I have a basic doubt in understanding the concept of equivalent inductance. When two inductor coils are in series the equivalent inductance is calculated as

\$ L_{eq} = L1+L2 \pm 2M \$ (depending on the dot convention)

where L1 and L2 are independent coil self inductance and M is the mutual inductance dependent on coupling between the coils k (coupling coefficient). M is calculated as

\$M = k \sqrt{L1. L2} \$

k can take values from 0 to 1, 0 for no coupling and 1 for perfect coupling between the coils.

So if two coils are placed adjacent to each other very far where there is no coupling between the coils then k = 0 and \$ L_{eq} = L1+L2 \$ because M = 0. If the coils are so far that L2 has no impact on L1 should we still consider L2 for calculating \$L_{eq} \$ ?

  • \$\begingroup\$ You are assuming, incorrectly, that there is no way of magnetically shielding the two inductors from each other or of orienting their magnetic fields such that they are orthogonal from each other and have a k=0. But both those techniques are actively used in circuits all the time. \$\endgroup\$
    – Dave
    Commented Jan 24, 2016 at 10:59
  • \$\begingroup\$ @Dave I am assuming a scenario where the two inductor coils are far apart that when we try to measure voltage across the farther coil there is no voltage induced in it. In this case i can take k=0 right? \$\endgroup\$ Commented Jan 24, 2016 at 14:08
  • \$\begingroup\$ Yes, at far distances you can take k=0. I never liked that explanation because because it assumes you have perfect 0 ohm conducting wires with no loss between the inductors. That's something that isn't practical and messes up beginners. \$\endgroup\$
    – Dave
    Commented Jan 25, 2016 at 0:28
  • \$\begingroup\$ at far distances when k = 0, how is Leq calculated? \$\endgroup\$ Commented Jan 25, 2016 at 2:17
  • \$\begingroup\$ Leq= L1+L2. They're still in series, so the electrons still must flow through both to complete the circuit, regardless of how far apart they are. Both pouyan and Tahmid have already answered that though. \$\endgroup\$
    – Dave
    Commented Jan 25, 2016 at 4:01

2 Answers 2


If coupling is not present as you mention:

Leq = L1 + L2

If L1 >> L2, Leq ~ L1

However, when L1 is not >> L2, you cannot just ignore L2 since it is still in series with L1.

  • \$\begingroup\$ So as per your answer if there is no coupling and irrespective of the position of the second coil the Leq is always L1+L2. I understand the case when L1 >> L2, if L1= L2=L then Leq is always 2L irrespective of the distance between the coils. please correct me if wrong \$\endgroup\$ Commented Jan 24, 2016 at 14:13
  • \$\begingroup\$ @SristiSravan, if there is an ac current through L2, it still generates a voltage, no matter how far away it is from L1. \$\endgroup\$
    – The Photon
    Commented Jan 24, 2016 at 17:00
  • \$\begingroup\$ ok I agree, I am not concerned on the voltage or current, I am trying to understand how Leq is calculated when k = 0 \$\endgroup\$ Commented Jan 25, 2016 at 2:18
  • \$\begingroup\$ how does the mutual inductance increase or decrease the self inductance of the coil, please can you explain? \$\endgroup\$ Commented Jan 25, 2016 at 2:18
  • \$\begingroup\$ When k=0, Leq = L1+L2. Mutual inductance is represented by the 2M term in the equation. If k=0, that term gets dropped. So, yes, to answer your question above, if there is no coupling, Leq is always L1+L2. \$\endgroup\$
    – Tahmid
    Commented Jan 25, 2016 at 8:32

As you can see on the schematic below, L2 is inductance of L2 itself, not effect of it on L1 inductance; So it can't be dropped


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ how does the mutual inductance increase or decrease the self inductance of the coil, please can you explain? \$\endgroup\$ Commented Jan 24, 2016 at 14:15

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