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    $\begingroup$ $J_i$ might have a different value then $J_c$ if the source and the conductor(resistor) have different cross sectional areas. I suspect the point is to try to make displacement current less mysterious by pointing out that there are other different kinds of current that you might not normally think of. You're not likely to ever hear about "impressed current" again. $\endgroup$
    – The Photon
    Commented Feb 8, 2017 at 5:22
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    $\begingroup$ Even if the cross sectional areas are different, KCL should still make J_i = J_c, right? $\endgroup$
    – Mark Betters
    Commented Feb 8, 2017 at 19:39
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    $\begingroup$ No, KCL makes $I_i=I_c$. Think about a thick pipe running into a skinny pipe. The total amount of water flowing out of the thick pipe has to equal the total amount of water flowing in to the skinny pipe. For that to happen, the current per mm^2 of cross section has to be more in the skinny pipe. $\endgroup$
    – The Photon
    Commented Feb 8, 2017 at 19:59

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It seems that this picture is from Balanis Advance Engr E&M. Here is a somewhat intuitive explanation taken from Harrington's book:

"In source-free media, We identify $J = J_c$ as the conduction current 'caused by' the field"

An antenna would be a good example for conduction current density, which has no source current density.

In your picture, the three currents are all equal. They have different names because they are different types of current density, but they would have been the same thing if this picture appeared in a circuits textbook.

As for $J_{ic}$. Physicists defined too many names and they got sick of tracking them. So they add the two so that they have one convenient quantity to work with.

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In the circuit shown in the figure, the elements are connected in series so that the current should be equal according to Kirchhoff's current law i.e. $J_i = J_c = J_d$. However they have different definitions as: $J_i$ is the impressed current or the external current source. If this current ($J_i$) passed through an element having free moving charges, it will be defined as conduction current ($J_c$). If this current passed through a dielectric element such as the dielectric material between the capacitor plate shown in the figure, the current will be defined as a displacment current ($J_d$). The latter ($J_d$) is due to the displacement of the bounded charges around the atom. The difference between $J_c$ and $J_d$ is that the first arises due to the free moving charges while the second arises due to the bounded charges.

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$$\nabla \times H=J_i+J_c+J_d $$

This means that area integral of magnetic field's curl equals line integral of current density. Impressed current is conduction current or displacement current, but it's more convenient to use impressed current to calculate.

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Nov 6, 2021 at 11:13
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I think the explanation in the book can be misleading. It is not possible to sum current densities that correspond to different points: the current density at a point on the wire cannot be summed with the current density at a point on the capacitor.

What he means is that there are several types of current density to take into account in the Maxwell Ampere equation. For example, if the dielectric of the capacitor is also a conductor, it will be necessary, at each point, to sum the conduction density and the polarization current density.

If we write Ampere's theorem in integral form, we have to look for the current that crosses the surface that is supported by Ampere's contour and in thius case, it will be the current in the wire if the surface cuts the wire, or the current in the capacitor if the surface cuts the capacitor.

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Kirchhoff's first circuital law is about currents $I$, not about current densities. Thus we have $I_i=I_c$, but it is not necessary to have $j_i = j_c$, because cross-section of the source and of the resistor need not be the same.

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