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I was reading this answer and I would like to understand why if the secondary is short circuited the primary winding will also reflect the short circuit despite the fact that they are isolated. Would the electromagnetic flux through the core be increased with a short circuit on the secondary? If the secondary is shorted I guess it would be as if no turns were present. How would that draw more current from the supply at which the primary is connected?

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  • \$\begingroup\$ Not exactly your question, but a dual one: electronics.stackexchange.com/questions/180910/… \$\endgroup\$
    – Eugene Sh.
    Sep 25, 2018 at 18:56
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    \$\begingroup\$ Because of the short circuit, the secondary coil will produce a counter magnetic field which eliminates (in an ideal case completeley) the field of the primary coil. \$\endgroup\$
    – Oldfart
    Sep 25, 2018 at 19:41
  • \$\begingroup\$ @Oldfart that is fundamentally incorrect or, at best misleading. \$\endgroup\$
    – Andy aka
    Sep 25, 2018 at 20:01
  • \$\begingroup\$ @vmms do you understand faradays law of induction? If you don’t then you need to study it before you’ll understand any answer to your current question. \$\endgroup\$
    – Andy aka
    Sep 25, 2018 at 20:03
  • \$\begingroup\$ If the secondary is shorted I guess it would be as if no turns were present. You had a valid partial answer in your question. Suggest research to tie all the pieces together. \$\endgroup\$
    – user105652
    Sep 25, 2018 at 20:31

2 Answers 2

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https://en.wikipedia.org/wiki/Transformer

First a small back step.

When Vp is applied to the primary winding, a primary current starts to flow and this causes a magnetic flux in the core, refer to the 'right hand rule' in physics texts for determining direction. But this increasing flux causes a voltage to be induced in the primary and is in opposition to the applied Vp. This is the origin of self inductance and why the primary doesn't appear to be a short circuit or very low value resistance as determined by the wire.

When the short circuit is applied, the secondary current produces a magnetic flux that is counter to the primary magnetic flux. This behaviour is described by Lenz's law. But it was the increasing flux that limited the primary current and now the secondary is reducing the self inductance of the primary. Hence an increased primary current results. In the extreme case of a secondary short, the secondary current will completely counter any flux in the core. Hence the primary is now operating as if there is no core i.e. it's just wire. Now the primary current will increase to whatever value dictated by Vp and the source resistance (not shown in diagram).

The secondary open circuit case is the same as no secondary, assuming no complexities like stray capacitance to allow unexpected currents to flow.

In an ideal transformer the secondary current doesn't change flux in the core (I'll avoid the extreme short as there is no secondary resistance so that case is a bit odd). In real transformers, the flux will reduce somewhat at high load. Due to significant currents flowing, there will be voltage drops across the windings. This means some of the primary voltage will be dropped across this resistance and not contribute to establishing a magnetising current. This concept is best seen when studying a transformer model with a discrete resistances.

Inductors saturate when excessive current is passed through them. Transformers saturate when excessive primary voltage is applied and not excessive secondary current. The exception being the incorrectly named fly-back transformer which is really a coupled inductor and works differently to normal transformer.

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A transformer uses feedback between secondary to primary, via the core flux, and requires some LOW series impedance between the primary voltage source and the voltage across the internal primary N*dphi/dT. Call this voltage the feedback-voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

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