The equivalent inductance of 2 identical inductance connected in parallel ( ignoring mutual coupling) is half of the orginal . This result is usually derived from the argument of identical voltage across parallel elements->Proof

I would like to know how can we derive the same result from a flux linkage point of view. Self inductance of any component is defined as the ratio of total flux linkage to the current which produces this flux. Isn't the flux linkage of the parallel combo same as the sum of flux linkages of individual inductors?

If so, wouldn't the total flux linkage be same as the original flux linkage with just one inductor. [ Reason: Each inductor carries half the current, hence half the flux linkage. Sum total of both the flux linkages remains the same]

  • \$\begingroup\$ Self inductance is not defined at the material level, but at the component level (that is, material + geometry). Check your inductance definition for the correct dimensions, it should be volt.seconds / amp, which as you say is flux per current, and multiplied by a dimensionless number of turns. \$\endgroup\$
    – Neil_UK
    Commented Sep 27, 2019 at 7:22
  • \$\begingroup\$ When placed in parallel, each inductor gets only half of the current. Because flux is proportional to current, the flux is halved too in each inductor. So what does the Maxwell–Faraday equation tell you about this? \$\endgroup\$
    – Bart
    Commented Sep 27, 2019 at 9:48
  • \$\begingroup\$ @Neil_UK Corrected the 'material' usage-> Replaced as 'component' \$\endgroup\$ Commented Sep 27, 2019 at 10:21

3 Answers 3


There are many different ways you can characterize inductance, including several different was that could be called 'flux-oriented', so this is hardly the only way to look at this, but it is my favorite due to the simplicity.

Inductance can be characterized as a measure of the ratio of magnetic flux and current. Just as the ways of characterizing inductance are many, so too are the ways you can represent the SI unit of inductance, the Henry. One of these ways is, you guessed it, \$\frac{Wb}{A} \$!

Webers per ampere. Webers of course being the unit of magnetic flux. This is the same unit used to represent flux linkage, like through a coil.

Now, consider a 20µH inductor. Ignoring things like saturation and temperature effects of course, this inductor will generate 20µWb of magnetic flux if one amp is conducting through its coil.

But, let's consider two 20µH inductors, connected in parallel (but physically distant enough that we can ignore any magnetic coupling between them). If we let 1 ampere flow through our circuit, this means the two parallel inductors will have to split this current between them. Since they are both 20µH, this means each inductor generates: $$500mA\cdot 20\mu H = 10\mu Wb$$

There is a very simple but maddeningly non-obvious subtlety here that I suspect might be the source of your question in the first place.

One might initially think that if each inductor is producing 10µWb of flux, then you have a total of 20µWb of flux between them, but that is not the case. You cannot add each inductor's flux together. Magnetic flux is magnetic field through an area, yet these two inductors, separated by enough distance that they do not interact magnetically, are absolutely not a single surface over which you can integrate the flux. Put another way, the flux of one inductor does not pass through, or 'link' the turns of the other inductor, so there is no flux linkage between them and you cannot add their individual fluxes together.

Indeed, there is a single flux value for any system of inductors that have no magnetic coupling between them connected in parallel. And in the example earlier, this is 10µWb. That is the flux for the entire system of two parallel inductors.

This becomes a little more intuitive if we imagine a different case, one where our two 20µH inductors are not only coupled, but ideally so - their coupling coefficient is 1. This means that our two coils are wound in such a way that all of the field lines passing through one coil also passes through the other. This is when we can actually add their individual fluxes together - when that flux links all the coils involved, not just one. When the inductors have mutual inductance. This is another way of saying the flux created from current through one coil will link not just the turn of that coil, but the turns of the mutually induced coil as well.

When there is no coupling, the flux through any of the inductors is the same, and that flux for any coil in isolation is equal to the total flux through all coils, since only their own flux is ever linked through any given coil.

Let's take a more complex example, one with a 1H inductor, a 9H inductor, and a 5H inductor in parallel. If we again conduct 1 ampere through these coils, each one completely uncoupled magnetically from the others, each inductor of course will have an inversely proportional amount of that ampere flowing through it. Because, no matter what, it takes time for the current to rise, and for the same voltage across each inductor, it will rise proportionally faster or slower.

Imagine 1V across our inductors. If we wait 1 second for 1A to be flowing through the 1H inductor, then the current flowing through the 9H and 5H inductors is proportionally less - 1V across a 5H inductor will only reach 200mA of current after 1 second. And 110mA for the 9H inductor.

These proportions will always hold due to their inductances slowing current rise times proportionally, and as they are in parallel, the voltage cross each one is the same.

So, our flux linkage is the flux seen by any single coil, and it is the same for all of them. Lets return to the total current of 1A, which yields currents of approximately 760mA, 152mA, and 84mA for the 1H, 5H, and 9H inductors respectively. To get the linkage, just multiply the current by the inductance. 760ma•1H = 760mWb. 152mA•5H=760mWb. 84ma•9H=760mWb.

Our flux linkage for the entire parallel system is 760mWb.

And characterizing inductance as the ratio of magnetic flux per ampere, well, we're only getting 760mWb of flux for our 1A of current. Rounding errors aside, if you calculate the parallel inductance for 1H, 5H, and 9H, you get 0.762H. Or exactly what we'd expect - it is saying we'll get 762mWb of flux for one amp input. And so we do.

In the case of two equal inductors in parallel, each inductor gets 500mA of our ampere, but the total flux is not added between them, because each inductor's flux only passes through half of our imaginary surface area. We have two inductors, and each one's flux only passes through one of them, not both. So the total flux, across both inductors, is what we get from a single 20µH inductor with 500mA flowing through it: 10µWb. Yet we are actually providing an entire ampere to get that 10µWb.

So now our inductance, the ratio of flux to current, is half that of either of the inductors when used individually.

  • \$\begingroup\$ Thanks for the elaborate answer \$\endgroup\$ Commented Sep 27, 2019 at 11:07
  • \$\begingroup\$ Nice illustration of the three unequal inductors. \$\endgroup\$
    – Neil_UK
    Commented Sep 27, 2019 at 14:00
  • \$\begingroup\$ Do you think you adequately cover the case for constant V_AC applied to first one then two parallel uncoupled inductors? \$\endgroup\$
    – Russell McMahon
    Commented Sep 28, 2019 at 2:11

When two identical inductors are paralleled, each gets half of the current. Flux is proportional to current, which means that flux is halved too. Now Faraday's law will tell you that voltage is proportional to rate of change of the flux, which is therefor halved too (assuming a fixed current).

So inductance is halved. In absence of magnetic coupling, the same equation applies for paralleling resistors.

  • \$\begingroup\$ Agreed, Flux generated in each inductor becomes half. But total flux linkage of the parallel combination remains the same. \$\endgroup\$ Commented Sep 27, 2019 at 10:08
  • \$\begingroup\$ @Divya K.S, the sum of the fluxes in the two inductors is still the same. But they are two separate components in parallel, so they have the same potential difference across them, And their fluxes are not linked in any way. So you just calculate with the flux of one component to get the potential difference. \$\endgroup\$
    – Bart
    Commented Sep 27, 2019 at 10:42
  • \$\begingroup\$ My turn :-) (but without the downvote that someone kindly gave me). The inductors are uncoupled. If the Voltage source is "stiff" (which is an entirely valid case) the current will double and inductance will be seen to have halved. ie it is NOT related to the per inductor current halving. \$\endgroup\$
    – Russell McMahon
    Commented Sep 27, 2019 at 12:33
  • \$\begingroup\$ @Russell, let me assure it was not me that downvoted you :). I based my answer on the pair being fed by a constant current source, because the current is what determines the magnetic flux (L times phi equals N times I). The induced voltage follows from Faraday's law. Of course you can reverse the reasoning by applying a stiff voltage, which will then determine the current, and the flux, but this is the same for each inductor and so the current doubles. I think it's a bit like a chicken and the egg story. Maxwell's laws are not very clear in what causes what. \$\endgroup\$
    – Bart
    Commented Sep 27, 2019 at 13:06
  • \$\begingroup\$ @Bart ZX80 eh - a youngster :-). But, yes, I had one, and a Spectrum, And a "phone/computer" at work at one stage that was a QL in disguise. ME thesis using an MC6802 for processing power from about 1978. Those were the days :-). \$\endgroup\$
    – Russell McMahon
    Commented Sep 27, 2019 at 13:12

This can be put far more formally, but this should help the intuitive grasp:

Inductance is related to "resistance to voltage".
For a given AC voltage at given frequency,
Impedance Z = 2.Pi.f.L
Current I = V/Z & Z_l = V/I

Place two uncoupled inductors in parallel and apply the same signal across the pair.
Current is now doubled.
V/I is halved.
So, Z is halved. Apparent (and actual) inductance is halved.

InductANCE is not related to what happens to the individual inductorS but is the result of the "black box" combination.

  • \$\begingroup\$ he was specifically asking for the flux point of view, not the component point of view. \$\endgroup\$
    – Neil_UK
    Commented Sep 27, 2019 at 7:21
  • \$\begingroup\$ @Neil_UK He was. That should provide what he needs. Let's see ... \$\endgroup\$
    – Russell McMahon
    Commented Sep 27, 2019 at 7:23
  • \$\begingroup\$ yes, I was toying with answering, but I can't get my head around dubunking the misconceptions, getting the dimensions right, this early in the morning. \$\endgroup\$
    – Neil_UK
    Commented Sep 27, 2019 at 7:27
  • \$\begingroup\$ @Russell McMahon, Apparent (and actual inductance is doubled? For parallel inductors, the same formula applies as for resistors (in absence of magnetic coupling). \$\endgroup\$
    – Bart
    Commented Sep 27, 2019 at 9:44
  • 1
    \$\begingroup\$ @Bart I WAS confused as to what your comment meant. Then I read my answer. Agh! :-). Thanks. Yes. Halved. mumble .... Edited. \$\endgroup\$
    – Russell McMahon
    Commented Sep 27, 2019 at 12:28

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