When it comes to inductance in circuits, it is a rule of thumb to minimize the loop area. This implies a return path (i.e. two wires). There is also self inductance of a single wire.

I am not sure how they play together: Do I need to add them? Or do I need to choose one, depending on the geometry? If so, which one to pick?

Example: Voltage source to sink, connected by two 1mm diameter wires in parallel (3mm apart) at a distance of 1m (one wire is supply, the other is return/ground).

A calculator for Parallel Wire Inductance tells me the inductance of this configuration is 705nH.

Another calculator for Wire Self Inductance tells me that the inductance (self inductance) of each wire is 1.51uH.

Note that if the two wires are completely adjacent to each other (distance 1mm instead of 3mm), 705nH becomes 0H which makes sense because the loop area is zero. Do we still need to consider self inductance in this case?

What is the correct equivalent circuit? Do I need to consider the first inductance ("loop area") or the second one ("self inductances") or both?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ There is no such thing as the self-inductance of a single wire. The inductance is only defined for a loop. But partial inductance can be defined for a single wire, in which case, the total self-inductance will be the sum of partial inductances of component wires in the loop. \$\endgroup\$
    – sarthak
    Feb 23, 2022 at 9:51
  • \$\begingroup\$ Is my circuit model then correct? Why/why not? I assume not, but I added the self-inds in the loop together with the inductance of two wires as you said. The point of my question is that it is unclear how these relate. \$\endgroup\$
    – divB
    Feb 23, 2022 at 10:03
  • \$\begingroup\$ Note that if wires are near each other, although these have self-inductance, they are "coupled" as a "transformer". Moreover, they are coupled capacitively ... \$\endgroup\$
    – Antonio51
    Feb 23, 2022 at 10:12
  • \$\begingroup\$ @Antonio51 There is also mutual inductance between the wires. Not only capacitive coupling. \$\endgroup\$
    – sarthak
    Feb 23, 2022 at 10:15
  • \$\begingroup\$ @divB read this: ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=1134841. I think this should clarify, if not let me know. \$\endgroup\$
    – sarthak
    Feb 23, 2022 at 10:16

3 Answers 3


There is no such thing as 'the inductance of a single straight wire'.

Inductance is exclusively a property of a current loop.

This a frequent misunderstanding propagated by the numerous online calculators that do suggest that one can associate a well defined inductance with a single straight current carrying wire. The magnetic field from a single straight current would extend infinitely far away and would require infinite energy, so inductance would be infinite as well.

In practice, the current must return somewhere, an only then, when current goes in a closed loop, loop inductance is well defined.

There is also the concept of net partial inductance. Here, the total loop inductance is distributed among all the segments of the loop in a sensible way. It is useful for reasoning what parts of a circuit contribute most to the loop inductance. But still, the net partial inductance of any component depends on the entire loop. Taking the wire segment out of the circuit would immediately invalidate its partial inductance.

  • \$\begingroup\$ Context of this question aside, I think ESD still encounters inductance during a discharge. \$\endgroup\$
    – DKNguyen
    Feb 25, 2022 at 15:24
  • \$\begingroup\$ Absolutely, but the ESD current is no different than any other signal. The current must return somewhere, eg. the earthing cable. \$\endgroup\$
    – polwel
    Feb 26, 2022 at 8:24
  • \$\begingroup\$ The "inductance" of a wire... is only a "means" of calculation, of course. .... But if we consider a "monopole" antenna, doesn't that make "sense"? \$\endgroup\$
    – Antonio51
    Feb 26, 2022 at 9:40
  • \$\begingroup\$ A monopole antenna is not a current going straight. The current returns (as displacement current) through the air. Sorry, but I am missing your point here entirely. Please elaborate :) \$\endgroup\$
    – polwel
    Feb 26, 2022 at 10:08
  • \$\begingroup\$ Where is the return "wire"? So, inductance is not the fact of a loop. What is then the inductance of some turns at the base or at the middle of the antenna? \$\endgroup\$
    – Antonio51
    Mar 1, 2022 at 7:58

What is the correct equivalent circuit? Do I need to consider the first inductance ("loop area") or the second one ("self inductances") or both?


The correct equivalent circuit for two 1mm dia wires spaced 3mm (between centers) and 1m long is a transmission line, characteristic impedance 211ohms, delay 3.33ns.

To calculate the impedance, see online calculators like:


or en.wikipedia.org/wiki/Twin-lead#Characteristic_impedance

If your frequencies are low so that 3.3ns is insignificant, then you can replace the transmission line with a lumped element equivalent, which is a series inductance of 703nH and a shunt capacitance of 15.8pF.

What are you trying to model?


But why does your answer not include 2x 1.51uH from the self inductance

There is no 'self inductance', there are only two parallel lines for which there is an exact solution, for example see

www.ece.rutgers.edu/~orfanidi/ewa/ch11.pdf section 11.5

This exact solution includes all the magnetic fields and gives you the inductance exactly.

The 'self inductance' you calculated from that website quoted a formula from Grover which comes from a paper by Rosa g3ynh.info/zdocs/refs/NBS/Rosa1908.pdf which is an attempt to calculate some form of inductance for an isolated wire. Why would you think your should add it to an exact solution for the two wire problem?

but I am explicitly interested in the inductance aspect only

with respect, if you have an easily accessible exact solution you should use it, you may decide to ignore the capacitance, but you it never hurts to know it's value.

To summarise, at low frequencies and for impedances where you can ignore the capacitance, the equivalent circuit is a single 703nH inductor.

If you are modelling a single wire above a conducting plane, then this is a little different - if the centre of the conductor is 'h' above the plane, then solve the 2 wire problem for 2 conductors spaced at 2h and halve the characteristic impedance.

  • \$\begingroup\$ Thanks. I am aware of T-lines but I am explicitely interested in the inductance aspect only. Modeling simple power supply line+ground. It seems 703nH is the same as my 705nH (calculated via parallel wire inductance eeweb.com/tools/parallel-wire-inductance). But why does your answer not include 2x 1.51uH from the self inductance (see also Antonio's answer)? How are 1.51uH and 705nH related? \$\endgroup\$
    – divB
    Feb 25, 2022 at 8:35
  • \$\begingroup\$ Something weird in formula (21) of Rosa1908 page 315. How can a "cm" unit (14816 cm) be used as "microHenry" one line later dividing by 1000 (14.816 uH)? \$\endgroup\$
    – Antonio51
    Feb 26, 2022 at 9:15
  • 1
    \$\begingroup\$ I think there were different conventions 112 years ago. If you look at the Biot-Savart law you will see a missing \$\mu_0/{4\pi} = 10^{-7}\$ compared to how we would write it today, which may explain things. I haven't studied the paper in great detail, I haven't found much use for these incremental inductances in the work I do, but it is interesting to see how they are derived. \$\endgroup\$
    – Tesla23
    Feb 26, 2022 at 22:33

Note that if the two wires are completely adjacent to each other (distance 1mm instead of 3mm), 705nH becomes 0H which makes sense because the loop area is zero.

Impossible, because the distance is "between" the centers, so it must be > ~ 1.01 mm. They can't "touch" each other.

What is the correct equivalent circuit?

EDIT: and insert from paper Rosa1908 p311, for a square loop ...
Thanks to @Tesla23.

enter image description here

These wires have self-inductance, they are also "coupled" as a "transformer". No capacitive effect in simulation. If yes, then this "system" is a "transmission line".

Here is what happens when coupling varies. Good coupling, no self?

Corrected inductance (external+internal) calculated as this.
It is why we twist each wire ...

enter image description here

  • \$\begingroup\$ It would be extremely helpful if you would use the numbers/setup from my question: Do your L1/L2 correspond to my 1.51uH (obtained from single wire self inductance)? Is the "parallel wire inductance" 705nH captured by your coupling factor? If yes, how are they related? \$\endgroup\$
    – divB
    Feb 23, 2022 at 20:55
  • \$\begingroup\$ In fact, when one calculates "inductance", there is two independant part: the "internal" inductance and the "external" inductance. Unless I am wrong, inductance is the sum of the two. See this coursehero.com/file/142424/Internal-and-External-Inductance and this nptel.ac.in/content/storage2/courses/108104051/chapter_1/… \$\endgroup\$
    – Antonio51
    Feb 23, 2022 at 21:08
  • \$\begingroup\$ See also this which is 1.51uH (internal+external) allaboutcircuits.com/tools/wire-self-inductance-calculator \$\endgroup\$
    – Antonio51
    Feb 23, 2022 at 21:15
  • \$\begingroup\$ The inductance of two wires is half of the inductance of a wire (in this case) because of negative coupling (k=~ -0.7) ... distance between wires, but not sure ... \$\endgroup\$
    – Antonio51
    Feb 23, 2022 at 22:07
  • 1
    \$\begingroup\$ One more thing makes sense now: Self inductance is the right modeling tool when there is no/negligible coupling. For example a large loop or a ground loop. If there is coupling, self inductance is reduced by the mutual coupling. The "loop area" is a proxy for coupling. The larger the loop area, the less coupling and the less self inductance gets reduced by the mutual inductance due to return current. When both wires are hypothetically completely together, loop area is zero and coupling cancels both self inductance exactly. Is this my right understanding? \$\endgroup\$
    – divB
    Feb 25, 2022 at 8:59

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