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Let's say that we have a transformer with two windings
a busy cat http://www.circuitstoday.com/wp-content/uploads/2011/11/Transformer-Circuit-Symbol.jpg
with nominal power \$S\$, nominal voltages \$V_1\$ and \$V_2\$ and windings \$N_1\$ and \$N_2\$ respectively. The equivalent impedance is \$R+jX\$ referred to the primary winding.
Now we rewire this transformer to become an autotransformer.

a busy cat http://www.vias.org/matsch_capmag/img/matsch_caps_magnetics-877.png

and the secondary has nominal load with lagging factor \$\cos \phi\$.

I am a little confused so my questions are: Does that mean that the voltage \$V_x\$ is equal to \$V_2\$? What about \$I_x\$ or \$I_H\$? How can we find the voltage \$V_H\$?

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I'm going to assume that the input voltage is Vh and the output is V2 / Vx.

You need to know the turns ratio of the transformer. You also need to know whether you have connected the windings in the buck configuration. I'm going to assume this is the case.

Let's work with some arbitrary numbers.

The transformer has a 120V primary and 12V secondary. This gives you a 10:1 turns ratio.

Let's further assume that the input voltage is 110 Vac - just to make the numbers easier.

The two windings are in series. You have x across the upper winding and 10x across the bottom winding. The total span is 10x + x = 11x

With 110 Vac input at Vh, you will have 100 Vac output at Vx.

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  • \$\begingroup\$ But given that there is an equivalent impedance (non ideal transformer) the output voltage isn't changed? \$\endgroup\$ – Theodore Samuel Jun 21 '15 at 10:15

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