If the wire of the secondary circut is thicker than that of the primary in a transformer, what type of transformer is this and why? There is this question in my physics book, and two teachers (a private teacher of a friend of mine and the school teacher) say that it's a step down transformer, while two other teachers say that it's niether of them, since a transformer's type is only determined by the number of turns.
I dont really know which one is correct and why, so if someone could explain id appreciate it.
 A: In a transformer the thicker wire is usually the one that carries a larger current. According to the simple formula relating the voltages and currents in a transformer $V_1 I_1=V_2 I_2$, you have the larger current at the lower voltage terminal. Thus if the primary voltage of this transformer is the mains voltage, then this is a step down transformer to a lower voltage. The ratio of the voltages of a transformer is given by the ratio of the number of turns of the primary and secondary coil. 
A: Practically speaking yes, this will almost certainly be a step-down transformer, but I would agree with those other teachers: it can't really be concluded from the wire thickness. It's perfectly possible to build a transformer in which the secondary coil has more windings, but nevertheless use thicker wire.
As John Rennie explained, this doesn't normally make sense, because then there would be more AC current in the thinner primary wire than in the secondary, though that could cope with more. However, that's not necessarily the only current that's relevant: there are many applications where the voltage stepping is not even the reason to use a transformer but to decouple voltages and/or currents. In particular, you could have a high DC current or low-frequency AC flowing on the secondary side, and use the transformer to modulate a much weaker, high-frequency control signal on top of it. In this case, you will need a thicker wire on the secondary side regardless of the winding ratio.
A: The thickness of the wire determines the maximum current the wire can carry without overheating. Thicker wire means a greater current.
With a transformer the power coming into the primary, $P_p = V_pI_p$, is the same as the power coming out of the secondary, $P_s = V_sI_s$,  (less a few resistive losses) and this means $VI$ is constant.


*

*For a step down transformer $V_p \gt V_s$ so $I_s \gt I_p$ - the current in the secondary is higher than the current in the primary so the secondary needs to be wound with thicker wire.

*For a step up transformer $V_s \gt V_p$ so $I_p \gt I_s$ - the current in the primary is higher than the current in the secondary so the primary needs to be wound with thicker wire.
A: As several others have correctly responded, the ratio of primary to secondary turns determines the ratio, NOT the cross-section (size) of the wire. As a side note, (aside from core losses) the POWER in the primary and secondary portions of a transformer will be equal. For example, in a 2-to-1 stepdown transformer: : Secondary 12V, 10A = Primary, 120V, 1A = 120W. On both the primary AND secondary. (Requiring MUCH 'heavier' wire on the secondary...)
So; fair to infer/assume that the transformer of interest is a step-down.
(Assuming manufacturer built it w/ suitable, lowest cost materials.))
A: A heavier secondary gives us a very strong clue that a higher-voltage, lower-current primary is coupled with a lower-voltage, higher-current secondary.
It's a clue, not a definite though.  We can imagine a situation where only a low current is required from the secondary of a step-down transformer, hence heavy wire would be unnecessary and wasteful.   It might be convenient to use the same light gauge wire for both.
A: In UPS the wires from transformer connected to the 12V battery are thicker(E72332 AWM 1015  12AWG 600V.) Normally it is a step down transformer to charge 12V battery. In case of power failure   it acts as a step up transformer to power the system. The wires are thinner (LL34978 18AWG 600V). Any comment. Thanks for your help.
