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.]].

I think your confusion lies in your first assumption. An *ideal* transformer doesn't even have windings, because it can't exist. Thus, it doesn't make sense to consider inductance, or leakage, or less than perfect coupling. All of these issues don't exist. An ideal transformer simply multiplies impedances by some constant. Power in will equal power out exactly, but the voltage:current ratio will be altered according to the turns ratio of the transformer.

For example, it is impossible to measure any difference between a 50Ω resistor, and a 12.5Ω resistor seen through an ideal transformer with a 2:1 turns ratio. This holds true for any load, including complex impedances.

^{simulate this circuit – Schematic created using CircuitLab}

Since an ideal transformer can't be realized, considering how it might work is a logical dead-end. It doesn't have to work because it is a purely theoretical concept used to simplify calculations.

The language you used in your first assumption is a description of the limiting case that defines an ideal transformer. Consider a simple transformer equivalent circuit:

^{simulate this circuit}

Of course, we can make a more complicated equivalent circuit according to how accurately we wish to model the non-ideal effects of a real transformer, but this one will do to illustrate the point. Remember also that XFMR1 represents an *ideal* transformer.

As the real transformer's winding resistance approaches zero, then R2 approaches 0Ω. In the limiting case of an ideal transformer where there is no winding resistance, then we can replace R2 with a short.

Likewise, as the leakage inductance approaches zero, L2 approaches 0H, and can be replaced with a short in the limiting case.

As the primary inductance approaches infinity, we can replace L1 with an open in the limiting case.

And so it goes for all the non-ideal effects we might model in a transformer. The ideal transformer has an infinitely large core that never saturates. As such, the ideal transformer even works at DC. The ideal transformer's windings have no distributed capacitance. And so on. After you've hit these limits (or in practice, approached them sufficiently close for your application for their effects to become negligible), you are left with just the ideal transformer, XFMR1.

## Best Answer

As long as the secondary remains open, it's basically not there magnetically or electrically. In that case the primary will look like a regular inductor.

However, transformers don't usually make good inductors. They are usually more lossy, it is often hard to get specs on just the inductance, and the frequency range may be not much more than the transformer was designed for. Above that frequency, you may get significant core losses.