Yes.
"Ideally, V=IR=(I)(0)=0" is true but a real power supply is not ideal. A real power supply has a non-ideal property called internal resistance and as you said; "But the wire does have a small resistance so there has to be some power generated". If you were to short the terminals together (not recommended), the power supply would get hot, because the most of the power would be dissipated by the internal resistance of the power supply and additional power would heat the wire used to short the terminals.
Under IDEAL conditions (Ideal source, ideal resistance, etc.) the source would have no internal resistance, the short circuit would have have zero resistance causing zero voltage drop across it and infinite current to flow through it. The power dissipated in the load would be undefined.
In the REAL world, the source has a limiting resistance built in (even batteries have an internal resistance), the load would also have some resistance so the voltage across it would be small (but not zero) and the current through it would be high (but not infinite). All these quantities would cause a high amount of power to be dissipated in the load.
Right, until the current blows a fuse or destroys the power supply. This can happen in less than a second.
If your power supply has an adjustable current limiting capability and it can supply 30 amps, then you can do what you want. Otherwise you need to use the right size resistor.
Can a power supply generate power if i short the two terminal together? Ideally, V=IR=(I)(0)=0. And since P=VI, P=(0)(I)=0. But the wire does have a small resistance so there has to be some power generated right? (I is high and v is low?) Thanks.
