How do I obtain an inductor from the given transformer in the image? ... So that the inductance of the resulting inductor must be maximum.

Connect the undotted end of one winding to the dotted end of the other.

eg P_{2} to S_{1} (or P_{1} to S_{2}) and use the pair as if they were a single winding.

(As per example in diagram below)

Using just one winding does NOT produce the required maximum inductance result.

The resulting inductance is greater than the sum of the two individual inductances.

Call the resultant inductance L_{t},

- L
_{t} > L_{p}
- L
_{t} > L_{s}
- L
_{t} > (L_{p} + L_{s}) !!! <- this may not be intuitive
- \$ L_t = ( \sqrt{L_p} + \sqrt{L_s}) ^ 2 \$ <- also unlikely to be intuitive.
- \$ \dots = L_p + L_s + 2 \times \sqrt{L_p} \times \sqrt{L_s} \$

Note that **IF** the windings were NOT magnetically linked (eg were on two separate cores) then the two inductances simply add and L_{sepsum} = L_{s} + L_{p}.

What will be the frequency behavior of the resulting inductor? Will it have a good performance at frequencies other than the original transformer was rated to run in.

**"Frequency behavior"** of the final inductor is not a meaningful term without further explanation of what is meant by the question and depends on how the inductor is to be used.

Note that "frequency behavior" is a good term as it can mean more than the normal term "frequency response" in this case.

For example, applying mains voltage to a primary and secondary in series, where the primary is rated for mains voltage use in normal operation will have various implications depending on how the inductor is to be used.Impedance is higher so magnetising current is lower so core is less heavily saturated. Implications then depend on application - so interesting. Will need discussing.

**Connecting the two windings together so that their magnetic fields support each other will give you the maximum inductance.**

When this is done

so the resultant inductance will be greater than the linear sum of the two inductances.

The requirement to get the inductances to add where there 2 or more windings is that the current flows into (or out of) all dotted winding ends at the same time.

**\$ L_{effective} = L_{eff} = (\sqrt{L_p} + \sqrt{L_s})^2 \dots (1) \$**

**Because:**

Where windings are mutually coupled on the same magnetic core so that all turns in either winding are linked by the same magnetic flux then when the windings are connected together they act like a single winding whose number of turns = the sum of the turns in the two windings.

ie \$ N_{total} = N_t = N_p + N_s \dots (2) \$

Now:
L is proportional to turns^2 = \$ N^2 \$

So for constant of proportionality k,

\$ L = k.N^2 \dots (3) \$

So \$ N = \sqrt{\frac{L}{k}} \dots (4) \$

k can be set to 1 for this purpose as we have no exact values for L.

So

From (2) above: \$ N_{total} = N_t = (N_p + N_s) \$

But : \$ N_p = \sqrt{k.L_p} = \sqrt{Lp} \dots (5) \$

And : \$ N_s = \sqrt{k.L_s} = \sqrt{L_s} \dots (6) \$

But \$ L_t = (k.N_p + k.N_s)^2 = (N_p + N_s)^2 \dots (7) \$

So

\$ \mathbf{L_t = (\sqrt{L_p} + \sqrt{L_s})^2} \dots (8) \$

Which expands to: \$ L_t = L_p + L_s + 2 \times \sqrt{L_p} \times \sqrt{L_s} \$

**In words:**

The inductance of the two windings in series is the square of the sum of the square roots of their individual inductances.

L_{m} is not relevant to this calculation as a separate value - it is part of the above workings and is the effective gain from crosslinking the two magnetic fields.

[[Unlike Ghost Busters - In this case you are allowed to cross the beams.]].

Apparently, there is only one "leakage inductance" at one side in the
figure above.

No, that is incorrect.

Every winding in a transformer does not couple 100% to each other winding and that is a fact. If some piece of text suggests that the leakage inductance is only attributable to one winding then that piece of text is at best misleading and, at worst blatantly wrong.

However, from the perspective of someone wishing to know how well two windings may couple then a single entity of leakage (a composite of both leakages) can be used to express that.

As for your 2nd question, you CANNOT short the secondary at the point you wish. This IS impossible - you can't take the equivalent circuit of the transformer and hack at it like that. The leakage measured is the composite leakage and this can be broken down into two components by using the turns ratio squared but even that is only an approximation; the magnetization inductance will alter the accuracy of this method slightly but, for all practical purposes, this method yields fairly accurate results.

## Best Answer

Shorting one winding and measuring the other gives you the

leakage inductance, which is a result of imperfect coupling between the two windings, and is a function of the physical construction of the transformer.Depending on your needs, it can have a practical application - zero voltage switching full-bridge converters often use the leakage inductance of the transformer to assist with the ZVS transitioning of the primary MOSFETs - but in general, transformers are designed to minimize this parameter by a variety of means (winding interleaving, etc.)