In order to 'get' the way a current transformer is dangerous, you need to think about the concept of a _current_ source.
We are all familiar with _voltaqe_ sources. An idea voltage source is some apparatus which maintains two terminals at constant voltage, no matter what the circumstances. If you connect a load to the voltage source, current will flow through the load. As the load impedance changes, the output current adjusts up and down to maintain a constant voltage drop across the load. As the load impedance is decreased, more current flows to the load, and more VA is delivered to the load. If the load happens to be resistive, than as you drop the load resistance, the power delivered increases. The result is that if you want to somehow 'turn off' the output of an ideal voltage source, you want to make the load impedance infinite: you want to open the circuit. Fuses work well for voltage sources. Note: _real_ voltage sources are never ideal, and have their own 'internal impedance', but the distribution transformers that we normally deal with are rather good approximations to ideal voltage sources.
But now imagine an ideal _current_ source. This is some apparatus that maintains a constant current flowing between its two output terminals, no matter what the circumstances. The terminal _voltage_ adjusts up and down to maintain the constant current flow through the load. As the load impedance is decreased, the _current_ stays exactly the same, but the _voltage_ drops, and _less_ VA is delivered to the load. If the load is resistive, then as you drop the resistance, the power delivered _decreases_. If you want to 'turn off' a constant current source, then you _short circuit_ the output. The output voltage falls to zero, and the power delivered similarly falls to zero. On no account do you want to _open_ a constant current circuit, because then the output voltage (and power) goes to infinity! Again, a _real_ current source is never ideal, and will have its 'internal conductance'. As the output voltage increases, the output current _will_ go down slightly. But there are a few situations where we see some pretty good approximations to current sources.
A CT is one such beastie. A CT is a _step up_ transformer. Like all step up transformers, as the output voltage is increased, the output current is decreased. Like any transformer, there is a primary coil and a secondary coil...the primary 'coil' is just the bus bar threading through the donut, and the secondary coil is wrapped around the donut. The voltage drop on the primary is related to the voltage drop on the secondary, by the turns ratio of this transformer. What makes the CT behave is unexpected ways is that the voltages on both primary and secondary side are _very_ small, perhaps 1V on the secondary side and 1/80V on the primary side (warning: numbers being pulled out of left but cheek).
The primary side is connected in series with the entire electrical system, and because the voltage drop on the primary side is so very small, it is the load on the electrical system that actually controls the current flowing on the primary side. The total voltage on the primary circuit may be 277V, with 276.99V dropped in the load, and 0.0125V dropped in the primary of the CT.
Now, imagine that the voltage on the secondary side of the CT is multiplied by a factor of 80. The secondary voltage drop has to increase, and the primary voltage drop needs to increase to match this. But the primary voltage drop is so very small, that as far as the over-all primary circuit is concerned, there is no change at all. On the secondary side we see 80V, on the primary side we see 1V, and the total primary circuit sees 276V dropped in the load, and 1V dropped on the primary of the CT. The voltage on the secondary side can go through the roof well before the primary side sees any significant change.
Now, if you actually open circuit CT, the voltage drop on the primary won't go to the full 277V, and the secondary voltage won't go to 22KV steady state, with no current flowing in the primary side; magnetic limitations will come into play to limit the final output voltage; and the inductance of the entire system will influence the transient response. There is potentially quite a bit of energy stored in the magnetic field in the donut, enough to strike an arc if you open the secondary. But the voltage will easily get high enough to cause significant damage. Remember, a CT is _not_ an ideal current source; just a damn fine approximation to a current source.
To protect the secondary of a CT, you don't want a fuse. Instead you want something that is the opposite of a fuse; something that will short circuit in the even that the potential gets too high. I don't know what these are called in the electrical distribution world; in the electronics world, the device is known as a 'crowbar circuit'.
-Jon
