Transformer Insulation Failed

A current transformer has a high voltage step-up winding ratio, N, which in use becomes a high current step-down ratio, 1/N. Current transformers are always fitted with a secondary current-sensing load resistor, R_S, typically 0.1 ohm, 1 ohm, etc., and the current-carrying wire passed through the primary appears to have an AC resistance of R_S / N², which is ideally a very small value, easily micro-ohms. In this fashion it causes little voltage drop Vp = Ip R_S / N², and wastes very little power P = Vp Ip, and has an insignificant effect on the system it's measuring.

If the CT's load resistor (or burden) ever becomes disconnected, and normal operating ac line current is passed through it, the CT simply becomes a high-voltage step-up transformer and all hell can break loose. A step-up ratio of 1:100 to 1:1000 or more is common, and this implies that at least 115vac*100, etc. = more than 10kV will appear across the secondary. But we know the secondary winding insulation (with a typical magnet-wire insulation rating of say 300 volts) will quickly fail as the voltage soars toward such a high value.

The photo below shows a bus-bar ground-fault current transformer with a melted section. It was said the secondary was left open. Properly treated, current transformers are not under much stress; they are not doing any real work. They should last forever.

It's imperative that a CT always have a connected load resistor. In the best case it should be permanently attached. (But this could mean its value cannot be easily changed, hence the manufacturer's motivation to leave connecting a load resistor up to the user. So user beware, DANGER! Don't forget to well-connect a load resistor! Soldering it permanently in place would be a good idea.)

Another caution is not to use too low a power rating for the secondary load current-sense resistor. You MUST NOT EVER let that resistor fail! Hell, now I'm thinking it should be two resistors in parallel.

In closing, here's another example of what can happen if you leave the secondary open.

In electrical switchgear and apparatus, that resistor is normally NOT discrete. It is part of the load or burden that is connected in the CT circuit: a meter, protective relay, current transducer, etc. You will not find it in the wiring diagrams or physically connected in the scheme. That is why it is imperative that the CT circuit be kept continuous; there should be no fuses, test switches, or other means of opening the circuit during operation. The utility for which I work even stipulates that wiring for CT circuits be wired with ring-type crimp terminals on the wires rather than fork terminals, to make it less likely that a loose screw lead to an open circuit and a failure. So while your explanation of what happens is correct, in electrical service apparatus you will not find such a resistor as a discrete device.