"Short-circuit inductance is the inductance when one of the primary winding or the secondary winding of the transformer is short-circuited and measured from the other winding."

I can't understand the part of the sentence in italic. What do they mean by "short-circuited" in this context? Even the image given in the Wiki page doesn't make much sense.

As far as I know, if an inductor is short-circuited it means there is connecting wire across it (i.e. the potential at both ends of inductor is same)

Source: https://en.wikipedia.org/wiki/Short-circuit_inductance

  • 2
    \$\begingroup\$ It means exactly what you think it does. The problem is that you're in the usual mindset of thinking about the voltage on the terminals, when what matters is not that (effectively zero) but rather the induced current flowing, and specifically how that reacts to change in the current flowing through the other coil to which it is magnetically coupled. \$\endgroup\$ – Chris Stratton Sep 24 '17 at 21:10
  • 1
    \$\begingroup\$ @ChrisStratton I can't understand what you're trying to say. It will be helpful if you could write an answer or explain with a diagram. \$\endgroup\$ – user133614 Sep 24 '17 at 21:12

Just in case you do not read and speak our ordinary Mathjargonish well, I give more visual explanation:

enter image description here

The wire at the right is the short circuit. The short circuit inductance is what the inductance meter shows.

This test gives some numerical data of how far the transformer is from an ideal one. In ideal transformer the short circuit inductance is =0. In practice it's greater. To actually get some useful info, the wire at the right should be removed and also check, how much the inductance is without the short circuit. Ideally it should be infinite.

ADDENDUM due the comment:

The derivation of the formula for the short circuit inductance unfortunately needs the phasors or differential equations. Here the formula is derived in the simplest case (=no losses taken into the account):

enter image description here

The short circuit inductance has taken the place of the inductance in inductor's general equation between voltage and current.

Without the short circuit in secondary one can measure L1. Measuring the short circuit inductance is a way to get the k.

  • \$\begingroup\$ Thanks a ton. I understand the concept now. But I could not understand how they got $$L_{sc1}=(1-k^2)L_1$$ I know that $$k$$ is the coupling coefficient. Could you please explain the derivation of the formula? \$\endgroup\$ – user133614 Sep 25 '17 at 6:24
  • \$\begingroup\$ @Blue ok. Takes few hours because I am on mobile now. \$\endgroup\$ – user287001 Sep 25 '17 at 6:34
  • \$\begingroup\$ @Blue the answer is augmented. \$\endgroup\$ – user287001 Sep 25 '17 at 9:03

Consider the real transformer model.

Each winding will have a "leakage inductance", which is the inductance of one winding which in not magnetically coupled to the other winding.

As you know, any impedance on the secondary can be "brought" to the primary by multiplying it by \$\alpha^2\$. Therefore you get the following equivalent schematics:

enter image description here

If you short the secondary (assuming \$R_P=0\$ - zero loss on copper - \$R_C=\infty\$ - zero loss on core- , and \$X_M=\infty\$ - infinite magnetizing inductance, ideal transformer), the short circuit inductance is simply the series of the two leakage inductances (the secondary leakage inductance is multiplied by \$\alpha^2\$):

$$L_{SC}=L_P+L'_S=L_P+L_S \cdot \alpha^2$$

(Because if the secondary is shorted, the primary will be shorted too).


The primary and secondary inductors are coupled by the core.

When the core is less than perfect, some of the flux just circulates around its own winding without being shared by the other. This is "leakage flux".

This has an equivalent primary series inductance when the secondary is shorted, and visa versa. Turns ratio² affects impedance from primary to secondary and Z(L)= ω L .

Transformers are defined by this non-ideal factor coupling ratio, k approaching 1 or leakage , σ approaching 0, where;

\${\displaystyle \sigma =1-k^{2}=1-{\frac {M^{2}}{L_{P}L_{S}}}} \$ and Mutal inductance, M is defined as

\${\displaystyle k=\left|M\right|/{\sqrt {L_{P}L_{S}}}} \$

  • \$\begingroup\$ How does one derive \$\sigma=1-k^2\$ ? \$\endgroup\$ – user133614 Sep 25 '17 at 6:31
  • \$\begingroup\$ it is already documented in Wiki. I just give you the answer. This is just a convenient definition to compare impedance vs inductance without units for leakage. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 25 '17 at 12:32

short the two poles of the secondary winding with a jumper wire, then measure the inductance across the poles of the primary winding.

that is how you measure the short-circuit inductance of the secondary, or Lsc2.

switch every instance of "primary" and "secondary", to find Lsc1.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy