Talk:Transformer

Referring to Ideal transformer equations, it is at best misleading to show that 'and for an isolation transformer a = 1'; saying so is true only as far as it goes. It is like a stopped clock that is right twice a day. An isolation transformer is a two-winding transformer that often has a turns ratio a = 1 and is not an autotransformer. But an isolation transformer can also be step-up type or step-down type. Cblambert (talk) 20:08, 29 January 2022 (UTC)

I suggest we specify that equatations are time-variant and are derived for the AC. AXONOV (talk) ⚑ 20:37, 23 July 2022 (UTC)

In the "Laminated steel core" section of the Wikipedia Transformer page the photograph reads "Laminated core transformer showing edge of laminations at top of photo". The picture is of a Shell type transformer. This is a confusion point as some will take it that it is a core type when it is not. I suggest changing either the photo out or changing the description to either. A shell core type transformer or in the description to "Laminated Shell transformer showing edge of laminations at top of photo". 69.201.21.27 (talk) 16:34, 3 October 2022 (UTC)

Core has a dual meaning. It is type of transformer construction. The magnetic material is also called the core whether the transformer is shell type or core type. Constant314 (talk) 19:58, 3 October 2022 (UTC)

I think that the discussion of the ideal transformer is a bit confusing. The point is that in the explanation given, it's not just the transformer by itself that is ideal, but also the impedance of the load shown on the far right.

For example, if we took $Z_L$ to infinity - an open circuit - I_S would obviously be zero, and i_p n_p − i_s n_s would not be zero. That's not the "fault" of the transformer.

I think that most people, reading a discussion of an "ideal transformer", would assume that it means a transformer with a lossless core and resistanceless wire, etc, so that it doesn't itself generate heat or create electromagnetic waves. As discussed it is a different definition and could be called out. Mgolden (talk) 15:11, 9 March 2026 (UTC)

If we took $Z_L$ to infinity - an open circuit - I_S would obviously be zero, and i_p n_p − i_s n_s would be zero, implying i_p=0. An ideal transformer is not only linear and lossless but also all the parasitic capacitances are zero, the leakage inductance is zero, and the magnetizing inductance is infinite. Constant314 (talk) 16:03, 9 March 2026 (UTC)

Why would i_p be zero? There is a current flowing in the primary, and it's 90 degrees behind the voltage. It's just a plain inductor. Mgolden (talk) 17:40, 9 March 2026 (UTC)

I see - it's because mu for the core is infinite. It might be helpful to add a sentence to explain the implications of this. Mgolden (talk) 18:51, 9 March 2026 (UTC)