7
\$\begingroup\$

I'm a newbie with electronic components and I never fully understood why transformers, while essentially being a short-circuit from a wire perspective, don't act like short circuits (i.e. they don't just blindly behave like a waterfall of electrons).

Why is that and how is it related to the "load" attached to the transformer itself? I'd prefer a 'layman's terms' explanation but I don't mind some math if necessary.

\$\endgroup\$
  • 6
    \$\begingroup\$ Because inductance! \$\endgroup\$ – Nick Johnson Jul 20 '15 at 15:17
  • 7
    \$\begingroup\$ Put DC on it, and watch the smoke of your long wire short circuit. \$\endgroup\$ – PlasmaHH Jul 20 '15 at 15:20
  • \$\begingroup\$ @PlasmaHH What? You need short the output/input and bring the input to nominal current to find characteristic of a transformer..... \$\endgroup\$ – MathieuL Jul 20 '15 at 15:30
  • 2
    \$\begingroup\$ @MathieuL: Did I ever say that I gave instructions on how to find characteristics of a transformer? All I gave was an instruction on how to see a transformer behave as a short. \$\endgroup\$ – PlasmaHH Jul 20 '15 at 15:31
  • 9
    \$\begingroup\$ You might wonder as well why capacitors are not open circuits... \$\endgroup\$ – Rol Jul 20 '15 at 15:51
13
\$\begingroup\$

As transformers are usually used with AC rather than with DC, there is what is known as inductance \$L\$, which is a property of a conductor to "resist" the changes in the current flowing in it due to the magnetic fields induced by that current (self-inductance). The magnetic field is "resisting" due to the fact that alternating magnetic field is in turn trying to induce current in the opposite direction. So when we speak of AC, it is an alternating current, i.e. constantly changing which will be resisted by such a conductor. The amount of magnetic field created by a conductor is relative to the density of the conductor windings, so a coil with many windings will create a stronger magnetic field, which in turn will resist more to the changes. In case of transformer, there is an additional coil "sharing" the magnetic field with the primary one, so the magnetic field trying to induce the current in this secondary coil as well. But when it is open, or connected to a load, it is "hard" to induce much current there, so it is "resisting" harder in the primary coil as well. This is pretty much of the intuitive understanding. If you want some math, you can easily find it.

\$\endgroup\$
0
\$\begingroup\$

Transformer don't act as short because they aren't perfect. In a nutshell:

First, the wire that are used cause Copper Loss, because of Joule Heating in wire.

Secondly, the material that are used in the metal struct that link both set of wire doesn't have an infinite permeability, therefore, some magnetic lines leak from the struct therefore affecting the transformer efficiency.

Finally, there is the mutual inductance effect that is cause by each bobine on the other one. Practically, this effect is neglect most of the time because the Mutual inductance got a massive value compare to the two other effect.

You can model these effects by using the Real model - Equivalent circuit.

\$\endgroup\$
  • 2
    \$\begingroup\$ The copper loss is not really any different from an other piece of wire which we would still call a short circuit. The key reason in the inductance and coupling with the secondary. \$\endgroup\$ – Tom Carpenter Jul 20 '15 at 15:33
  • \$\begingroup\$ @TomCarpenter Check my link, coupling with the secondary is modal by X_M coupling inductance and R_M is the Iron loss, these value are a lot higher than the value X_p/R_p (self/copper loss from primary) and X_s/R_s (self/copper loss from secondary) that are under 10 ohms most of the time. The effect of mutual can be neglected in most case. \$\endgroup\$ – MathieuL Jul 20 '15 at 15:39
-2
\$\begingroup\$

I think what he is asking is why when two different phases touch they short but when two phases are connected to a transformer together they do not. It is because the coils create resistance which like your load determines current; if the primary coils become shorted to themselves then the primary will short out.

\$\endgroup\$
  • 2
    \$\begingroup\$ The answer accepted six months ago is worth reading as it explains the theory correctly. (Your's doesn't.) Welcome to EE.SE. \$\endgroup\$ – Transistor Mar 2 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.