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Above is a transformer with its primary and secondary windings. I just wrote my understanding about a transformer very shorty before my question:

Neglecting losses, we can write the voltage and power unity equations as:

  • Vs = (Ns/Np) * Vp
  • Vp * Ip = Vs * Is

It seems like; as long as the Vp and (Ns/Np) ratio are the same, whatever the load R is Vs will be the same. Only the current drawn will change.

And if the above argument is true the power dissipated in secondary part is:

Ps = (Vs^2)/R

And if R goes to zero or should I say the secondary winding is shorted, Ps goes to infinity which means this secondary winding would burn.

I have the following questions:

1-) Since there is power unity i.e. Pp = Ps; would that mean if the secondary winding is shorted, would the primary winding burn as well?(I'm asking because the interaction between the windings is electromagnetic which could be a different phenomenon)

2-) If the conclusion is primary winding would burn as well, is that enough to add a fuse to only before the primary winding but not secondary winding?

  • \$\begingroup\$ Ideal transformers don't burn. It will deliver however much power the load requires for as long as the load requires it. Real transformers have current and or power limits specified. Some are even designed so that they are inherently power limited (no matter what type of load is applied to the secondary). \$\endgroup\$ – mkeith May 28 '16 at 22:02
  • \$\begingroup\$ I see, what you mean; maybe I should have asked that windings have resistance as well \$\endgroup\$ – user16307 May 28 '16 at 22:06
  • \$\begingroup\$ I changed the question, nor the transformer is not ideal but magnetic losses are winding resistance losses are neglected. I mean I didnt write equations just simple equations. \$\endgroup\$ – user16307 May 28 '16 at 22:08

Under real-world conditions no transformer is 100% efficient at converting power. If the secondary is shorted then the primary 'sees' more watts being dissipated than the secondary. In fact the primary always 'sees' more power dissipated regardless of the load.

Many fused transformers have fuses on the primary only, and they are usually the slow-blow type because of inrush currents when the transformer is turned on. Most 50HZ to 60HZ transformers are only 40% to 60% efficient at converting power, so for a given known maximum continuous load the transformer is likely over-rated by 50%.

Some transformers have short-circuit protection, usually those called 'wall-packs'. So called power transformers and industrial transformers have a fused primary.

Those with extreme high power/high voltage may have secondary fuses as well. I have seen pole mounted transformers explode from a lightning strike maybe 50 yards from me, only to see a huge fuse explode about 100 yards away on another pole. For major power distribution, it pays to fuse both sides of a transformer.

  • 3
    \$\begingroup\$ Larger 50/60 Hz Power transformers have efficiencies well into the 90s at full load .Your 40 to 60% would only be true for very small transformers or very light load. \$\endgroup\$ – Autistic May 28 '16 at 22:54
  • \$\begingroup\$ @Autistic. Granted that is true, especially those with ferrite cores. SMPS transformers can approach 95% efficiency or better if synchronous rectifies are used. But the OP was focused on the short-circuit penalties of standard transformers. I saw no need for esoteric details, though another OP might ask for them one fine day. \$\endgroup\$ – user105652 May 29 '16 at 2:13

In any transformer, (or any other load connected to the voltage source) always place the fuse before the unit.

In step down transformers, shorting the secondary will stress the primary winding more than the stress taken by the secondary. The reason for this is since both windings are in tight magnetic coupling, a step current increment in the secondary will have a step increment of current in the primary following the inverse of the Np/Ns ratio.

But since in the step down transformer the primary is made from longer and thinner wire, its resistance is higher and dissipate higher power and heats up much quickly. Thus in case your transformer is the stepdown type, protect the primary is sufficient to prevent your transformer from catching fire.

If your transformer is step up type, its more appropriate to have fuse on both sides of the transformer. Derating the fuse according to the winding current limits will protect the transformer. Fuse on the secondary normally is used to protect the load connected to the transformer, not to protect the transformer itself.

  • \$\begingroup\$ When the primary and secondary windings have the same copper volume which is normal from a materials optimisation viewpoint the copper losses will be the same in the primary and the secondary .This is true when magnetisation current is small which is generally the case .So both the primary and the secondary heat up . \$\endgroup\$ – Autistic May 28 '16 at 23:00
  • \$\begingroup\$ True, but insulation breakdown due to temperature rise is more common in the primary side of the stepdown transformer because the higher voltage operation. Thus the prime side is more failure prone. \$\endgroup\$ – soosai steven May 29 '16 at 0:46

The maximum amount of current a transformer can handle is dependent on several things, the resistance of the windings, the reactance of the windings (current through an inductor cannot change instantaneously and eventually you reach a point where the current just can't ramp up fast enough to supply the load).

The current is also dependent on the quality of the coupling between the coils, see, there is a magnetic equivalent of resistance (called reluctance), it acts like a series inductor which limits the current (old fashioned fluoro ballasts are just a series inductor). Old arc welders would control the peak output current but changing the coupling between the primary and secondary windings (they'd be moved away from each other, the greater separation increases the reluctance of the magnetic circuit).

  • \$\begingroup\$ Because a transformer core will tend to saturate more on no-load than when loaded your statements about core saturation seem inappropriate considering the OP is asking about heavily loaded situations. \$\endgroup\$ – Andy aka May 29 '16 at 9:40
  • \$\begingroup\$ I was under the impression that saturation under heavy load is exactly how some arc welders limit current (which wikipedia lists as 'saturable-core transformers'), as the construction of an arc welding transformer is not that dissimilar to most ordinary transformers, I drew the conclusion that excessive current = excessive core flux = potential core saturation. I'm not sure how a transformer is more likely to saturate under light loads, I thought you need lots of current to generate strong magnetic fields and you need strong fields to saturate something or have I missed something? \$\endgroup\$ – Sam May 29 '16 at 23:17
  • \$\begingroup\$ Yes you have missed something. Ampere-turns caused by secondary load currents are totally cancelled by ampere-turns (due to secondary load currents) in the primary. If you still have problems understanding this look at my profile and try and find answers given by me on this. \$\endgroup\$ – Andy aka May 30 '16 at 1:08
  • \$\begingroup\$ I henceforth retract that part of my answer \$\endgroup\$ – Sam May 31 '16 at 11:20

Using any ideal transformer or one with low leakage inductance and resistance, you will need a fuse or your transformer will burn with shorted secondary.

Having a transformer which is inherently protected against short circuit (very few are), you have enough leakage inductance to limit the current to a safe level which the transformer can withstand continuously.


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