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 P2 to S1 (or P1 to S2) 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 Lt,
- Lt > Lp
- Lt > Ls
- Lt > (Lp + Ls) !!! <- 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 Lsepsum = Ls + Lp.
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.
Lm 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 have 1 kVA, single phase, 250/125 V transformer.
If you can adjust the primary voltage so that the input current is 4 amps RMS when the secondary is shorted, the secondary current will be 8 amps for a 2:1 step-down transformer.
NB - you will probably find that the easiest way to do this is use the output of a variac and raise the primary voltage carefully whilst noting the primary current. Typically the variac output will only need to be around 10V RMS to get full primary current with secondary shorted out.
The primary-secondary turns ratio determines both step-down voltage ratio and step-up current ratio.
Best Answer
Just in case you do not read and speak our ordinary Mathjargonish well, I give more visual explanation:
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):
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.