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In a short-circuit test where the secondary winding is short circuited, how come the primary voltage isn't zero (in real life) when it should ideally be zero?

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The voltage on the primary side shouldn't ideally be zero as you claim.

If you short the secondary winding of a transformer then all the power you put into the primary winding will be dissipated in the internal resistance of the secondary winding.

Lets take the following example: We have a 1:1 transformer with an internal resistance in the secondary winding of 1 Ohm. If we apply 1V AC to the primary winding it will cause 1V to be applied across the 1 Ohm resistance of the secondary winding (because it is shorted) and hence 1A will run in the secondary winding. Since the transformer is 1:1 the same current (1A) will also run in the primary winding.

However as Rohat correctly states in his reply: If the transformer was ideal (ie. no internal resistances etc.) then the current in the primary winding would TEND TOWARDS infinity. And it follows that if the output impedance of your source is not zero ohm, then the voltage will drop to zero, but not because of your transformer, your transformer just wants to draw infinite current, it is your source which is not able to deliver it.

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In an ideal transformer, input power equals to output power: \$v_{pri} \cdot i_{pri} = v_{sec} \cdot i_{sec}\$

If you short the secondary then the secondary voltage will be zero, thus the secondary current will be infinite: \$i_{sec} = v_{pri} \cdot i_{pri} / 0 = \infty \$.

Since the primary is driven from a source with zero (or nearly-zero) output impedance, the primary voltage cannot be zero. Instead, the primary current will reach infinity.

When it comes to practice, the non-zero impedances show themselves: the source's output impedance, the windings' DC resistances, finite magnetizing inductances, etc. They limit the output current according to Ohm's law.

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