Please correct me (anyone) if this is too simplistic or down right incorrect:
The system contains inductance which stores a charge. In the case of a fault, the inductance contributes energy in the form of DC current. As with any stored charge / energy, this decays according to a time constant.
I would respectfully disagree. To me it sounds like the idea being portrayed is that there is initially energy stored in the inductor at the time of the fault, and the fault allows this initially stored energy to decay with the L/R time constant.
I would like to show a counterexample which proves this wrong.
Start with an assumption that we have an ideal power supply feeding through an inductance (perhaps a transformer) to a line which has no loads prior to the fault.
The initial energy stored in the inductance prior to the fault is zero. Now close in a low-resistance representing a fault on the transformer secondary at the moment the voltage is zero. We get the maximum dc offset. It didn't come from the energy initially stored in the inductor, because that energy was zero in this case.
At the risk of being really obnoxious, let me try another explanation expanding on what others have already said:
1 – ASSUME for simplicity the post-fault circuit has no resistance.
2 – ASSUME for simplicity the prefault current is negligible, i.e. i(0-) = 0.
3 - The total post-fault response consists of a steady state sinusoidal response plus a transient response.
4 - The current cannot jump when the fault is applied. i(0+)=i(0-) (since it is flowing through an inductor).
5 – Putting together 2 and 4, we see that the total response at t=0+ is 0. Itotal(0+)=0.
6 – Putting together 5 and 3, Itotal(0+) = Itransient(0+) + Isteadystate(0+) = 0
7 – Examining 6, the only way we can have NO transient is if Isteadystate(0+) = 0
8 - Isteadystate(0+) = 0 occurs only if we close the fault at a phase angle where the steady state solution for current is 0. This occurs only when the voltage is at its peak (since current lags voltage by 90 degrees in steady state inductive circuit).
9 – Examining 7 and 8, if you apply the fault at any time other than the maximum or the minimum of the supply voltage, you will have a transient component of the current (the exponentially decaying dc component). That transient current occurs as a direct result of the characteristic of an inductor v=L di/dt and we can calculate it directly from the corresponding integral equation i = Integral v(t) dt + i(0)
It is not a dissipation of energy stored before the fault. It is a transient response. In general we expect a transient response when the system configuration changes. The transient response in a simple L/R system will always be of the form I(0)*exp(-t*L/R)
I am a hearty supporter of the idea that there is not only one right answer. There were good answers given by many others. Mine is probably not the simple answer some people are looking for and mine is not without equations as was requested in the original post. My apologies. Feel free to comment. Otherwise I will shut up.
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