Would it be better to have the switches on the output side of the transformer ?
whats "iron losses"
I guess you forgot the link...
The magnetic field that creates current in the wire of the secondary also creates currents in the iron core of the tranformer. These currents warm the metal core and thus waste energy.
The old engineering saying relates to this "Eddy Currents steals the Joules
Is that the case with one of these though.
God forbid, if the resident transformer expert has his wires crossed
Whilst it is not primarily a wire-wound transformer (although it probably contains a small one of those - albeit probably ferrite-, rather than iron-, cored), I very much doubt that the device draws zero power when there is no load - so it will have 'losses' (wasted power), even if not 'iron' ones.Is that the case with one of these though.
Is that the case with one of these though.
Yes, but, at least in theory, 'ferrite losses' ought to be extremely small, since one of the main features of ferrite cores is that they are essentially non-conductive. However, as I said, even if there were no losses associated with wire-wound components, I would not expect one of those boxes of electronics to consume zero power when unloaded, so there almost certainly would be some 'losses'.Assuming it has a small high frequency magnetic transformer then yes. But it might be "ferrite losses" instead of iron losses.
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