Dspark,
Happy to assist!!!
In this scheme, the setup is very much an Autotransformer, which is derived with the
"Test Jumper" connection between primary and secondary.
So the results will be either:
<OL TYPE=A>
[*]Primary Voltage + Secondary Voltage = Output Voltage; for "In-Phase" additive polarity, or
[*]Primary Voltage - Secondary Voltage = Output Voltage; for "Out-Of-Phase" subtractive polarity.
</OL>
*** NOTE *** Output Voltage is what's measured at the Voltmeter "E"As you have said here, the primary can and is used as a reference voltage.
If the transformer was rated for 200 VAC on the primary, with a secondary voltage of 20 VAC - this would need to be kept within tolerances in order to correctly reflect the Impedance [Z] of the secondary's load.
If 1/2 the voltage, or 100 VAC was applied to the primary, the secondary will be effected - possibly as far as 1/2 the secondary voltage, or only 10 VAC available in the secondary.
This applies to all transformers.
A Power-type transformer will work in the basic fashion of a typical Impedance matching transformer - which "reflects" a different connected Impedance to the supplying source, than the actual connected Impedance.
Example:
If the connected load on the secondary would be 10 ohms Z, and it was rated for either 10 VAC or just 1 amp [or both], and the voltage available is 100 VAC, we cannot simply connect the 10 Z load to the 100 VAC supply and expect the smoke to remain inside!!!, so we connect it thru a Transformer - which makes the 10 Z secondary load appear as a 1000 Z load to the power supply.
Now the power [in Volt-Amps] can be applied as needed; 10 VA on the primary [supply], 10 VA on the secondary [load]. 100% power transfer - except minimal transformer losses and such.
Thank you for the reply!!!
Scott SET
