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enter image description here

Source of the figure.

Setting \$R_{2}\$ and \$L_{2}\$ to zero and assuming the primary circuit is connected to an AC source and the secondary circuit is left unloaded, then the current passing through the primary of the ideal transfomer XFMR1 is going to be zero even when the current passing through L1 isn't of zero value, right?

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    \$\begingroup\$ If it has finite permeability, what are the effective geometric parameters (Ae, le, lg) and in what way is that still an "ideal" transformer? You have a contradiction here. \$\endgroup\$
    – Tim Williams
    Commented Jul 9 at 12:49
  • \$\begingroup\$ Also is this not just a repost or rephrase of your previous question?: electronics.stackexchange.com/questions/718628/… which currently has no answers so you are welcome to edit it into a different question if you like. \$\endgroup\$
    – Tim Williams
    Commented Jul 9 at 12:50
  • \$\begingroup\$ In my previous question I was asking about the properties of the inductor in XFMRI. For example, does it have a reactance? \$\endgroup\$
    – Jack
    Commented Jul 9 at 12:53

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In this model, you are correct. Note that physically, R2, L1 and L2 are the primary. Breaking them out separately is a modeling convenience.

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  • \$\begingroup\$ By setting R2 and L2 to zero keeping only L1 means I am just taking into consideration the finite permeability of the iron core. That means the transfomer isn't an ideal transfomer in such a case, is it? \$\endgroup\$
    – Jack
    Commented Jul 9 at 11:55
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    \$\begingroup\$ @Jack Which transformer? Real physical transformers aren't ideal. You've drawn a representation of a model of a non-ideal transformer. That model embeds a symbol which could sensibly represent an ideal transformer. \$\endgroup\$
    – John Doty
    Commented Jul 9 at 15:06

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