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As known the transformer's equivalent circuit can be developed as below:

enter image description here

Assume the \$V_{1}\$ is the voltage applied to primary side, and \$V_{2}\$ is the voltage developed at the secondary load.

Now, let the primary side open circuit, and apply a voltage to the secondary. My colleague argued that, because the primary side current is zero, the leakage reactance at the secondary will not exist, that is, the \$X_{i2}\$ will be zero! But I think it will be still there, because there is always some flux that does not flow through the transformer core.

Which one of the above points is right?

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Leakage inductance always exists. It is present on the primary winding and the secondary winding. It represents the small percentage of inductance that doesn't magnetically couple to the other coil.

Think about a transformer that has 1:1 turns ratio - which is primary and which is secondary. Of course it doesn't matter so things can't magically be different on one winding compared to the other.

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  • \$\begingroup\$ @diverger are you happy with this answer or do you need clarification? \$\endgroup\$ – Andy aka Jun 23 '17 at 14:50

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