Well said, Jon.
I thought iwire answered fungguy's original question very well, but it's always educational to delve a little deeper into theory.
If I may summarize:
1. The secondary current in a CT is a fixed fraction of the primary current. As Larry pointed out, 1/80 is common.
2. Ideally, the meter and wiring on the secondary side have minimal impedance. That way, even though several amps may be flowing, the voltage is negligible and the wasted power is minimized.
3. The minimized impedance on the secondary is also "reflected" into the primary, so the voltage drop is negligible.
We return now to the discussion of minutiae:
Remember that the primary is in series with the load, so that even if the primary impedance was _zero_ the current would still be reasonably limited.
I didn't understand that statement, but I agree with everything else.
As to the voltage present at the open secondary of a CT, I just don't believe the "hot-stick" levels being suggested. That voltage will be a function of the secondary turns count and the magnetic flux magnitude, and nothing else. As Jon mentioned, CTs have relatively small cores that are easily saturated, so the flux magnitude won't be nearly high enough to induce multi-kV voltages on the secondary.
Just how much voltage can be dropped across that big wire or bus passing through the CT with the open secondary? I'd be surprised to see more than a few tenths of a volt, so the secondary voltage can only be 80 times that, on the order of a few tens of volts.
What
will happen is that the saturated core will dissipate a lot of heat, which certainly could damage the windings. That's probably the real reason never to leave the secondary open.