Final answer:
The maximum secondary current of 1600 A in an ideal transformer is the theoretical limit assuming 100% efficiency and no energy losses. The actual current will depend on the load resistance but cannot exceed this maximum without violating conservation of energy.
Step-by-step explanation:
You are correct that the 1600 A on the secondary side of an ideal transformer represents a theoretical maximum current, based on the assumption that the transformer is 100% efficient and that there are no energy losses. If you connect a load with very low resistance, like a nail, to the secondary winding, the current will approach this maximum value. However, with a higher resistance load, the current will be lower. It is true that the current cannot exceed 1600 A, because that would violate the principle of conservation of energy, assuming the transformer has no losses and is operating under ideal conditions.
The relationship between primary and secondary coils in a transformer is given by the equations:
- Vp/Vs = Np/Ns (voltage ratio equals turns ratio)
- Ip/Is = Ns/Np (inverse current ratio equals turns ratio)
- Pp = Ps (primary power equals secondary power, assuming 100% efficiency)
So indeed, the maximum secondary current is determined by the transformer's turns ratio and the primary current, as long as the transformer operates within its efficiency limits.