[Amps] HV transformer and Variac

[Larry Benko]

Gerald and others,

To add to the auto-transformer (variac) discussion I did a quick 
simulation in LTSpice to make sure my memory was correct before making a 
fool out of myself. :-)

There are 2 currents in the variac: the magnetizing current and the 
current due to the load.

1.) The magnetizing current is generally set to be a couple of percent 
of the maximum load current.  If we view the variac as 2 coupled 
inductors in series the magnetizing current is in phase in the 2 
inductors and the junction of the 2 inductors is the voltage out point.

2.) The current due to the load  does not add in one of the windings as 
people often believe.  The current due to the load is out of phase in 
the the 2 series coupled inductors.

To make this clearer here are a couple of examples assuming an AC input 
voltage of 120VACrms and a 10 ohm load.  The magnetizing current is 
150mArms (~2H total variac inductance) and the windings have .001 ohm of 
resistance.

1.) Assume a tap point of 90%, the output is 120*.9 = 108Vrms and the 
load current is 10.8Arms.  The inductor that connects to the AC input 
has a current of 9.73Arms and the current in the inductor that connects 
to AC common is 1.13Arms.  The inductor currents are OUT OF PHASE due to 
the transformer action and they sum to 10.8A.

2.) Assume a tap point of 50%, the output is 120*.5 = 60Vrms and the 
load current is 6.0Arms.  The inductor that connects to the AC input has 
a current of 3.02Arms and the current in the inductor that connects to 
AC common is 3.01Arms.  The inductor currents are OUT OF PHASE due to 
the transformer action.

Notice how the currents in the winding of the transformer are less than 
the load current which might seem impossible.  If anyone wants the 
simple LTSpice circuit email me.

Larry


On 7/23/2015 8:14 AM, Gerald Williamson via Amps wrote:
> Hi Ros, to split hairs here, one would say the autotransformer would loose
> very little efficiency. As you know, when there is copper wire and there is
>   current, there will be a loss of current squared times resistance.
>   
> The majority of the current would be in the turns of the autotransformer
> between the input and output terminals so the loss would be related to the
> input  to output voltage ratio.
>   
> That is my personal understanding and subject to correction.
>   
> 73,
> Gerald K5GW
>   
>