It sounds like you're dealing with a SPICE simulator.

In SPICE, you define a transformer as a coupled inductor, with the following ratio:

\$\dfrac{N_1}{N_2} = \sqrt{\dfrac{L_1}{L_2}}\$

There's also a coupling factor K that comes into play.

The magnetizing inductance of a transformer is the inductance you measure across the primary winding with the secondaries open-circuit (floating). It's a function of the core material and geometry, air gap and number of turns.

Your \$L_1\$ and \$L_m\$ should be one and the same.

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.]].

## Best Answer

The induced magnetic field is proportional to ampere-turns, that's current times number of turns. Electrical energy is converted to magnetic energy in the core and back to electrical. The core must be big enough to hold that without saturating. For a 100 VA transformer you want to transfer more energy magnetically than for a 10 VA transformer. The 100 VA is larger because it has more turns to build up a stronger field, and also needs a bigger core to avoid it saturating.