How does a shunt save an electrical device? Suppose there is circuit having an ideal DC voltage source (a high one) and a device having resistance R , therefore the current flowing in the circuit is $\frac {V}{R}$.It has been told that this amount of current will fry the device.Now my teacher said , "To protect the device , we connect a resistance of very low value parallel to the device , now most of the current in the circuit will pass through this resistance since current got an easier path to flow ,therefore the device is safe". I find this reasoning false.The voltage being constant , no matter how many parallel resistors I connect across it , the current will always be independently determined by  $I_i =\frac {V}{R_i}$ where 'i' is indicating branch.So no matter how low shunt resistance is the voltage across the device is same so current will be same , there is no current diversion here  , the shunt will draw a separate amount of current for itself.So is this use of shunt wrongly explained by my teacher ?
 A: To reduce the current through the device to be protected, the voltage across the device must be reduced.  Since an ideal DC voltage source has a fixed voltage across independent of the current through, the voltage across the device can only be reduced by adding an additional device in series forming a voltage divider.
It's true that adding a parallel resistor will form a current divider but it's also true that this doesn't affect the current through the protected device but, rather, the current through the DC voltage source.
Now, there are practical shunt protection circuits that are 'normally open'.  When the voltage across exceeds a threshold (due to some type of fault condition), these protection devices become (effectively) short circuits which should cause a fuse to blow or a breaker to trip.
A: Yes, your teacher is wrong where an ideal voltage source is concerned because such a source is capable of delivering an unlimited current to an unlimited number of circuit elements all at once. 
No, your teacher is right because ideal voltage sources do not exist. A battery has an internal resistance. A dynamo is driven by a mechanical power source of limited capacity. Whatever the mechanism, at some point increased current means decreased voltage. 
(Of course if the power source has a fuse then it is even simpler: once the current has increased past a certain point, the voltage goes to zero and stays there).
With a non-ideal voltage source, a protective shunt makes sense. In the “internal resistance of battery” case, it will form a potential divider with that internal resistance, taking the supply voltage to a very low level. In the “fused power supply” case, it will blow the fuse. 
