Hey guys,

This has boggled my mind for a bit: How come a plugged in transformer does not consume power? Here's a link to a circuit diagram of a phone charger:
http://circuitdigest.com/sites/default/files/circuitdiagram/Cell-Phone-Charger-Circuit.gif
Obviously as its plugged in, the primary coil circuit of the transformer is completed with the wall which means ac power is flowing through it regardless of whats going on with the secondary coil. What am I missing here? Is it consuming power if it is plugged in and not used? Any clarification would be great!

 

Of course it consumes power.
The primary or other SMPS device is presenting a load, albeit small to the service power supply.
Max.

 

So then how does a plugged in phone charger that is not connected to a phone not consume power? I mean plugging it in to the wall is basically shorting the circuit. Is it not?

 

So then how does a plugged in phone charger that is not connected to a phone not consume power? I mean plugging it in to the wall is basically shorting the circuit. Is it not?

It most definitely is not.
A transformer with no load is not a short, instead it presents a high impedance (high inductance) to the line.
Why would you think it's a short?
For an ideal transformer the input power equals the output power so, no output power equals no input power.
A real transformer has some small loses in maintaining the magnetizing inductance so it will draw a small amount of power (typically a few percent of its rated power).
Suggest you read up on how transformers work.

 

They do consume power. I recall a study, that concluded saving a ridiculous amount of power by unplugging wall-warts.

 

How come a plugged in transformer does not consume power?

an ideal transformer is essentially an impedance transformer: it "maps" the secondary windings impedance onto's the primary side. An unloaded transformer has an infinite impedance on its seconday winding. Thus infinite impedance onto the primary winding -> no power consumption.

In reality, a transformer isn't perfect so it does consume a little bit of power, even when unloaded.

 

It most definitely is not.
A transformer with no load is not a short, instead it presents a high impedance (high inductance) to the line.
Why would you think it's a short?
For an ideal transformer the input power equals the output power so, no output power equals no input power.
A real transformer has some small loses in maintaining the magnetizing inductance so it will draw a small amount of power (typically a few percent of its rated power).
Suggest you read up on how transformers work.

The reason I think it's a short is because one end of the primary coil is in contact with one terminal of the outlet and the other end of the primary coil is in contact with the other terminal of the outlet. I understand that there is an impedence but is that really enough to prevent the power consumption of a short? Also this then means that a primary coil with low impedence will act as more of a short. Is this not true?

 

How about a tiny bit of deductive reasoning. Transformers are plugged into the power grid everywhere, the power grid has not been shorted out because of these transformers. So what would make you think an unloaded transformer is a short?

 

Actually, for a very short period of time (microseconds) the transformer is just a piece of copper wire.
Once current flow starts a magnetic field is established in the iron core.
This creates a "resistance" to current flow. This kind of resistance is called impedance.
It is the magnetic field that prevents the short circuit you are envisioning

 

Hey guys,

This has boggled my mind for a bit: How come a plugged in transformer does not consume power? Here's a link to a circuit diagram of a phone charger:
http://circuitdigest.com/sites/default/files/circuitdiagram/Cell-Phone-Charger-Circuit.gif
Obviously as its plugged in, the primary coil circuit of the transformer is completed with the wall which means ac power is flowing through it regardless of whats going on with the secondary coil. What am I missing here? Is it consuming power if it is plugged in and not used? Any clarification would be great!

It does consume some power, just no where near as much as when in use. The key is the characteristic impedance of an inductor. Impedance changes with the density of the magnetic field which depends on the current. As it starts drawing current impedance decreases so primary current increases. (in 100 words or less)

 

It is not a short for the same reason an inductor across and AC voltage is not a short. The current into the coil is inversely proportional to the frequency of the AC signal and the inductance, 1/(2*pi*f*L). Since a transformer has a very high magnetization inductance the current through the primary coil, with the secondary coil open, is quite small. And most of the energy that goes into the primary is returned to the mains since an inductor can not dissipate energy, just momentarily store the energy and return it to whence it came. There will be small IR losses due to resistance in the wire, magnetization of the core and eddy currents, so a few watts will be lost as heat.

 

Thank you all for your replies! This makes a lot more sense now.

 

To the OP please read this:
http://www.allaboutcircuits.com/tex...chpt-9/mutual-inductance-and-basic-operation/

Resistive load on secondary has voltage and current in-phase.

At first, one might expect this secondary coil current to cause additional magnetic flux in the core. In fact, it does not. If more flux were induced in the core, it would cause more voltage to be induced voltage in the primary coil (remember that e = dΦ/dt). This cannot happen, because the primary coil’s induced voltage must remain at the same magnitude and phase in order to balance with the applied voltage, in accordance with Kirchhoff’s voltage law. Consequently, the magnetic flux in the core cannot be affected by secondary coil current. However, whatdoes change is the amount of mmf in the magnetic circuit.

Magnetomotive force is produced any time electrons move through a wire. Usually, this mmf is accompanied by magnetic flux, in accordance with the mmf=ΦR “magnetic Ohm’s Law” equation. In this case, though, additional flux is not permitted, so the only way the secondary coil’s mmf may exist is if a counteracting mmf is generated by the primary coil, of equal magnitude and opposite phase. Indeed, this is what happens, an alternating current forming in the primary coil—180° out of phase with the secondary coil’s current—to generate this counteracting mmf and prevent additional core flux. Polarity marks and current direction arrows have been added to the illustration to clarify phase relations: (Figure below)

Flux remains constant with application of a load. However, a counteracting mmf is produced by the loaded secondary.

If you find this process a bit confusing, do not worry. Transformer dynamics is a complex subject. What is important to understand is this: when an AC voltage is applied to the primary coil, it creates a magnetic flux in the core, which induces AC voltage in the secondary coil in-phase with the source voltage. Any current drawn through the secondary coil to power a load induces a corresponding current in the primary coil, drawing current from the source.

Notice how the primary coil is behaving as a load with respect to the AC voltage source, and how the secondary coil is behaving as a source with respect to the resistor. Rather than energy merely being alternately absorbed and returned the primary coil circuit, energy is now being coupled to the secondary coil where it is delivered to a dissipative (energy-consuming) load. As far as the source “knows,” its directly powering the resistor. Of course, there is also an additional primary coil current lagging the applied voltage by 90o, just enough to magnetize the core to create the necessary voltage for balancing against the source (the exciting current).

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