But, how can that happen it would leave no voltage for the
resistance??
Andy's given the correct answer to this question in the comments. This is the complete analysis.
Call the left winding primary and the right winding secondary. See that the secondary winding is open and thus there is zero secondary current.
Since this is an ideal transformer, the primary current must also be zero and thus, by Ohm's Law, the voltage across the \$2 \Omega\$ resistor, in series with the primary, is zero.
By KVL, the voltage across the primary is \$-1V\$ so the voltage across the secondary is \$-4V\$.
Finally, the open circuit voltage is \$V_{ab} =V_{th} = 1 - 4 = -3V = 3\angle 180 V\$
To find the Thevenin impedance directly, zero the independent voltage source and see that the \$2 \Omega\$ resistor is across the primary which, reflected to the secondary, appears as \$2 \cdot 4^2 = 32 \Omega\$. Thus, the Thevenin Impedance is \$Z_{th} = 32 \Omega\$.
For example , 5v * 44 = 220v , and then 220v * 44 = 9680v
Paraphrasing the above quote: Using two cascaded transformers, the first produces 220V and this feeds into the second that magnifies this by a further 44 times to get 9680 V
It won't work so don't even bother trying it. There are two reasons: -
- The magnetic core of the 2nd transformer will heavily saturate and fry (if enough energy could be supplied by the 1st transformer)
- The breakdown voltage between consecutive turns (and layers) of a winding intended for 220 VAC might be 1500 V but certainly won't be ~10kV.
Enough said.
Best Answer
We'll you can't really uncover the inductance value of an unknown winding by just knowing a formula. The normal tests to do this are an open secondary test and a shorted secondary test.
The open circuit test requires a measurement of primary current for an applied voltage. The impedance calculated from V and I is usually regarded as indicative of the primary mag inductance.
A short circuit secondary test with a smaller input voltage whilst measuring the primary current will reasonably hint at the leakage inductance of both windings together. Then, knowing the transformer turns ratio, you can start to put leakage inductance values on the primary and secondary.
If it's a HF transformer you need to adopt a different approach that measures resonant frequencies of the primary and secondary and, because leakage capacitance is significantly more dominant now, it's a bit trickier doing the estimations.