【正文】 he person will still receive a painful shock. Or suppose a 120 V eletric toster tumbles into a swimming pool. The heating elements and contacts will produce a hazardos leakage current throughout the pool ,ven if the frame of the toaster is securely grounded. Devices have been developed that will cut the source of power as soon as such accidents occur. These groundfault circuit breaker will typically trip in 25 ms if the leakage current exceeds 5 mA. How do these protective devices operate ? A small current transformer surrounds the live and neutral wires as shown in . The secondary is connected to a sensitive electronic detetor that can trigged a circuit breaker CB that is in series with the 120 V line. Under normal conditions the current Iin the line conductor is exactly equal to thecurrent Iin the neutral ,and so the net current (II) flowing through the hole in the toroidal core is zero. Consequently, no flux is produced in the core, the induced voltage E is zero, and breaker CB does not trip. Suppose now that a fault current I leaks directly from the live wire to ground . This could happen if someone touched a live terminal (). A fault current I would also be produced if the insulation broke down between a motor and its grounded enclosure . Under any of these conditions, the net current flowing through the hole of the CT is no longer zero but equal to I or I. A flux is set up and a voltage E is induced, which trips CB. Because an imbalance of only 5 mA has to be detected, the core of the transformer must be very permeable at low flux densities. Supermalloy is often used for this purpose because it has a relative permeability of typically 70000 at a flux density of only 4mT . rapid conductor heating: the It factor It sometimes happens that a current far greater than normal flows for a brief period in a conductor . The Ilosses are than very large and the temperature of the conductor can rise several hundred degrees in a fraction of a second . For example, during a severe shortcircuit, intense currents can flow in conductors and cables before the circuit is opened by the fuse or circuit breaker.Furthermore, the heat does not have time to be dissipated to the surroundings and so the temperature of the conductor increases very rapidly. What is the temperature rise under these condition?Suppose the conductor has a mass m, a resistance R, and a thermal heat capacity c. Moreover, suppose the current is I and that it flows for a period t that is typically less than 15 seconds. The heat generated in the conductor is given by Form ,we can calculate the temperature rise for a given value of :hence from which If follows that for a given conductor the temperature rise depends upon the I factor . It is well known that high temperature damage the insulation that covers a conductor. The I factor is ,therefore ,very