I think this must be the most naive question that I am going to ask. I now doubt on my knowledge of the electrical engineering that I have learned throughout my acadmics.

We have a situation, wherein the switchboard is subjected to heat run test (to determine temperature rise). The busbars are shorted on the one end and fed with constant current source on the other. The constant current is an FLU (feeder loading unit). As per a paper (unpublished) in our company, the above setup was simulated using an FEA program that when the busbars are arranged horizontally as shown in the picture, the busbars in the middle (to be precise Y2) tend to carry less current. This is because (as it is stated) of skin effects and proximity effects. Due to these losses, the conductor tend to get more hot when subjected to full loading (100% RDF).

Let us consider that, the busbars are carrying 3000 A AC current. In DC system, the busbars carry similar current, because skin effect and proximity effect is absent. So Y1 and Y2 both will carry 1500 A each.
In AC, due to skin and proximity effect the Y2 shall carry less current than Y1. So it has been noted in paper that Y1 was found to be carrying 1800 A, while Y2 was carrying 1200 A.

Now, as we all have learned, that magnitude of current flowing through the conductor will not change because of the skin or proximity. But their distribution pattern inside the conductor will be affected. So, if this is the case, then why was the current was found to be lesser in Y2. I can easily established through Jmag (current density) plot, that the current is more concentrated at periphery than at the core. But, should the current magnitude change?

Also one should experience voltage drop and not the reduction in current due to increase in losses. I would like to reiterate that this is a constant current source.

Please confirm if my understanding is correct and the thing noted in paper is incorrect or please tell me where I am making a mistake.

This is the side view of the bus chamber of a switchboard

( Image from here )

enter image description here

  • \$\begingroup\$ Isn't this because the middle one is warmer? \$\endgroup\$
    – Jeroen3
    Commented Oct 28, 2023 at 21:02
  • \$\begingroup\$ What are the lines? I see no color or position legend. What is black, is that neutral? Ground? Is it carrying current during the test? What frequency, and what dimensions are the bars? \$\endgroup\$ Commented Oct 28, 2023 at 21:22
  • 2
    \$\begingroup\$ Could you label which ones are which (Y? 1, 2?)? \$\endgroup\$ Commented Oct 29, 2023 at 7:29
  • 1
    \$\begingroup\$ @TimWilliams, it's R-Y-B as in red-yellow-blue. There's no mention of this in the question though. I assumed X-Y-Z! \$\endgroup\$
    – Transistor
    Commented Jan 14 at 19:02
  • 1
    \$\begingroup\$ TL;DR: You've got yourself an air-core transformer :) \$\endgroup\$ Commented Jan 14 at 19:52

2 Answers 2


So, we have an unequal distribution of currents between the two conductors in a seemingly symmetric, w.r.t. current magnitudes, arrangement of busbars (conductors, wires). The rigorous explanation of this effect requires a solution of the Maxwell equations, but I'm going to use a workhorse of electronics.SE, a circuit simulator.

For this problem, the conductor (busbar) should be modeled with the component 'inductor', because, as you'll see, the decisive property of the conductor for modelling your arrangement is inductance. As we are not requested to examine the details of current partitioning within the red and blue busbars, a single inductor, LR, models a combined red busbars and LB models a combined blue busbar. The yellow busbars we model with two components, LY1 and LY2. In accordance with OP's description (am I right?), Y1 is adjacent to the R busbar, Y2 is adjacent to the B busbar. The effect is due to mutual inductances of busbars, so we introduce coupling directives: KRY1 LR LY1 0.167 models the mutual inductance of the red busbar and the yellow busbar Y1; KY2B LY2 LB 0.167 models the mutual inductance of the blue busbar and the yellow busbar Y2; KY1Y2 LY1 LY2 0.67 models the mutual inductance of the two yellow busbars.

power line model

Selection of inductances and serial resistances is quite arbitrary and directed only by the desire to have numerical values close to those specified in the OP post. In particular, in a real switchboard the inductance of any of two combined busbars, either R or B, is different than that of any of two yellow busbars, either Y1 or Y2, and also the neutral busbar may contribute into current distribution.

As for selection of LR/LY1 (LB/LY2) ratios in simulation, we can adjust a quantity of interaction between R(B) and Y1(Y2) by changing the coupling KRY1(KBY2) to account for the arbitrary selection of these ratios. After all, the simulation has been built just to show that only selection of current source phases -- which ones are leading/lagging-behind the others by the three-phase's nominal 120 degree -- is dictated by the requirement for an Y1 current to be greater than an Y2 current.

Notice that the other selection of parameter values, especially the series resistance, may noticeably shift the Y1/Y2 relative phases and, with the sum of Y1/Y2 AC currents still being 3000A (KCL), their amplitudes would no longer add to the same 3000A*SQRT(2).

We run the tran analysis for 0.1 s of simulation time, six full periods of alternate currents.


As you see in the plot, when the red busbar current leads the yellow busbar current by 120 degree and the blue busbar current lags behind the yellow busbar current by the same 120 degree, the Y1 AC RMS current is approx. 1.8KA and the Y2 AC RMS current is approx. 1.2KA. The inductance coupling works as a sort of "constructive interference" for the red-yellow pair and a sort of "destructive interference" for the yellow-blue pair of conductors.

DISCLAIMER: As many details, like coupling of non-adjacent conductors, are omitted from this model, it cannot be used for quantitative estimation of current partitioning in wires of electric power lines. Its only purpose is to indicate the possible source of the OP confusion. There is an asymmetry in a seemingly symmetric switchboard: if the red busbar current leads/lags-behind the yellow busbar current, then the blue busbar current lags-behind/leads the yellow busbar current.


I have been dealing with similar setups and found that current asymmetry often is more likely due to different resistance. When thousands of amps is passing, connections an and centimeters count in the equation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.