Most household appliances operate with a step-down transformer if I'm correct where they decrease the voltage. Consequently, the current will be relatively high in the secondary winding, but this is true for an unloaded, shorted transformer. But would the currents also be high in the appliance's circuitry as well? Having a huge difference between the number of windings in the primary and the secondary would mean a high current in the load or not?

Background: Welders work on the principle of step-down transforming because it will result in a current high enough to melt metal (the metal being part of the circuit). Household appliances also use step-down transformers but of course they don't melt during operation.

  • \$\begingroup\$ Your initial assumptions are incorrect and as a result, the rest of your question makes little sense. \$\endgroup\$ – brhans Apr 20 '17 at 13:54
  • \$\begingroup\$ That would be damaging, wouldn't it? Not if you design everything such that it can handle it, duh ! The rest of what you assume is indeed also very wrong, it is not how things work. Maybe you should educate yourself by reading here: hardwaresecrets.com/anatomy-of-switching-power-supplies \$\endgroup\$ – Bimpelrekkie Apr 20 '17 at 13:55
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    \$\begingroup\$ Why would the current in the second winding be high? It is higher than in the primary, but not higher than the stuff on the secondary winding needs. Some designs require a minimal load, otherwise the output voltage would rise above the specification. \$\endgroup\$ – Arsenal Apr 20 '17 at 13:57
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    \$\begingroup\$ Both. Phone chargers are built using switch-mode powers supplies - not simple transformers, and the current in the secondary of a transformer is dependent on the voltage it produces and the circuit it's delivering power to. \$\endgroup\$ – brhans Apr 20 '17 at 13:59
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    \$\begingroup\$ Your understanding of a resistor is also flawed. You get just as much current out of one end as you put into the other so it doesn't reduce the current, just the voltage. \$\endgroup\$ – Finbarr Apr 20 '17 at 15:09

A transformer with an open circuit secondary produces only a voltage i.e. No secondary current flows until a load is connected. That is what happens in a transformer. Yes, there may be a small primary current with no secondary load but that doesn't manifest as a secondary current.

  • \$\begingroup\$ I think I know what confused me. The inversely proportional current is valid for a short circuited secondary coil. So V1/V2=I2/I1 is valid with no load but the current running in the secondary coil (its ends connected). As soon as we put a load on the secondary, the traditional rules will apply: we have a voltage (on the secondary) and current will appear on the load according to all the properties of the load circuitry. Is this kind of correct? \$\endgroup\$ – stevie Apr 20 '17 at 19:29
  • \$\begingroup\$ @stevie yes that's correct. \$\endgroup\$ – Andy aka Apr 20 '17 at 20:45

The main behavior inside a transformer is a FEEDBACK system. The Flux in the core, that flux common to both primary and secondary, is regulated.

The regulation (requiring negative feedback, where forward and reverse are subtracted) occurs in the core.

With no secondary load, the primary current is very small. As the secondary begins to demand current, the increased secondary current generates a flux that reduces the core flux.

That reduced core flux allows the primary to provide MORE flux by increasing the primary current.

Result? The secondary current is only high if the secondary load demands high current.

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    \$\begingroup\$ So you mean the secondary load and thus current is totally on-demand? \$\endgroup\$ – stevie Apr 20 '17 at 15:26
  • \$\begingroup\$ @stevie: In general, a load will draw whatever current it requires. The rated current on a power source is the maximum the source is designed to supply - the source doesn't force that current through the load. \$\endgroup\$ – Peter Bennett Apr 20 '17 at 18:11
  • \$\begingroup\$ Your answer appears confusing. With pretty much all normal transformers, the core magnetic flux is maximum under no load conditions and falls a little bit as the secondary is loaded. The ampere turns of the secondary due to the load are cancelled by the ampere turns increase on the primary due to that secondary load. Core flux (magnetisation flux) only reduces due to load currents causing a volt drop across the primary leakage components. \$\endgroup\$ – Andy aka Apr 20 '17 at 19:09

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