If the primary side of a transformer is a closed circuit why doesn't current flow until the secondary's circuits are closed?

That is a GREAT question...

Resistance?

Well, if I can jump in with a guess, the primary impedance is probably greater with an open secondary because the magnetic field in the transformer core can't "bleed off" as secondary current. The magnetic field is bigger and pushes against almost all the current flow in the primary.

Transformers work because a changing (AC) electric current creates a magnetic field (or flux) around the conductors, and a changing magnetic field creates an electric current. In a transformer, the primary AC current creates a magnetic field in the core which is then transferred to the secondary (or secondaries.)

If there isn't a closed secondary circuit, the magnetic flux can only work on the primary. At least, I'm pretty sure this, and I have heard the phrase "counter-EMF", which refers to a voltage created inside inductors which opposes an applied voltage. Considering how efficient most transformers are, I'd assume that with an open secondary, they could also create a massive inductive load on the primary and hold off most of the potential current.

Definitely a great question. It deserves a great answer - wish I was able to give one
I'll keep it short and sweet..

The transformer will "idle" with either no secondary load connected or the feeders to the secondary not connected.

At idle, the primary will draw only the current needed to magnetize the core, plus some current from losses in the windings, the core and external from the circuitry. You feel the power developed from these losses as heat at the transformer.

Small losses in the secondary coils will also be applied.

The reason that current in the primary is so low [almost zero] is that the magnetic flux in the core is stable [no flow], which results in an almost equal Inductive Reactance and Counter EMF imposed to the primary current, which results in nearly no flow of input current [or input KVA].
When a load is connected to the secondary, the stability of the core is changed, due to the change of XL in the secondary [the load allows current to flow out of the coil].
This results in a change in the core's flux [now is flowing], which ends up changing the XL on the primary windings.
Current increases until a steady state level is accomplished, then it remains at that level [until some value on either primary or secondary - or both -changes].

At startup, there is a high current draw on the primary - even if the secondary is not connected. It stabilizes over a period of time [usually by no more than 2 seconds].
This is similar to the inrush of an AC Induction motor, which by the way, is just a transformer with a rotating secondary winding!!

John-Tx has the idea!! Good job!

Hope this is understandable

I'll get to your other questions ASAP, so don't think I have ignored them.

Will submit that theory message before this weekend - or die ....

Scott SET

Ooh, I forgot to include the most important and simple reason why this happens;

There's nothing on the secondary side to run!
The transformer is simply an energy transfering thingee, which changes [increases or decreases] output levels, as opposed to input levels. Lower output voltages have higher output currents per transmitted KVA.

Simply, the transformer is a transfer medium between a load and the supply. Since their is no connected load in your situation, there is nothing to transfer energy to - hence no power is required to be drawn from the supply.
This is referring to a "Perfect" situation - one with no losses.
With a connected load to the device, the power supply is now going to deliver energy. The net input energy to the primary is in apparent power form [KVA]. Output energy can be in either True power form [Wattage, or KW], or be included inside of apparent power [Volt-Amps, or KVA].

Same thing goes for an AC induction motor.
At idle [shaft is spinning freely with nothing connected], the only input is magnetizing current [VARs] and losses. A non-loaded induction motor can spin freely and draw nearly zero current from the supply.
Connect some type of load to the shaft, and the motor begins to transfer energy to it, via the rotating shaft.
Now the power supply will reflect the load and deliver the power needed to do the work to the input of the motor - through total KVA input.
So once again, if there's nothing to run - no power is needed.

Scott SET

Are there a set amount of windings to create this situation, 1000? 100,000? If this is a complicated extensive engineering explanation, don't feel a need to elaborate, just curious if there is a simple answer to this. Do more windings create a more efficient x-fmr?

I've been told that the efficiency of a transformer can be used to calculate the amperage that is flowing through primary side, was that a correct statement?

Steve T,

There needs to be a "Happy Medium" for the total windings that allows enough windings to create a good magnetic flux in the core, while not too many to cause increased resistance. Since this mainly applies to the Primary, the secondary will be effected, since the winding ratio needs to reflect the desired output voltage[s].

This stuff can also apply to the second part of your question - transformer efficency.
The Primary winding[s] will have lossed power from all the combined losses of the core and the windings, but fortunately these do not exceed 5% in the worst cases.
Laminated cores have helped reduce major losses.
Silicon Steel used for the core laminations also helps.
Iron is bad, because it's "Reluctant" to lose it's built up magnetic field. Steel loses the magnetic field rather quickly when KVA is removed [through the zero line of the sine wave].

Primary to Secondary power transfer percentages is what the %Z [Impedance Percentage] will reflect. It's mostly used in the field as part of a Short Circuit Calculation.
I'll leave out the rest of those details now, but feel free to ask if you would like to know more.

Some transformers, such as Reactor coils [used in HID Ballasts] will work "against" the direction of flux in the core with the current that's flowing. This is how they limit the current that flows through the lamp.

Connecting Secondaries in Series Subtractive will cause the output voltage to be lower than the rated voltage. This is done by placing the coils out of phase polarity with each other and with the primary.

One simple field connection done to a typical Isolated Transformer, is to make it an Autotransformer for slight voltage correction.
Isolated Transformer means any Transformer that has separate / non-connected Primary and Secondary windings, such as a typical transformer, whereas an Autotransformer has a common connected input and output.
The familiar term is: Buck / Boost Transformer.

To do this, one side of the Primary is connected to one side of the Secondary. The connections are done with "Additive Polarity" - or in other words, the relative phase positions are "In Phase" or additive.
This connection makes the output voltage higher than the input. The total voltage at the input [primary] added to the rated voltage of the secondary winding[s] is the output voltage.

To "Buck" or reduce the voltage, the input KVA is brought in through the Secondary winding.

A Choke circuit can be made by connecting the transformer in the "Boost" method, except reversing the relative polarities so they are "out of Phase". This will have a floating voltage [according to the output], but will limit the flow of current - like a ballast.

Transformers and other types of Inductors are really interesting Animals, so I would suggest finding a few detailed books on them.

These same books will help de-mistify a common magnetic ballast, plus other transformers.

Scott SET