Why the curren on the primary of a transformer depends on the secondary's load ?

[opablo]

I'm suspecting this may be a newbie question but I -have- to ask...

Can somebody explain in simple terms what are the electromagnetic effects that makes a transformer under small/no load to consume less current on the input ?

[IanB]

Well, why does a big heavy car consume more fuel than a small light car? Why does the same car consume more fuel if you load it down with many people and lots of luggage?

If you put a pan full of water on the stove top, why does it take more heat to boil it faster?

How can you get something for nothing?

[IanB]

I see you have asked the question the other way round. Why does the transformer not consume much current when it is not loaded?

It's because the unloaded transformer acts like an inductor and resists the flow of AC current.

When you put a load on the secondary it reduces the effective inductance of the primary and causes more current to flow.

[opablo]

wow... so under load the inductive reactance on the primary gets modified (lowered) because the inductance on the primary (the Henries folks) get modified (lowered).... so... A coil changes it's inductance ? like what happens when you insert and remove a ferrite inside a coil ?

[c4757p]

The effective inductance does get reduced. If you measure the inductance of the primary of a transformer while shorting the secondary, you would ideally measure zero. (In reality you will get a small number, the leakage inductance - the part of the inductance that is not coupled between the windings.) Inductance basically tells you how much energy the coil stores, so if you're losing some to other windings, it's not storing as much.

IanB, I think your explanation addresses why it's mathematically necessary for it to happen, but I think this was already understood (conservation of energy is basic). I think the question was more about the exact mechanism by which it happens. I'll take a stab at it, but I'm not the best at explaining things like this, so feel free to take over 

As far as electromagnetic effects, I think a simple way to explain it is this: As current through a coil increases, magnetic field strength increases, and if a field increases, current increases (induction). If two coils are coupled so that they share a magnetic field, then when the current through one coil increases, the field shared by both increases, inducing a current in the other, and pulling some of the energy out of the field. Because of this lost energy, it takes more current to maintain the same field strength. The more heavily the secondary is loaded, the more current it will take.

Obviously that deals with change of current, but sinusoidal currents are special in that the rate of change is itself a sinusoid.

(Google Chrome's spell check thinks "sinusoid" is not a word but "sinusoidal" is perfectly cromulent...  )

[IanB]

The only precise way to understand what is going on is to look at the mathematical equations that describe the operation of the  transformer in terms of the electric currents, the magnetic fields and the properties of the core. This is way outside my field so I can't really do that.

Intuitively we might consider changing magnetic fields and back-EMF. With an inductor (or the primary of a transformer with the secondary open circuit) a changing current in the primary induces a magnetic field in the core that generates an opposing voltage in the winding (a back-EMF) that tries to stop the current from changing. It generates a voltage on the secondary as well, but because the secondary is open circuit no current can flow there.

Now suppose the secondary is connected to a load (or even shorted out). Now when the current changes in the primary, the induced voltage can cause current to flow in the secondary. The current flowing in the secondary will generate its own magnetic field, and I think this magnetic field will be opposite in polarity to the magnetic field of the primary. The two magnetic fields will tend to cancel out, and make it as if the core wasn't there. It will be as if the primary winding was wound around something non-magnetic, which will of course reduce the size of the back-EMF and therefore reduce the resistance to current flow.

Any EE's out there feel free to correct me...

[amspire]

I'm suspecting this may be a newbie question but I -have- to ask...

Can somebody explain in simple terms what are the electromagnetic effects that makes a transformer under small/no load to consume less current on the input ?

One simple way to think of it is this.

A transformer usually has low DC resistance, so if you applied 110V DC, you would end up with so much current flowing, something would melt or explode.

So the thing that controls the AC current in a transformer is the fact that the back emf generated in the coils almost exactly matches the applied voltage.  This has to be the case regardless of current flowing.

Voltage is proportional to the rate of change of the magnetic field in the transformer, so this means that with a fixed AC input to a transformer, the amount of sinusoidal magnetic field in the core also has to be nearly constant regardless of load. This is important to realize - going from no load on the secondary to full load, the size of the sinusoidal magnetic field has to remain constant in an ideal transformer. If you are applying 110V AC, then the amount of field that always has to be in the magnetic core is the amount that generates 110V AC in the primary to oppose the applied 110V AC. Don't worry about the imperfections of a real transformer.

Just to restate this, if your multimeter had a range that could measure the size of the alternating magnetic field in a transformer, then with a constant AC voltage applied, the measurement on the field strength in the transformer would be constant regardless of the transformer load. Regardless of the amount of current leaving the secondary.

Constant AC voltage in -> Constant alternating magnetic field strength in transformer core. No exceptions.

When you start drawing current from a transformer secondary, then there is current now flowing in the secondary but it is flowing out of the secondary, instead of into the transformer as is happening on the primary. This current in the secondary generates a magnetic field as well but in the opposite direction to the field generated by the primary so it actually reduces the magnetic field in the core.

But if you remember above, I said that the size of the sinusoidal magnetic field has to remain constant, so the primary current has to increase until the field is back up to its required level. With a 1:1 transformer, the current in the primary has to increase exactly by the amount of current drawn from the secondary to cancel out the field reduction caused by the secondary. 

Hope that makes sense.  Once you understand this basic idea, then it is not that hard to start to understand the non-ideal aspects of a transformer like winding resistance, leakage inductance and the reduction of core permeability as the field increases.

Richard.

[Kremmen]

I'm suspecting this may be a newbie question but I -have- to ask...

Can somebody explain in simple terms what are the electromagnetic effects that makes a transformer under small/no load to consume less current on the input ?

In simple terms - well... This is one of the questions where a proper answer is inordinately long compared to the length of the question. Nevertheless, the electromagnetic effects in play here:

First we need the concept of MagnetoMotor Force or MMF for short.
MMF is defined as the intensity of the magnetic field penetrating a closed path. Such a closed path being e.g. a transformer winding. Properly this is formulated in Ampere's law, taking the path integral around the path thus arriving at the intensity. The end result of that integration is what interests us now, and the result is simply:

MMF = Ni, where

N = number of turns in the coil
i = current in the coil conductor

So, instead of being a "true" force in the mechanical sense, the unit of MMF is ampere-turns. This is the "force" driving magnetic induction.

Next, the magnetic flux density B can be calculated. Again, we want to integrate the magnetic field intensity H over the surface area penetrated by that field, multiplied by the magnetic permeability of the medium. The density of the magnetic flux is an important quantity in determining the requirements for the magnetic core material.

Now finally, the voltage resulting from the MMF in a conductor (coil) can be determined from Faraday's law stating that the voltage over a closed path is the product of the flux density over the area enclosed by said path, times the rate of change of the flux density. The resulting voltage in the closed path is known as the ElectroMotor Force or EMF.

Further, Lenz's law talks about the directions of the quantities such that a time-varying flux induces currents in the conductor opposing the original flux. This is the mechanism causing inductive reactance in a conductor.

On to transformers.
Ampere's law states that the sum of MMFs over a closed loop is 0. Thus in an ideal transformer where the primary and secondary share the same magnetic loop, it follows that MMF_pri + MMF_sec = 0. Recalling the unit of MMF we can also write N₁i₁ + N₂i₂ = 0. From Faraday's law we can rewrite the previous in the form:

U₁ = N₁ * dPHI/dt,  U₂ = N₂ * dPHI/dt where

U₁ = primary EMF
U₂ = secondary EMF
PHI = magnetic flux linking the coils

Eliminating common terms, we can write simply U₁/N₁ = U₂/N₂ and from there

U₁/U₂ = N₁/N₂; I₁/I₂ = N₂/N₁

So, the answer to your specific question was embedded in the middle, in Lenz's law that introduces the opposing magnetic induction. The magnetic flux in the transformer core couples the coils together. Faraday's law states the resulting Electromotor forces (coil voltages), and Ampere's law the Magnetomotor forces (magnetic flux) In case there is a load connected to a secondary coil, the coil will be a consumer of magnetic flux energy. Balancing the equations results in increase of primary current until MMFs over the core loop again cancel out. When the load is disconnected, Lenz's law balances the primary MMF with little primary current.

That's it in a nutshell.

Edit: subscripts for readability.