Asymmetrical Fault Current - DC Component

There is a lot of stuff on the web, here is one link: http://www.powerstudies.com/articles/ImportanceofX-over-RRatios.pdf

thanks for the reference but ive already read that. It does state what happens but not why! I gave it some thought and all i could come up with was at the instant of a fault, if the circuit is purely reactive (high x/r), this means that the existing current waveform is at a peak during the zero crossing of the voltage waveform. Is it this value of high current that is added to the fault current initially to look as if theres aa dc offset initially when a fault occurs?

sorry this was my post, forgot to login!

This is a question that can not be easily explained from its physical or logical point of view. However, it can be explained with mathematical concept. When the differential equation for RL circuit is solved for the current, it gets two components: one is sinusoidal component and the other one is decaying DC component. See this current value from the equation 2.2 in this link: http://www.sayedsaad.com/Protection/files/short_circuit/1_short_circuit.htm. The first part of the equation is sinusoidal (varying sinusoidaly with time) and the second part is decaying DC component (exponentially decaying with time).

The initial value of DC components depends on the time (value of α) the circuit is closed (or the fault occurs).

- MS

Remember,even a single wire has an inductive and capacitive value,and they oppose a change in current/voltage respectively,and will exhibit maximum effect when the sine wave crosses zero.