DC Component in Short Circuit

check http://www.energy.siemens.com/us/pool/us/services/power-transmission-distribution/medium-voltage-product-services/tech-topics-application-notes/techtopics72.pdf

If the current lies to one side of zero, then it is DC (which is possible for up to half a cycle of what would otherwise be AC).

These may also help:

http://cr4.globalspec.com/thread/61831/Asymmetrical-Fault-Current-DC-Component

http://www.aeusp.com/fundamentals/dshe.asp

- MS

Well! Any electrical system would definitely have network components such as resistances, inductances & capacitances. The resistive network components would dissipate energy in the form of heat, whereas the reactive components would store energy. Remember! Energy stored in an inductor is 1/2 L I(square) and the energy stored in a capacitor is 1/2 C V (square). Till the moment of fault, current would have been flowing in these system components and the resistive components would have dissipated their I(square) R t Enegry in the form of heat. But, the reactive components (both inductive and the capacitive reactances) would have stored energy. And, at the instant of fault, the voltage applied to these network components becomes zero, post-fault point. This is akin to switching OFF supply to these reactances. When the supply to a reactance is switched OFF, the energy stored in the reactor is DC in nature (at the instant of switch OFF, one terminal would be more positive with respect to the other terminal). So, these reactances would act as DC accumulators. As the fault is also a short circuit at the terminals of these reactances, they would discharge thro the short circuit. As the stored energy in these reactances is DC in nature, the discharge current would also be DC in nature. But, as these are finite sources, their current feed into the fault would last only for a few cycles, the time constant decided by the system R/X ratio. Hence, they are transient in nature.

Hope it is clear!