Ashok, your previous statement about "entire input power is consumed by the device" is correct, but it doesn't describe the situation completely. I agree with Leszek to an extent. But the situation is a little more complicated than that.
I guess one way to look at the problem is to merge the transformer into the power supply (conceptually). Then, you can just look at the Thevenin equivalent of the power supply + transformer combination. Better transformers will produce a lower series Thevenin impedance than poorer ones. But, that only indirectly addresses the question.
With open circuit conditions, the transformer will still dissipate energy. There are eddy current losses, field losses, etc. Since transformers are one of the most efficient (energy transforming) machines built by man, (large industrial transformers can achieve greater than 99% efficiency) you can expect such open circuit losses to be very low, but non-zero. The open circuit "load" will be reflected back to the power supply and it will only deliver a small amount of power to cover dissipative losses.
Now, since there is no "load", efficiency is mathematically zero. That is, eta = energy delivered to load / energy supplied. This is pretty unsatisfying, I think. Perhaps another way to look at an open circuit configuration is to view the transformer as an energy storage device. It's a high inductance coil, after all. So, perhaps looking at quality factor (Q) might be a little more informative. Q = energy stored / energy dissipated. If the transformer is well designed, we can expect high Q values.
With a short circuit condition, the situation is a bit different. In this case, there really will be induced current in the secondary and passing through the short circuit "load". Still, if we imagine a "perfect" zero impedance load, no power will be delivered to it (again). And, eta will again be zero. Wow, this is even more unsatisfying than before.
Clearly, the transformer will blow up if the source can supply enough energy. I mean, when one of those power transformers on a light pole blows ... wow, big flash! sparks ... molten copper ... very impressive! Of course, it's never a good idea to short circuit a power supply.
Q, for the sort circuit case can still be calculated, but the value is drastically different. We still have similar energy storage, even more than the O.C. case due to fields induced in the secondary coil by secondary current. But, energy dissipation is now dominated by resistive losses of the wires. These losses are normally much greater than eddy current losses. So even though energy storage is up, energy losses are much greater than the O.C. case. Thus, the effective Q of the short circuited transformer will be much lower than the O.C. case.
So, I guess there are several ways of looking at the problem, and some are more informative than others.
Let me a box having two terminals in which a watt meter is connected with proper voltmeter and ammeter connection with the required supply system in this case consumed power shows the watt meter reading which is equal to input power than what is the efficiency of the box.
During O.C test, we get to know about the Iron or constant losses in the core of the transformer. While during S.C test, we get to know about the Copper or variable losses in the windings of the transformer. In both the cases losses occur. Hence, during these tests efficiency of transformer is > 0 & < 100.
Generally, when Iron loss = Copper loss, efficiency is maximum.